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First Order Partial Differential Equations
P. Sam Johnson
March 5, 2020
P. Sam Johnson First Order Partial Differential Equations March 5, 2020 1/63
Overview
We are concerned in the course with partial differential equations with one
dependent variable z and mostly two independent variables x and y.
We discuss the following in three lectures.
Formation of partial differential equations by either the elimination of
arbitrary constants or by the elimination of arbitrary functions.
Solution of partial differential equation by direct integration.
Standard types of first order partial differential equations.
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Introduction
Many physical and engineering problem can be interpreted by the ideas in
differential equations. Newton’s governing equations describing mechanical
systems, Maxwell’s equations describing electro dynamical phenomena and
Schrodinger’s equations describing the aspects of quantum mechanics are
some examples to be mentioned.
Especially, propagation of waves, heat conduction and many other physical
phenomena can be explained by partial differential equations.
A partial differential equation is the one in which there is one
dependent variable which is the function of two or more independent
variables with its partial derivatives. Most of the mathematical models in
partial differential equation have two independent variables. Convention is
that we take the variable z as a dependent variable and x and y as
independent variables so the relation is z = f (x, y).
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Introduction
In general, initial conditions or boundary conditions are given to describe
the model completely. The order of a partial differential equation is the
order of the highest partial differential coefficient occurring in it.
By a solution of a partial differential equation, we mean the expression of
the form z = f (x, y) which upon proper partial differentiation, coincides
with the given partial differential equation on the same domain. Implicit in
this idea of solution is the stipulation that z possesses as many continuous
partial derivatives as required by the partial differential equation.
For example, a solution to a second order equation should have two
continuous partial derivatives so that it makes sense to calculate the
derivatives and substitute them in the given equation. Whereas the
solution of an ordinary differential equation involves arbitrary constants,
the general solution of a partial differential equation involves arbitrary
functions.
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Formation of partial differential equations
The partial differential equation of a given relation can be formed by either
the elimination of the arbitrary constants or by the elimination of arbitrary
functions depending upon whether the given relation involved by the
arbitrary constants or arbitrary functions.
We concentrate the partial differential equation formed by relation having
one dependent variable namely z and two independent variables namely x
and y. This relation may be of arbitrary functions or arbitrary constants.
We use the following notions for the convenience which are used globally.
∂z
∂x
= p,
∂z
∂y
= q,
∂2z
∂x2
= r,
∂2z
∂x∂y
= s,
∂2z
∂y2
= t.
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Formation of partial differential equations by eliminating
arbitrary constants
Let f (x, y, z, a, b) = 0 be a given relation with two arbitrary constants a
and b. Differentiating the above equation partially with respect to x and y
we get
∂f
∂x
+
∂f
∂z
∂z
∂x
= 0 =⇒
∂f
∂x
+ p
∂f
∂z
= 0
∂f
∂y
+
∂f
∂z
∂z
∂y
= 0 =⇒
∂f
∂y
+ q
∂f
∂z
= 0.
Eliminating a and b between the above three equations, we get a partial
differential equation of the form
F(x, y, z, p, q) = 0.
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Examples
Example 1.
1. Form the partial differential equation by eliminating the constants a
and b from z = (x + a)(y + b). [Answer : z = pq]
2. Form the partial differential equation by eliminating the constants a
and b from z = axn + byn. [Answer: nz = px + qy]
3. Form the partial differential equation by eliminating the constants a
and b from z = (x2 + a2)(y2 + b2). [Answer: 4xyz = pq]
4. Form the partial differential equation by eliminating the constants a
and b from z = ax + by +
√
a2 + b2.
[Answer: z = px + qy +
p
p2 + q2]
5. Form the partial differential equation by eliminating the constants
from log(az − 1) = x + ay + b. [Answer: q(z − p) = p]
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Examples
Example 2.
1. Form the partial differential equation of the surface
(x − a)2 + (y − b)2 = z2 cot2 α, where a and b are constants.
[Answer: p2 + q2 = tan2 α]
2. Find the partial differential equation of all spheres whose center lies
on z-axis. [Hint : If the equation of the sphere whose center lies on
z-axis is given by x2 + y2 + (z − c)2 = r2, where c and r are
constants, then PDE is xq = yp.]
3. Find the partial differential equation of all spheres whose radius is
unity and center lies on xy plane. [Hint : If the equation of the sphere
whose radius is unity and center lies on xy plane is given by
(x − a)2 + (y − b)2 + z2 = 1, where a and b are constants, then
z2(p2 + q2 + 1) = 1.]
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Examples
Example 3.
1. Find the partial differential equation of all planes passing through the
origin.
[Hint : The equation of the plane which passes through origin is given
by ax + by + cz = 0, where a, b and c are constants, then
z = px + qy]
2. Find the partial differential equation of all planes whose sum of x, y, z
intercepts is unity.
[Hint : The equation of the plane with a, b, c as the x, y, z intercepts
is x
a + y
b + z
c = 1. Answer: z = p x + q y + pq
p+q−pq .]
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Formation of partial differential equations by eliminating
arbitrary functions
Formation of partial differential equations by the elimination of arbitrary
functions from the given relation is explained in the following examples.
The elimination of one arbitrary functions from the given relation gives
partial differential equations of first order and elimination of two arbitrary
functions from the given relation gives partial differential equations of
second or higher order.
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Examples
Example 4.
1. Form the partial differential equation by eliminating the arbitrary
function f from z = f (x + y). [Answer: p = q]
2. Form the partial differential equation by eliminating f from
z = f (x2 + y2). [Answer: py = qx]
3. Form the partial differential equation by eliminating f from
z = x + y + f (xy). [Answer: px − qy = x − y]
4. Form the partial differential equation by eliminating f from
z = f (x2 + y2 + z2). [Answer: yp = xq]
5. Form the partial differential equation by eliminating f from
z = x2 + 2f

1
y + log x

. [Answer: px + qy2 = 2x2]
6. Form the partial differential equations by eliminating the arbitrary
functions from z = f1(x) + f2(y). [Answer: ∂2z
∂y∂x = 0]
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Examples
Example 5.
1. Form the partial differential equations by eliminating the arbitrary
functions from z = yf1(x) + f2(y). [Answer: ∂z
∂x = y ∂2z
∂y∂x ]
2. Form the partial differential equation by eliminating the f and φ from
z = f (y) + φ(x + y + z). [Answer: r(1 + q) = s(1 + p)]
3. Form a partial differential equation from the relation by eliminating
the arbitrary functions f1 and f2 from z = f1(x + 2y) + f2(x − 2y).
[Answer: 4∂2z
∂x2 = ∂2z
∂y2 .]
4. Form a PDE from the relation z = f1(ax + by) + f2(cx + dy).
[Answer: bd ∂2z
∂x2 − (ad + bc) ∂2z
∂x∂y + ac ∂2z
∂y2 = 0]
5. Form a partial differential equation from the relation
z = f1(2x + 3y) + f2(4x + 5y). [Answer: 15∂2z
∂x2 − 22 ∂2z
∂x∂y + 8∂2z
∂y2 = 0]
6. Form a partial differential equation by eliminating the functions from
z = f1(x2 + 3y) + f2(x2 − 3y) [Answer : 9∂2z
∂x2 − 9
x
∂z
∂x − 4x2 ∂2z
∂y2 = 0]
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Formation of partial differential equations by elimination of
arbitrary function φ from φ(u, v) = 0 where u and v are
functions of x, y and z.
Let
φ(u, v) = 0 (1)
be the given relation with u and v are functions of x, y and z.
Differentiating (1) partially with respect to x and y we get
∂φ
∂u
∂u
∂x
+
∂φ
∂v
∂v
∂x
= 0 (2)
∂φ
∂u
∂u
∂y
+
∂φ
∂v
∂v
∂y
= 0. (3)
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Formation of partial differential equations by elimination
Now eliminating φ between (2) and (3) we get the required partial
differential equation as
∂u
∂x
∂v
∂x
∂u
∂y
∂v
∂y
= 0.
Example 6.
1. Form a partial differential equation by eliminating φ from
φ

x
y , y
z

= 0. [Answer: z = xp + yq is the required solution.]
2. Form a partial differential equation by eliminating φ from
φ(x + y + z, x2 + y2 + z2) = 0. [Answer:(y − z)p + (z − x)q = x − y]
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Solution of partial differential equations
Consider the relation
z = ax + by. (1)
Eliminating the arbitrary constants a and b we get
z = px + qy. (2)
Also consider the relation
z = x f
y
x

. (3)
Eliminating the arbitrary function f we get
z = px + qy.
From these we conclude that both (1) and (3) are solutions of the same
equation (2) and for a single partial differential equation we have more
than one solutions.
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Types of solutions
1. A solution which contains the number of arbitrary constants is equal
to the number of independent variables is called complete solution
or complete integral.
2. A complete solution in which if a particular value is given to arbitrary
constant is called particular solution.
3. A solution which contains the maximum possible number of arbitrary
functions is called a general solution.
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Solution of partial differential equation by direct integration
Simple partial differential equations can be solved by direct integration.
The method of solving such equations explained as follows.
Example 7.
Solve ∂z
∂x = 0.
Solution. The given equation is
∂z
∂x
= 0.
This shows that z is a function of the independent x. Integrating with
respect to x
z = f (y)
is the solution of the given partial differential equation.
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Solution of partial differential equation by direct integration
Example 8.
Solve ∂2
z
∂x∂y
= 0.
Solution. The given equation can be written as
∂2z
∂x∂y
=
∂
∂x

∂z
∂y

= 0.
Integrating with respect to x
∂z
∂y
= f1(y).
Now integrating with respect to y
z =
Z
f1(y)dy + f2(x)
= g1(y) + f2(x).
Hence the solution is z = g1(y) + f2(x).
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Examples
Example 9.
Solve ∂2
z
∂x∂y
= x
y
+ c.
Solution. The given equation can be written as
∂2z
∂x∂y
=
∂
∂x

∂z
∂y

=
x
y
+ c.
Integrating with respect to x
∂z
∂y
=
x2
2y
+ cx + f1(y).
Now integrating with respect to y
z =
x2
2
log y + cxy +
Z
f1(y)dy + f2(x)
=
x2
2
log y + cxy + g1(y) + f2(x).
Hence the solution is z = x2
2
log y + cxy + g1(y) + f2(x).
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Examples
Example 10.
Solve ∂2
z
∂x2 = cos x.
Solution. The given equation is
∂2z
∂x2
= cos x.
Integrating with respect to x
∂z
∂x
= sin x + f1(y).
Again integrating with respect to x
z = − cos x + xf1(y) + f2(y).
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Example
Example 11.
Solve ∂2
z
∂y∂x
= e−x cos y given that z = 0 when x = 0 and ∂z
∂x
= 0 when y = 0.
Solution. The given equation can be written as
∂2z
∂y∂x
=
∂
∂y

∂z
∂x

= e−x
cos y.
Integrating with respect to y, ∂z
∂x
= e−x sin y + f1(x).
Applying ∂z
∂x
= 0 when y = 0, we get f1(x) = 0, hence ∂z
∂x
= e−x sin y.
Now integrating with respect to x
z = −e−x
sin y + f2(y).
Applying z = 0 when x = 0
0 = − sin y + f2(y) ⇒ f2(y) = sin y.
Hence
z = −e−x
sin y + sin y.
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Standard types of first order partial differential equations
Type 1. Consider the partial differential equation of the form
F(p, q) = 0. (1)
Here the variables x, y, z do not occur explicitly. Let
z = ax + by + c. (2)
be a solution of the given equation. Then
∂z
∂x
= p = a,
∂z
∂y
= q = b.
Substituting these values of p and q in the equation (1) we get
F(a, b) = 0. Solving for b we get b = f (a) then
z = ax + f (a)y + c (3)
is the complete solution of the given equation.
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Type 1
To get the general solution put c = g(a)
z = ax + f (a)y + g(a) (4)
and differentiating with respect to a
0 = x + f 0
(a)y + g0
(a). (5)
Eliminating a between (4) and (5) we get the general solution.
To get the singular solution, differentiating (2) with respect to c we get
0 = 1, which is not possible. Hence there is no singular solution.
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Examples
Example 12.
Solve p2 + p − q2 = 0.
Solution. The given equation is of the form F(p, q) = 0. The complete solution is
z = ax + by + c (1)
where a2 + a − b2 = 0. Solving for b we get b2 = a2 + a, so b =
√
a2 + a. Hence the
complete solution is
z = ax +
p
a2 + ay + c. (2)
Differentiating (2) with respect to c we get 0 = 1, which is not possible. Hence there is no
singular solution. To get the general solution put c = g(a)
z = ax +
p
a2 + ay + g(a) (3)
and differentiating with respect to a, we have 0 = x + 1+2a
2
√
a2+a
y + g0(a). Eliminating a between
(3) and (4) we get the general solution.
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Examples
Example 13.
Solve p2 + q2 = 9.
Solution. The given equation is of the form F(p, q) = 0. The complete solution is
z = ax + by + c (1)
where a2 + b2 = 9. Solving for b we get b2 = 9 − a2, so b =
√
9 − a2. Hence the complete
solution is
z = ax +
p
9 − a2y + c (2)
Differentiating (2) with respect to c we get 0 = 1, which is not possible. Hence there is no
singular solution. To get the general solution put c = g(a)
z = ax +
p
9 − a2y + g(a) (3)
and differentiating with respect to a
0 = x +
−2a
2
√
9 − a2
y + g0
(a). (4)
Eliminating a between (3) and (4) we get the general solution.
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Examples
Example 14.
Solve pq + p + q = 0.
Solution. The given equation is of the form F(p, q) = 0. The complete solution is
z = ax + by + c (1)
where ab + a + b = 0. Solving for b we get b(1 + a) = −a, so b = − a
1+a
. Hence the complete
solution is
z = ax −
a
1 + a
y + c. (2)
Differentiating (2) with respect to c we get 0 = 1, which is not possible. Hence there is no
singular solution. To get the general solution put c = g(a)
z = ax −
a
1 + a
y + g(a) (3)
and differentiating with respect to a
0 = x −
1
(1 + a)2
y + g0
(a). (4)
Eliminating a between (3) and (4) we get the general solution.
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Type 2
Consider the partial differential equations of the form
z = px + qy + F(p, q).
This equation is called the Clairut’s equation.
Let
z = ax + by + c
be a solution of the given equation. Then
∂z
∂x
= p = a,
∂z
∂y
= q = b.
Substituting these values of p and q in the given equation, we get
z = ax + by + F(a, b).
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Type 2 : Example
Example 15.
Solve z = px + qy + pq.
Solution. The given equation is of the form z = px + qy + F(p, q). The complete solution is
z = ax + by + ab. (1)
To find the singular solution differentiating (1) partially with respect to a
0 = x + b ⇒ b = −x
and differentiating (1) partially with respect to b we get
0 = y + a ⇒ a = −y.
Substituting these values in (1) we get
z = −(y)x − (x)y + (−y)(−x)
z = −xy − xy + xy
x = −xy
z + xy = 0.
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Example
Example 16.
Solve z = px + qy + p2 + q2
Solution. The given equation is of the from z = px + qy + F(p, q). The complete solution is
z = ax + by + a2
+ b2
(1)
To find the singular solution differentiating (1) partially with respect to a
0 = x + 2a ⇒ a = −
x
2
and differentiating (1) partially with respect to b we get, 0 = y + 2b ⇒ b = −y
2
. Substituting
these values in (1) we get
z = −
x
2

x −
y
2

y +

−
x
2
2
+

−
y
2
2
z = −
x2
2
−
y2
2
+
x2
4
+
y2
4
z = −
x2
4
−
y2
4
4z = −(x2
+ y2
).
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Example
Example 17.
Solve z = px + qy + p2 + pq + q2.
Solution. The given equation is of the form z = px + qy + F(p, q).
The complete solution is
z = ax + by + a2
+ ab + b2
. (1)
To find the singular solution differentiating (1) partially with respect to a and b we get,
x + 2a + b = 0 and y + a + 2b = 0.
Solving we get a = 1
3
(y − 2x) and b = 1
3
(x − 2y). Substituting these values of a and b in (1) we
get
z =
1
3
(y − 2x)x +
1
3
(x − 2y)y +
1
9
(y − 2x)2
+
1
9
(y − 2x)(x − 2y) +
1
9
(x − 2y)2
9z = 3x(y − 2x) + 3y(x − 2y) + (y − 2x)2
+ (y − 2x)(x − 2y) + (x − 2y)2
9z = −3x2
+ 3xy − 3y2
.
Hence 3z + x2 − xy + y2 = 0.
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Example
Example 18.
Solve z = px + qy +
p
1 + p2 + q2.
Solution. The given equation is of the form z = px + qy + F(p, q).
The complete solution is
z = ax + by +
p
1 + a2 + b2. (1)
To find the singular solution differentiating (1) partially with respect to a and b we get
x = − a
√
1+a2+b2
and y = − b
√
1+a2+b2
. Hence
p
1 − x2 − y2 = 1
√
1+a2+b2
. From x and y, we
get a = − x
√
1−x2−y2
and b = − y
1−x2−y2 . Substituting the values in (1) we have
z = −
x2
p
1 − x2 − y2
−
y2
p
1 − x2 − y2
+
1
p
1 − x2 − y2
z =
1 − x2 − y2
p
1 − x2 − y2
z2
= 1 − x2
− y2
.
Thus x2 + y2 + z2 = 1.
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Type 3.
This type of partial differential equation can be further classified into three categories. We
discuss each case as follows.
Case I. Consider the partial differential equation of the form
F(x, p, q) = 0.
Let us assume that q = a. Then
F(x, p, a) = 0.
Solving for p we get
p = φ(x, a).
Since z is a function of x and y
dz =
∂z
∂x
dx +
∂z
∂y
dy
= pdx + qdy
= φ(x, a)dx + qdy
z =
Z
φ(x, a)dx +
Z
ady + c
z =
Z
φ(x, a)dx + ay + c.
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Examples
Example 19.
Solve pq = x.
Solution. The given equation is of the form F(x, p, q) = 0.
Let us assume that q = a. Then pa = x ⇒ p = x
a
.
dz = pdx + qdy
=
x
a
dx + a dy
z =
Z
x
a
dx +
Z
ady + c
z =
x2
2a
+ ay + c.
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Example
Example 20.
Solve q = px + p2.
Solution. The given equation is of the form F(x, p, q) = 0.
Let us assume that q = a. then
p2
+ px − a = 0
p =
−x ±
√
x2 + 4a
2
dz = pdx + qdy
=
−x ±
√
x2 + 4a
2
!
dx + a dy
z =
Z
x
2
dx ±
1
2
Z p
x2 + 4a dx +
Z
ady + c
=
x2
4
±
1
2

x
2
p
x2 + 4a +
4a
2
sinh−1

x
2
√
a

+ ay + c
=
x2
4
±

x
4
p
x2 + 4a + a sinh−1

x
2
√
a

+ ay + c.
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Case II
Consider the partial differential equations of the form
F(y, p, q) = 0.
Let us assume that p = a. Then
F(y, a, q) = 0.
Solving for q we get
q = φ(y, a)
Since z is a function of x and y
dz =
∂z
∂x
dx +
∂z
∂y
dy
= pdx + qdy
= adx + φ(y, a)dy
z =
Z
adx +
Z
φ(y, a)dy + c
z = ax +
Z
φ(y, a)dy + c.
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Examples
Example 21.
Solve q2 = yp3.
Solution. The given equation is of the form F(y, p, q) = 0.
Let us assume that p = a. Then
q2
= ya3
= 0
q = a
3
2
√
y.
dz = pdx + qdy
= a dx + a
3
2
√
ydy
z =
Z
a dx +
Z
a
3
2
√
y dy + c
x = a x +
2
3
a
3
2 y
3
2 + c.
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Example
Example 22.
Solve p = qy + q2.
Solution. The given equation is of the form F(y, p, q) = 0.
Let us assume that p = a. Then
q2
+ qy − a = 0
q =
−y ±
p
y2 + 4a
2
dz = pdx + qdy
= a dx +
−y ±
p
y2 + 4a
2
!
dy
z =
Z
adx +
Z
y
2
dy ±
1
2
Z p
y2 + 4a dy + c
= ax +
y2
4
±
1
2

y
2
p
y2 + 4a +
4a
2
sinh−1

y
2
√
a

+ c
= ax +
y2
4
±

y
4
p
y2 + 4a + a sinh−1

y
2
√
a

+ c
P. Sam Johnson First Order Partial Differential Equations March 5, 2020 37/63
Case III
Consider the partial differential equations of the form
F(z, p, q) = 0.
Let us assume that q = ap. Then
F(z, p, ap) = 0.
Solving for p we get
p = φ(z, a)
q = ap = aφ(z, a)
Since z is a function of x and y
dz =
∂z
∂x
dx +
∂z
∂y
dy
= pdx + qdy
= φ(z, a)dx + aφ(z, a)dy
Z
dz
φ(z, a)
=
Z
dx +
Z
ady + c
Z
dz
φ(z, a)
= x + ay + c.
P. Sam Johnson First Order Partial Differential Equations March 5, 2020 38/63
Examples
Example 23.
Solve 1 + z3 = z4pq.
Solution. The given equation is of the form F(z, p, q) = 0. Let us assume that q = ap. Then
1 + z3 = z4p2a gives p = 1
√
a
√
1+z3
z2 and q = a 1
√
a
√
1+z3
z2 . So dz = pdx + qdy gives
Z
dz =
1
√
a
Z √
1 + z3
z2
dx + a
1
a
Z √
1 + z3
z2
dy
Z
z2
√
1 + z3
dz =
1
√
a
Z
dx + a
Z
dy + c

2
3
p
1 + z3 =
1
√
a
(x + ay + c)
4
9
(1 + z3
) =
1
a
(x + ay + c)2
4 a(1 + z3
) = 9(x + ay + c)2
.
P. Sam Johnson First Order Partial Differential Equations March 5, 2020 39/63
Examples
Example 24.
Solve z = p2 + q2.
Solution. The given equation is of the form F(z, p, q) = 0. Let us assume that q = ap. Then
z = p2 + a2p2 gives p = 1
√
1+a2
√
z and q = a
√
1+a2
√
z. So dz = pdx + qdy gives
Z
dz =
1
√
1 + a2
Z
√
zdx +
a
√
1 + a2
Z
√
zdy
Z
dz
√
z
=
1
√
1 + a2
Z
dx + a
Z
dy + c

2
√
z =
1
√
1 + a2
(x + ay + c)
4z =
1
1 + a2
(x + ay + c)2
.
P. Sam Johnson First Order Partial Differential Equations March 5, 2020 40/63
Examples
Example 25.
Solve p(1 − q2) = q(1 − z).
Solution. The given equation is of the form F(z, p, q) = 0. Let us assume that q = ap. Then
p(1 − a2
p2
) = ap(1 − z)
(1 − a2
p2
) = a(1 − z)
a2
p2
= 1 − a + az
p =
1
a
√
1 − a + az
q =
√
1 − a + az
dz = pdx + qdy
Z
dz =
1
a
Z
√
1 − a + az dx +
Z
√
1 − a + az dy
Z
dz
√
1 − a + az
=
1
a
Z
dx + a
Z
dy + c

2
a
√
1 − a + az =
1
a
(x + ay + c)
4(1 − a + az) = (x + ay + c)2
.
P. Sam Johnson First Order Partial Differential Equations March 5, 2020 41/63
Example 26.
Solve q2 = z2p2(1 − p2).
Solution. The given equation is of the form F(z, p, q) = 0. Let us assume that q = ap. Then
a2
p2
= z2
p2
(1 − p2
)
a2
= z2
(1 − p2
)
z2
p2
= z2
− a2
p =
1
z
p
z2 − a2
q = a
1
z
p
z2 − a2.
dz = pdx + qdy
Z
dz =
Z
1
z
p
z2 − a2dx +
Z
a
1
z
p
z2 − a2dy
Z
z dz
√
z2 − a2
=
Z
dx + a
Z
dy + c

p
z2 − a2 = (x + ay + c)
z2
− a2
= (x + ay + c)2
z2
= a2
+ (x + ay + c)2
.
P. Sam Johnson First Order Partial Differential Equations March 5, 2020 42/63
Example
Example 27.
Solve p3 + q3 = z3.
Solution. The given equation is of the form F(z, p, q) = 0. Let us assume that q = ap. Then
p =
1
(1 + a3)
1
3
z
q = a
1
(1 + a3)
1
3
z
dz = pdx + qdy
Z
dz =
Z
1
(1 + a3)
1
3
z dx +
Z
a
1
(1 + a3)
1
3
z dy
(1 + a3
)
1
3
Z
dz
z
=
Z
dx + a +
Z
dy + c

(1 + a3
)
1
3 log z = (x + ay + c).
P. Sam Johnson First Order Partial Differential Equations March 5, 2020 43/63
Example
Example 28.
Solve z2(p2 + q2 + 1) = 1.
Solution. The given equation is of the form F(z, p, q) = 0. Let us assume that q = ap. Then
p =
1
√
1 + a2
√
1 − z2
z
q =
a
√
1 + a2
√
1 − z2
z
.
dz = pdx + qdy
Z
dz =
Z
1
√
1 + a2
√
1 − z2
z
dx +
Z
a
1
√
1 + a2
√
1 − z2
z
dy
Z
z dz
√
1 − z2
=
1
√
1 + a2
Z
dx + a
Z
dy + c

−
p
1 − z2 =
1
√
1 + a2
(x + ay + c).
P. Sam Johnson First Order Partial Differential Equations March 5, 2020 44/63
Example
Example 29.
Solve 2p2 − qz = z2.
Solution. The given equation is of the form F(z, p, q) = 0. Let us assume that q = ap. Then
p = z
a ±
√
a2 + 8
2
!
q = a z
a ±
√
a2 + 8
2
!
.
dz = pdx + qdy
Z
dz =
Z
z
a ±
√
a2 + 8
2
!
dx +
Z
a z
a ±
√
a2 + 8
2
!
dy
Z
dz
z
=
a ±
√
a2 + 8
2
!
(
Z
dx + a
Z
dy + c)
log z =
a ±
√
a2 + 8
2
!
(x + ay + c).
P. Sam Johnson First Order Partial Differential Equations March 5, 2020 45/63
Example
Example 30.
Solve p(1 + q) = q z.
Solution. The given equation is of the form F(z, p, q) = 0. Let us assume that q = ap. Then
p(1 + ap) = zpa
p[(1 + az) + ap] = 0
p = 0,
1
a
(az − 1)
q = a
1
a
(az − 1).
dz = pdx + qdy
Z
dz =
1
a
Z
(az − 1)dx + a
1
a
Z
(az − 1)dy
Z
1
az − 1
dz =
1
a
Z
dx + a
Z
dy + c

1
a
log(az − 1) =
1
a
(x + ay + c)
log(az − 1) = (x + ay + c).
P. Sam Johnson First Order Partial Differential Equations March 5, 2020 46/63
Type 4
Consider the partial differential equation of the form
F1(x, p) = F2(y, q).
Let us assume that
F1(x, p) = a
F2(y, q) = a.
Solving for p and q we get
p = φ1(x, a)
q = φ2(y, a).
Since z is a function of x and y
dz =
∂z
∂x
dx +
∂z
∂y
dy
= pdx + qdy
= φ1(x, a)dx + φ2(y, a)dy
z =
Z
φ1(x, a)dx +
Z
φ2(y, a)dy + c.
P. Sam Johnson First Order Partial Differential Equations March 5, 2020 47/63
Examples
Example 31.
Solve p + q = x + y.
Solution. The given equation is of the form F1(x, p) = F2(y, q).
p + q = x + y = a(say)
p − x = a ⇒ p = x + a
q − y = a ⇒ q = y + a
dz = pdx + qdy
z =
Z
(x + a)dx +
Z
(y + a)dy + c
z =
(x + a)2
2
+
(y + a)2
2
+ c.
P. Sam Johnson First Order Partial Differential Equations March 5, 2020 48/63
Examples
Example 32.
Solve p2 + q2 = x + y.
Solution. The given equation is of the form F1(x, p) = F2(y, p).
p2
+ q2
= x + y = a(say)
p2
− x = a ⇒ p =
√
x + a
y − q2
= a ⇒ q =
√
y − a
dz = pdx + qdy
z =
√
x + adx +
Z
√
y − ady + c
z =
2
3
(x + a)
3
2 +
2
3
(y − a)
3
2 + c.
P. Sam Johnson First Order Partial Differential Equations March 5, 2020 49/63
Example
Example 33.
Solve p2 + q2 = x2 + y2.
Solution. The given equation is of the form F1(x, p) = F2(y, q).
From p2 + q2 = x2 + y2 = a2(say),
we have,
p2
− x2
= a2
⇒ p =
p
x2 + a2,
y2
− q2
= a2
⇒ q =
p
y2 − a2
dz = pdx + qdy
z =
Z p
x2 + a2dx +
Z p
y2 − a2dy + c
z =
x
2
p
x2 + a2
a2
2
sinh−1 x
a
+
y
2
p
y2 − a2 −
a2
2
cosh−1 x
a
+ c.
P. Sam Johnson First Order Partial Differential Equations March 5, 2020 50/63
Example
Example 34.
Solve q2x(1 + y2) = py2.
Solution. The given equation can written in form F1(x, p) = F2(y, q).
q2
x(1 + y2
) = py2
q2 1 + y2
y2
=
p
x
= a2
(say)
p
x
= a2
⇒ p = a2
x
q2 1 + y2
y2
= a2
⇒ q =
ay
p
1 + y2
dz = pdx + qdy
z =
Z
a2
x dx +
Z
ay
p
1 + y2
dy + c
z = a2 x2
2
+ a
p
1 + y2 + c.
P. Sam Johnson First Order Partial Differential Equations March 5, 2020 51/63
Type 5
Consider the partial differential equations of the form
F(xm
p, yn
q) = 0 (1)
and
F(z, xm
p, yn
q) = 0, (2)
where m and n are constants.
Case I. For m 6= 1 and n 6= 1 put
X = x1−m
⇒
∂X
∂x
= (1 − m)x−m
, Y = y1−n
⇒
∂Y
∂y
= (1 − n)y−n
P =
∂z
∂x
=
∂z
∂X
∂X
∂x
= P(1 − m)x−m
⇒ P(1 − m) = pxm
(3)
Q =
∂z
∂y
=
∂z
∂Y
∂Y
∂y
= Q(1 − n)y−n
⇒ Q(1 − n) = qyn
(4)
where P = ∂x
∂X
and Q = ∂z
∂Y .
Using expressions (3) and (4), the equations (1) and (2) can be
transformed into equations F(P, Q) = 0 and F(z, P, Q) = 0, respectively. These equations are
easily solvable by standard method.
P. Sam Johnson First Order Partial Differential Equations March 5, 2020 52/63
Case 2.
For m = 1 and n = 1 the equation is of the form F(xp, yp) = 0. To solve this put
X = log x ⇒
∂X
∂x
=
1
x
Y = log y ⇒
∂Y
∂y
=
1
y
p =
∂z
∂x
=
∂z
∂X
∂X
∂x
= P
1
x
⇒ P = px (5)
q =
∂z
∂y
=
∂z
∂Y
∂Y
∂y
= Q
1
y
⇒ Q = qy (6)
where P = ∂z
∂X
and Q = ∂z
∂Y
.
Substituting the values of (5) and (6) in (1) and (2) we get transformed equatons F(P, Q) = 0
and F(z, P, Q) = 0, respectively. These equations are easily solvable by standard method.
P. Sam Johnson First Order Partial Differential Equations March 5, 2020 53/63
Examples
Example 35.
Solve xp + yq = 1.
Solution. The given equation is of the form F(xmp, ynq) = 0. Here m = 1 and n = 1. Put
X = log x and Y = log y, then xp = P yp = Q. Hence the equation becomes
P + Q = 1
which is of the form F(P, Q) = 0.
Let z = aX + bY + c
be the solution with
a + b = 1 ⇒ b = 1 − a
z = aX + (1 − a)Y + c
z = a log x + (1 − a) log y + c.
P. Sam Johnson First Order Partial Differential Equations March 5, 2020 54/63
Example
Example 36.
Find the solution of px2 + qy2 = z2.
Solution. The given equation is of the form F(xmp, ynq, z) = 0. Here m = 2 and n = 2. Put
x1−2 = X and y1−2 = Y , then x2p = −P, y2q = −Q. Hence the equation becomes
P + Q = −z2. Putting Q = aP, we get P + aP = −z2 ⇒ P = − z2
(1+a)
and Q = − a z2
(1+a)
.
dz = PdX + QdY
= −
z2
(1 + a)
dX −
a
(1 + a)
z2
dY
−
Z
dz
z2
=
1
(1 + a)
Z
dX + a
Z
dY

1
z
=
1
(1 + a)
(X + aY ) + c
1
z
=
1
(1 + a)

1
x
+ a
1
y

+ c.
P. Sam Johnson First Order Partial Differential Equations March 5, 2020 55/63
Example
Example 37.
Solve x2y2pq = z2.
Solution. The given equation is of the form F(xmp, ynq, z) = 0. Here m = 2 and n = 2. Put
x1−2 = X and y1−2 = Y , then x2p = −P and y2q = −Q. Hence the equation becomes
PQ = z2. Putting Q = a2P, we get P2a2 = z2 ⇒ P = z
a
and Q = a2 z
a
= az.
dz = PdX + QdY
=
1
a
zdX + azdY
Z
dz
z
=
1
a
Z
dX + a
Z
dY

log z =
1
a
(X + aY ) + c
log z =
1
a

1
x
+ a
1
y

+ c.
P. Sam Johnson First Order Partial Differential Equations March 5, 2020 56/63
Example
Example 38.
Solve p
x2 + q
y2 = z.
Solution. The given equation is of the form F(xmp, ynq, z) = 0. Here m = −2 and n = −2.
Put x1−(−2) = x3 = X and y1−(−2) = y3 = Y , then x−2p = 3P and y−2q = 3Q. Hence the
equation becomes 3P + 3Q = z. Putting Q = aP, we get 3P + 3aP = z ⇒ P = z
3(1+a)
and
Q = a z
3(1+a)
= a
3(1+a)
z.
dz = PdX + QdY
=
1
3(1 + a)
zdX +
a
3(1 + a)
zdY
Z
dz
z
=
1
3(1 + a)
Z
dX + a
Z
dY

log z =
1
3(1 + a)
(X + aY ) + c
log z =
1
3(1 + a)
(x3
+ ay3
) + c.
P. Sam Johnson First Order Partial Differential Equations March 5, 2020 57/63
Example
Example 39.
Solve x4p2 − yzq = z2.
Solution. The given equation can be written as (x2p)2 − (yz)q = z2. Putting x1−2 = X and
log y = Y , we get x2p = P and yp = Q. Hence the equation becomes P2 − Qz = z2. This is
the form F(P, Q, z) = 0. Putting Q = aP, we get
P2
− azP = z2
⇒ P =
z(a ±
√
a2 + 4)
2
, and Q = a
z(a ±
√
a2 + 4)
2
.
dz = PdX + Qdy
=
z(a ±
√
a2 + 4)
2
dX + a
z(a ±
√
a2 + 4)
2
dY
Z
dz
z
=
(a ±
√
a2 + 4)
2
Z
dX + a
Z
dY

log z =
(a ±
√
a2 + 4)
2
(X + aY ) + c
log z =
(a ±
√
a2 + 4)
2

1
x
+ a log y

+ c.
P. Sam Johnson First Order Partial Differential Equations March 5, 2020 58/63
Type 6.
Consider the partial differential equations of the form
F(zm
p, zm
q) = 0 (1)
and
F1(x, zm
p) = F2(y, zm
q) (2)
where m is a constant.
Case I For m 6= −1 put Z = zm+1, we get
∂Z
∂x
=
∂Z
∂z
∂z
∂x
⇒ P = (m + 1)zm
p ⇒
P
m + 1
= p (3)
∂Z
∂y
=
∂Z
∂z
∂z
∂y
⇒ Q = (m + 1)zm
q ⇒
Q
m + 1
= q, (4)
where P = ∂Z
∂x
and Z = ∂Z
∂y
.
Substituting (3) and (4) in (1) and (2) we get transformed equations of the form F(P, Q) = 0
and F1(x, P) = F2(y, Q) and this can be easily solvable by standard method.
P. Sam Johnson First Order Partial Differential Equations March 5, 2020 59/63
Type 6
Case 2. For m = −1 put
Z = log z
∂Z
∂x
=
∂Z
∂z
∂z
∂x
⇒ P =
1
z
p ⇒ P = z−1
p (5)
∂Z
∂y
=
∂Z
∂z
∂z
∂y
⇒ Q =
1
z
q ⇒ Q = z−1
q, (6)
where P = ∂Z
∂x
and Q = ∂Z
∂y.
Substituting (5) and (6) in (1) and (2) we have transformed equations of the form F(P, Q) = 0
and F1(x, P) = F2(y, Q) and this can be easily solvable by standard method.
P. Sam Johnson First Order Partial Differential Equations March 5, 2020 60/63
Examples
Example 40.
Solve z2(p2 + q2) = x2 + y2.
Solution. The given equation is of the form F1(x, zmp) = F2(y, zmq). Putting Z = z2 we get
P = 2zp ⇒ P
2
= zp and Q = 2zq ⇒ Q
2
= zq. Hence the equation becomes

P
2
2
+

Q
2
2
= x2
+ y2
⇒ P2
+ Q2
= 4x2
+ 4y2
⇒ P2
− 4x2
= −Q2
+ 4y2
= 4a2
(say)
P2
− 4x2
= 4a2
and − Q2
+ 4y2
= 4a2
dZ = Pdx + Qdy
Z
dZ = 2
Z p
x2 + a2dx + 2
Z p
y2 − a2dy
Z = 2

x
2
p
x2 + a2 +
a2
2
sinh−1
x
a

+ 2

y
2
p
y2 − a2 −
a2
2
cosh−1
y
a

z2
= x
p
x2 + a2 + a2
sinh−1
x
a

+ y
p
y2 − a2 − a2
cosh−1
y
a

.
P. Sam Johnson First Order Partial Differential Equations March 5, 2020 61/63
Example
Example 41.
Solve (z2p + x)2 + (z2q + y)2 = 1.
Solution. The given equation is of the form F1(x, zmp) = F2(y, zmq). Putting Z = z3, we get
P = 3z2p ⇒ P
3
= z2p and Q = 3z2q ⇒ Q
3
= z2q. Hence the equation becomes

P
3
+ x
2
+

Q
3
+ y
2
= 1 ⇒ P = 3(a − x)
1 −

Q
3
+ y
2
= a2
⇒ Q = 3(
p
1 − a2 − y)
dZ = Pdx + Qdy
Z
dZ =
Z
3(a − x)dx +
Z
3(
p
1 − a2 − y)dy
z = 3

ax −
x2
2

+ 3

(
p
1 − a2)y −
y2
2

+ c
z3
= 3

ax −
x2
2

+ 3

(
p
1 − a2)y −
y2
2

+ c.
P. Sam Johnson First Order Partial Differential Equations March 5, 2020 62/63
References
T. Amaranath, An Elementary Course in Partial Differential
Equations, Second Edition, Narosa Publishing House, 1997.
Ian Sneddon, Elements of Partial Differential Equations, McGraw-Hill
Book Company, 1957.
P. Sam Johnson First Order Partial Differential Equations March 5, 2020 63/63

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  • 1. First Order Partial Differential Equations P. Sam Johnson March 5, 2020 P. Sam Johnson First Order Partial Differential Equations March 5, 2020 1/63
  • 2. Overview We are concerned in the course with partial differential equations with one dependent variable z and mostly two independent variables x and y. We discuss the following in three lectures. Formation of partial differential equations by either the elimination of arbitrary constants or by the elimination of arbitrary functions. Solution of partial differential equation by direct integration. Standard types of first order partial differential equations. P. Sam Johnson First Order Partial Differential Equations March 5, 2020 2/63
  • 3. Introduction Many physical and engineering problem can be interpreted by the ideas in differential equations. Newton’s governing equations describing mechanical systems, Maxwell’s equations describing electro dynamical phenomena and Schrodinger’s equations describing the aspects of quantum mechanics are some examples to be mentioned. Especially, propagation of waves, heat conduction and many other physical phenomena can be explained by partial differential equations. A partial differential equation is the one in which there is one dependent variable which is the function of two or more independent variables with its partial derivatives. Most of the mathematical models in partial differential equation have two independent variables. Convention is that we take the variable z as a dependent variable and x and y as independent variables so the relation is z = f (x, y). P. Sam Johnson First Order Partial Differential Equations March 5, 2020 3/63
  • 4. Introduction In general, initial conditions or boundary conditions are given to describe the model completely. The order of a partial differential equation is the order of the highest partial differential coefficient occurring in it. By a solution of a partial differential equation, we mean the expression of the form z = f (x, y) which upon proper partial differentiation, coincides with the given partial differential equation on the same domain. Implicit in this idea of solution is the stipulation that z possesses as many continuous partial derivatives as required by the partial differential equation. For example, a solution to a second order equation should have two continuous partial derivatives so that it makes sense to calculate the derivatives and substitute them in the given equation. Whereas the solution of an ordinary differential equation involves arbitrary constants, the general solution of a partial differential equation involves arbitrary functions. P. Sam Johnson First Order Partial Differential Equations March 5, 2020 4/63
  • 5. Formation of partial differential equations The partial differential equation of a given relation can be formed by either the elimination of the arbitrary constants or by the elimination of arbitrary functions depending upon whether the given relation involved by the arbitrary constants or arbitrary functions. We concentrate the partial differential equation formed by relation having one dependent variable namely z and two independent variables namely x and y. This relation may be of arbitrary functions or arbitrary constants. We use the following notions for the convenience which are used globally. ∂z ∂x = p, ∂z ∂y = q, ∂2z ∂x2 = r, ∂2z ∂x∂y = s, ∂2z ∂y2 = t. P. Sam Johnson First Order Partial Differential Equations March 5, 2020 5/63
  • 6. Formation of partial differential equations by eliminating arbitrary constants Let f (x, y, z, a, b) = 0 be a given relation with two arbitrary constants a and b. Differentiating the above equation partially with respect to x and y we get ∂f ∂x + ∂f ∂z ∂z ∂x = 0 =⇒ ∂f ∂x + p ∂f ∂z = 0 ∂f ∂y + ∂f ∂z ∂z ∂y = 0 =⇒ ∂f ∂y + q ∂f ∂z = 0. Eliminating a and b between the above three equations, we get a partial differential equation of the form F(x, y, z, p, q) = 0. P. Sam Johnson First Order Partial Differential Equations March 5, 2020 6/63
  • 7. Examples Example 1. 1. Form the partial differential equation by eliminating the constants a and b from z = (x + a)(y + b). [Answer : z = pq] 2. Form the partial differential equation by eliminating the constants a and b from z = axn + byn. [Answer: nz = px + qy] 3. Form the partial differential equation by eliminating the constants a and b from z = (x2 + a2)(y2 + b2). [Answer: 4xyz = pq] 4. Form the partial differential equation by eliminating the constants a and b from z = ax + by + √ a2 + b2. [Answer: z = px + qy + p p2 + q2] 5. Form the partial differential equation by eliminating the constants from log(az − 1) = x + ay + b. [Answer: q(z − p) = p] P. Sam Johnson First Order Partial Differential Equations March 5, 2020 7/63
  • 8. Examples Example 2. 1. Form the partial differential equation of the surface (x − a)2 + (y − b)2 = z2 cot2 α, where a and b are constants. [Answer: p2 + q2 = tan2 α] 2. Find the partial differential equation of all spheres whose center lies on z-axis. [Hint : If the equation of the sphere whose center lies on z-axis is given by x2 + y2 + (z − c)2 = r2, where c and r are constants, then PDE is xq = yp.] 3. Find the partial differential equation of all spheres whose radius is unity and center lies on xy plane. [Hint : If the equation of the sphere whose radius is unity and center lies on xy plane is given by (x − a)2 + (y − b)2 + z2 = 1, where a and b are constants, then z2(p2 + q2 + 1) = 1.] P. Sam Johnson First Order Partial Differential Equations March 5, 2020 8/63
  • 9. Examples Example 3. 1. Find the partial differential equation of all planes passing through the origin. [Hint : The equation of the plane which passes through origin is given by ax + by + cz = 0, where a, b and c are constants, then z = px + qy] 2. Find the partial differential equation of all planes whose sum of x, y, z intercepts is unity. [Hint : The equation of the plane with a, b, c as the x, y, z intercepts is x a + y b + z c = 1. Answer: z = p x + q y + pq p+q−pq .] P. Sam Johnson First Order Partial Differential Equations March 5, 2020 9/63
  • 10. Formation of partial differential equations by eliminating arbitrary functions Formation of partial differential equations by the elimination of arbitrary functions from the given relation is explained in the following examples. The elimination of one arbitrary functions from the given relation gives partial differential equations of first order and elimination of two arbitrary functions from the given relation gives partial differential equations of second or higher order. P. Sam Johnson First Order Partial Differential Equations March 5, 2020 10/63
  • 11. Examples Example 4. 1. Form the partial differential equation by eliminating the arbitrary function f from z = f (x + y). [Answer: p = q] 2. Form the partial differential equation by eliminating f from z = f (x2 + y2). [Answer: py = qx] 3. Form the partial differential equation by eliminating f from z = x + y + f (xy). [Answer: px − qy = x − y] 4. Form the partial differential equation by eliminating f from z = f (x2 + y2 + z2). [Answer: yp = xq] 5. Form the partial differential equation by eliminating f from z = x2 + 2f 1 y + log x . [Answer: px + qy2 = 2x2] 6. Form the partial differential equations by eliminating the arbitrary functions from z = f1(x) + f2(y). [Answer: ∂2z ∂y∂x = 0] P. Sam Johnson First Order Partial Differential Equations March 5, 2020 11/63
  • 12. Examples Example 5. 1. Form the partial differential equations by eliminating the arbitrary functions from z = yf1(x) + f2(y). [Answer: ∂z ∂x = y ∂2z ∂y∂x ] 2. Form the partial differential equation by eliminating the f and φ from z = f (y) + φ(x + y + z). [Answer: r(1 + q) = s(1 + p)] 3. Form a partial differential equation from the relation by eliminating the arbitrary functions f1 and f2 from z = f1(x + 2y) + f2(x − 2y). [Answer: 4∂2z ∂x2 = ∂2z ∂y2 .] 4. Form a PDE from the relation z = f1(ax + by) + f2(cx + dy). [Answer: bd ∂2z ∂x2 − (ad + bc) ∂2z ∂x∂y + ac ∂2z ∂y2 = 0] 5. Form a partial differential equation from the relation z = f1(2x + 3y) + f2(4x + 5y). [Answer: 15∂2z ∂x2 − 22 ∂2z ∂x∂y + 8∂2z ∂y2 = 0] 6. Form a partial differential equation by eliminating the functions from z = f1(x2 + 3y) + f2(x2 − 3y) [Answer : 9∂2z ∂x2 − 9 x ∂z ∂x − 4x2 ∂2z ∂y2 = 0] P. Sam Johnson First Order Partial Differential Equations March 5, 2020 12/63
  • 13. Formation of partial differential equations by elimination of arbitrary function φ from φ(u, v) = 0 where u and v are functions of x, y and z. Let φ(u, v) = 0 (1) be the given relation with u and v are functions of x, y and z. Differentiating (1) partially with respect to x and y we get ∂φ ∂u ∂u ∂x + ∂φ ∂v ∂v ∂x = 0 (2) ∂φ ∂u ∂u ∂y + ∂φ ∂v ∂v ∂y = 0. (3) P. Sam Johnson First Order Partial Differential Equations March 5, 2020 13/63
  • 14. Formation of partial differential equations by elimination Now eliminating φ between (2) and (3) we get the required partial differential equation as ∂u ∂x ∂v ∂x ∂u ∂y ∂v ∂y = 0. Example 6. 1. Form a partial differential equation by eliminating φ from φ x y , y z = 0. [Answer: z = xp + yq is the required solution.] 2. Form a partial differential equation by eliminating φ from φ(x + y + z, x2 + y2 + z2) = 0. [Answer:(y − z)p + (z − x)q = x − y] P. Sam Johnson First Order Partial Differential Equations March 5, 2020 14/63
  • 15. Solution of partial differential equations Consider the relation z = ax + by. (1) Eliminating the arbitrary constants a and b we get z = px + qy. (2) Also consider the relation z = x f y x . (3) Eliminating the arbitrary function f we get z = px + qy. From these we conclude that both (1) and (3) are solutions of the same equation (2) and for a single partial differential equation we have more than one solutions. P. Sam Johnson First Order Partial Differential Equations March 5, 2020 15/63
  • 16. Types of solutions 1. A solution which contains the number of arbitrary constants is equal to the number of independent variables is called complete solution or complete integral. 2. A complete solution in which if a particular value is given to arbitrary constant is called particular solution. 3. A solution which contains the maximum possible number of arbitrary functions is called a general solution. P. Sam Johnson First Order Partial Differential Equations March 5, 2020 16/63
  • 17. Solution of partial differential equation by direct integration Simple partial differential equations can be solved by direct integration. The method of solving such equations explained as follows. Example 7. Solve ∂z ∂x = 0. Solution. The given equation is ∂z ∂x = 0. This shows that z is a function of the independent x. Integrating with respect to x z = f (y) is the solution of the given partial differential equation. P. Sam Johnson First Order Partial Differential Equations March 5, 2020 17/63
  • 18. Solution of partial differential equation by direct integration Example 8. Solve ∂2 z ∂x∂y = 0. Solution. The given equation can be written as ∂2z ∂x∂y = ∂ ∂x ∂z ∂y = 0. Integrating with respect to x ∂z ∂y = f1(y). Now integrating with respect to y z = Z f1(y)dy + f2(x) = g1(y) + f2(x). Hence the solution is z = g1(y) + f2(x). P. Sam Johnson First Order Partial Differential Equations March 5, 2020 18/63
  • 19. Examples Example 9. Solve ∂2 z ∂x∂y = x y + c. Solution. The given equation can be written as ∂2z ∂x∂y = ∂ ∂x ∂z ∂y = x y + c. Integrating with respect to x ∂z ∂y = x2 2y + cx + f1(y). Now integrating with respect to y z = x2 2 log y + cxy + Z f1(y)dy + f2(x) = x2 2 log y + cxy + g1(y) + f2(x). Hence the solution is z = x2 2 log y + cxy + g1(y) + f2(x). P. Sam Johnson First Order Partial Differential Equations March 5, 2020 19/63
  • 20. Examples Example 10. Solve ∂2 z ∂x2 = cos x. Solution. The given equation is ∂2z ∂x2 = cos x. Integrating with respect to x ∂z ∂x = sin x + f1(y). Again integrating with respect to x z = − cos x + xf1(y) + f2(y). P. Sam Johnson First Order Partial Differential Equations March 5, 2020 20/63
  • 21. Example Example 11. Solve ∂2 z ∂y∂x = e−x cos y given that z = 0 when x = 0 and ∂z ∂x = 0 when y = 0. Solution. The given equation can be written as ∂2z ∂y∂x = ∂ ∂y ∂z ∂x = e−x cos y. Integrating with respect to y, ∂z ∂x = e−x sin y + f1(x). Applying ∂z ∂x = 0 when y = 0, we get f1(x) = 0, hence ∂z ∂x = e−x sin y. Now integrating with respect to x z = −e−x sin y + f2(y). Applying z = 0 when x = 0 0 = − sin y + f2(y) ⇒ f2(y) = sin y. Hence z = −e−x sin y + sin y. P. Sam Johnson First Order Partial Differential Equations March 5, 2020 21/63
  • 22. Standard types of first order partial differential equations Type 1. Consider the partial differential equation of the form F(p, q) = 0. (1) Here the variables x, y, z do not occur explicitly. Let z = ax + by + c. (2) be a solution of the given equation. Then ∂z ∂x = p = a, ∂z ∂y = q = b. Substituting these values of p and q in the equation (1) we get F(a, b) = 0. Solving for b we get b = f (a) then z = ax + f (a)y + c (3) is the complete solution of the given equation. P. Sam Johnson First Order Partial Differential Equations March 5, 2020 22/63
  • 23. Type 1 To get the general solution put c = g(a) z = ax + f (a)y + g(a) (4) and differentiating with respect to a 0 = x + f 0 (a)y + g0 (a). (5) Eliminating a between (4) and (5) we get the general solution. To get the singular solution, differentiating (2) with respect to c we get 0 = 1, which is not possible. Hence there is no singular solution. P. Sam Johnson First Order Partial Differential Equations March 5, 2020 23/63
  • 24. Examples Example 12. Solve p2 + p − q2 = 0. Solution. The given equation is of the form F(p, q) = 0. The complete solution is z = ax + by + c (1) where a2 + a − b2 = 0. Solving for b we get b2 = a2 + a, so b = √ a2 + a. Hence the complete solution is z = ax + p a2 + ay + c. (2) Differentiating (2) with respect to c we get 0 = 1, which is not possible. Hence there is no singular solution. To get the general solution put c = g(a) z = ax + p a2 + ay + g(a) (3) and differentiating with respect to a, we have 0 = x + 1+2a 2 √ a2+a y + g0(a). Eliminating a between (3) and (4) we get the general solution. P. Sam Johnson First Order Partial Differential Equations March 5, 2020 24/63
  • 25. Examples Example 13. Solve p2 + q2 = 9. Solution. The given equation is of the form F(p, q) = 0. The complete solution is z = ax + by + c (1) where a2 + b2 = 9. Solving for b we get b2 = 9 − a2, so b = √ 9 − a2. Hence the complete solution is z = ax + p 9 − a2y + c (2) Differentiating (2) with respect to c we get 0 = 1, which is not possible. Hence there is no singular solution. To get the general solution put c = g(a) z = ax + p 9 − a2y + g(a) (3) and differentiating with respect to a 0 = x + −2a 2 √ 9 − a2 y + g0 (a). (4) Eliminating a between (3) and (4) we get the general solution. P. Sam Johnson First Order Partial Differential Equations March 5, 2020 25/63
  • 26. Examples Example 14. Solve pq + p + q = 0. Solution. The given equation is of the form F(p, q) = 0. The complete solution is z = ax + by + c (1) where ab + a + b = 0. Solving for b we get b(1 + a) = −a, so b = − a 1+a . Hence the complete solution is z = ax − a 1 + a y + c. (2) Differentiating (2) with respect to c we get 0 = 1, which is not possible. Hence there is no singular solution. To get the general solution put c = g(a) z = ax − a 1 + a y + g(a) (3) and differentiating with respect to a 0 = x − 1 (1 + a)2 y + g0 (a). (4) Eliminating a between (3) and (4) we get the general solution. P. Sam Johnson First Order Partial Differential Equations March 5, 2020 26/63
  • 27. Type 2 Consider the partial differential equations of the form z = px + qy + F(p, q). This equation is called the Clairut’s equation. Let z = ax + by + c be a solution of the given equation. Then ∂z ∂x = p = a, ∂z ∂y = q = b. Substituting these values of p and q in the given equation, we get z = ax + by + F(a, b). P. Sam Johnson First Order Partial Differential Equations March 5, 2020 27/63
  • 28. Type 2 : Example Example 15. Solve z = px + qy + pq. Solution. The given equation is of the form z = px + qy + F(p, q). The complete solution is z = ax + by + ab. (1) To find the singular solution differentiating (1) partially with respect to a 0 = x + b ⇒ b = −x and differentiating (1) partially with respect to b we get 0 = y + a ⇒ a = −y. Substituting these values in (1) we get z = −(y)x − (x)y + (−y)(−x) z = −xy − xy + xy x = −xy z + xy = 0. P. Sam Johnson First Order Partial Differential Equations March 5, 2020 28/63
  • 29. Example Example 16. Solve z = px + qy + p2 + q2 Solution. The given equation is of the from z = px + qy + F(p, q). The complete solution is z = ax + by + a2 + b2 (1) To find the singular solution differentiating (1) partially with respect to a 0 = x + 2a ⇒ a = − x 2 and differentiating (1) partially with respect to b we get, 0 = y + 2b ⇒ b = −y 2 . Substituting these values in (1) we get z = − x 2 x − y 2 y + − x 2 2 + − y 2 2 z = − x2 2 − y2 2 + x2 4 + y2 4 z = − x2 4 − y2 4 4z = −(x2 + y2 ). P. Sam Johnson First Order Partial Differential Equations March 5, 2020 29/63
  • 30. Example Example 17. Solve z = px + qy + p2 + pq + q2. Solution. The given equation is of the form z = px + qy + F(p, q). The complete solution is z = ax + by + a2 + ab + b2 . (1) To find the singular solution differentiating (1) partially with respect to a and b we get, x + 2a + b = 0 and y + a + 2b = 0. Solving we get a = 1 3 (y − 2x) and b = 1 3 (x − 2y). Substituting these values of a and b in (1) we get z = 1 3 (y − 2x)x + 1 3 (x − 2y)y + 1 9 (y − 2x)2 + 1 9 (y − 2x)(x − 2y) + 1 9 (x − 2y)2 9z = 3x(y − 2x) + 3y(x − 2y) + (y − 2x)2 + (y − 2x)(x − 2y) + (x − 2y)2 9z = −3x2 + 3xy − 3y2 . Hence 3z + x2 − xy + y2 = 0. P. Sam Johnson First Order Partial Differential Equations March 5, 2020 30/63
  • 31. Example Example 18. Solve z = px + qy + p 1 + p2 + q2. Solution. The given equation is of the form z = px + qy + F(p, q). The complete solution is z = ax + by + p 1 + a2 + b2. (1) To find the singular solution differentiating (1) partially with respect to a and b we get x = − a √ 1+a2+b2 and y = − b √ 1+a2+b2 . Hence p 1 − x2 − y2 = 1 √ 1+a2+b2 . From x and y, we get a = − x √ 1−x2−y2 and b = − y 1−x2−y2 . Substituting the values in (1) we have z = − x2 p 1 − x2 − y2 − y2 p 1 − x2 − y2 + 1 p 1 − x2 − y2 z = 1 − x2 − y2 p 1 − x2 − y2 z2 = 1 − x2 − y2 . Thus x2 + y2 + z2 = 1. P. Sam Johnson First Order Partial Differential Equations March 5, 2020 31/63
  • 32. Type 3. This type of partial differential equation can be further classified into three categories. We discuss each case as follows. Case I. Consider the partial differential equation of the form F(x, p, q) = 0. Let us assume that q = a. Then F(x, p, a) = 0. Solving for p we get p = φ(x, a). Since z is a function of x and y dz = ∂z ∂x dx + ∂z ∂y dy = pdx + qdy = φ(x, a)dx + qdy z = Z φ(x, a)dx + Z ady + c z = Z φ(x, a)dx + ay + c. P. Sam Johnson First Order Partial Differential Equations March 5, 2020 32/63
  • 33. Examples Example 19. Solve pq = x. Solution. The given equation is of the form F(x, p, q) = 0. Let us assume that q = a. Then pa = x ⇒ p = x a . dz = pdx + qdy = x a dx + a dy z = Z x a dx + Z ady + c z = x2 2a + ay + c. P. Sam Johnson First Order Partial Differential Equations March 5, 2020 33/63
  • 34. Example Example 20. Solve q = px + p2. Solution. The given equation is of the form F(x, p, q) = 0. Let us assume that q = a. then p2 + px − a = 0 p = −x ± √ x2 + 4a 2 dz = pdx + qdy = −x ± √ x2 + 4a 2 ! dx + a dy z = Z x 2 dx ± 1 2 Z p x2 + 4a dx + Z ady + c = x2 4 ± 1 2 x 2 p x2 + 4a + 4a 2 sinh−1 x 2 √ a + ay + c = x2 4 ± x 4 p x2 + 4a + a sinh−1 x 2 √ a + ay + c. P. Sam Johnson First Order Partial Differential Equations March 5, 2020 34/63
  • 35. Case II Consider the partial differential equations of the form F(y, p, q) = 0. Let us assume that p = a. Then F(y, a, q) = 0. Solving for q we get q = φ(y, a) Since z is a function of x and y dz = ∂z ∂x dx + ∂z ∂y dy = pdx + qdy = adx + φ(y, a)dy z = Z adx + Z φ(y, a)dy + c z = ax + Z φ(y, a)dy + c. P. Sam Johnson First Order Partial Differential Equations March 5, 2020 35/63
  • 36. Examples Example 21. Solve q2 = yp3. Solution. The given equation is of the form F(y, p, q) = 0. Let us assume that p = a. Then q2 = ya3 = 0 q = a 3 2 √ y. dz = pdx + qdy = a dx + a 3 2 √ ydy z = Z a dx + Z a 3 2 √ y dy + c x = a x + 2 3 a 3 2 y 3 2 + c. P. Sam Johnson First Order Partial Differential Equations March 5, 2020 36/63
  • 37. Example Example 22. Solve p = qy + q2. Solution. The given equation is of the form F(y, p, q) = 0. Let us assume that p = a. Then q2 + qy − a = 0 q = −y ± p y2 + 4a 2 dz = pdx + qdy = a dx + −y ± p y2 + 4a 2 ! dy z = Z adx + Z y 2 dy ± 1 2 Z p y2 + 4a dy + c = ax + y2 4 ± 1 2 y 2 p y2 + 4a + 4a 2 sinh−1 y 2 √ a + c = ax + y2 4 ± y 4 p y2 + 4a + a sinh−1 y 2 √ a + c P. Sam Johnson First Order Partial Differential Equations March 5, 2020 37/63
  • 38. Case III Consider the partial differential equations of the form F(z, p, q) = 0. Let us assume that q = ap. Then F(z, p, ap) = 0. Solving for p we get p = φ(z, a) q = ap = aφ(z, a) Since z is a function of x and y dz = ∂z ∂x dx + ∂z ∂y dy = pdx + qdy = φ(z, a)dx + aφ(z, a)dy Z dz φ(z, a) = Z dx + Z ady + c Z dz φ(z, a) = x + ay + c. P. Sam Johnson First Order Partial Differential Equations March 5, 2020 38/63
  • 39. Examples Example 23. Solve 1 + z3 = z4pq. Solution. The given equation is of the form F(z, p, q) = 0. Let us assume that q = ap. Then 1 + z3 = z4p2a gives p = 1 √ a √ 1+z3 z2 and q = a 1 √ a √ 1+z3 z2 . So dz = pdx + qdy gives Z dz = 1 √ a Z √ 1 + z3 z2 dx + a 1 a Z √ 1 + z3 z2 dy Z z2 √ 1 + z3 dz = 1 √ a Z dx + a Z dy + c 2 3 p 1 + z3 = 1 √ a (x + ay + c) 4 9 (1 + z3 ) = 1 a (x + ay + c)2 4 a(1 + z3 ) = 9(x + ay + c)2 . P. Sam Johnson First Order Partial Differential Equations March 5, 2020 39/63
  • 40. Examples Example 24. Solve z = p2 + q2. Solution. The given equation is of the form F(z, p, q) = 0. Let us assume that q = ap. Then z = p2 + a2p2 gives p = 1 √ 1+a2 √ z and q = a √ 1+a2 √ z. So dz = pdx + qdy gives Z dz = 1 √ 1 + a2 Z √ zdx + a √ 1 + a2 Z √ zdy Z dz √ z = 1 √ 1 + a2 Z dx + a Z dy + c 2 √ z = 1 √ 1 + a2 (x + ay + c) 4z = 1 1 + a2 (x + ay + c)2 . P. Sam Johnson First Order Partial Differential Equations March 5, 2020 40/63
  • 41. Examples Example 25. Solve p(1 − q2) = q(1 − z). Solution. The given equation is of the form F(z, p, q) = 0. Let us assume that q = ap. Then p(1 − a2 p2 ) = ap(1 − z) (1 − a2 p2 ) = a(1 − z) a2 p2 = 1 − a + az p = 1 a √ 1 − a + az q = √ 1 − a + az dz = pdx + qdy Z dz = 1 a Z √ 1 − a + az dx + Z √ 1 − a + az dy Z dz √ 1 − a + az = 1 a Z dx + a Z dy + c 2 a √ 1 − a + az = 1 a (x + ay + c) 4(1 − a + az) = (x + ay + c)2 . P. Sam Johnson First Order Partial Differential Equations March 5, 2020 41/63
  • 42. Example 26. Solve q2 = z2p2(1 − p2). Solution. The given equation is of the form F(z, p, q) = 0. Let us assume that q = ap. Then a2 p2 = z2 p2 (1 − p2 ) a2 = z2 (1 − p2 ) z2 p2 = z2 − a2 p = 1 z p z2 − a2 q = a 1 z p z2 − a2. dz = pdx + qdy Z dz = Z 1 z p z2 − a2dx + Z a 1 z p z2 − a2dy Z z dz √ z2 − a2 = Z dx + a Z dy + c p z2 − a2 = (x + ay + c) z2 − a2 = (x + ay + c)2 z2 = a2 + (x + ay + c)2 . P. Sam Johnson First Order Partial Differential Equations March 5, 2020 42/63
  • 43. Example Example 27. Solve p3 + q3 = z3. Solution. The given equation is of the form F(z, p, q) = 0. Let us assume that q = ap. Then p = 1 (1 + a3) 1 3 z q = a 1 (1 + a3) 1 3 z dz = pdx + qdy Z dz = Z 1 (1 + a3) 1 3 z dx + Z a 1 (1 + a3) 1 3 z dy (1 + a3 ) 1 3 Z dz z = Z dx + a + Z dy + c (1 + a3 ) 1 3 log z = (x + ay + c). P. Sam Johnson First Order Partial Differential Equations March 5, 2020 43/63
  • 44. Example Example 28. Solve z2(p2 + q2 + 1) = 1. Solution. The given equation is of the form F(z, p, q) = 0. Let us assume that q = ap. Then p = 1 √ 1 + a2 √ 1 − z2 z q = a √ 1 + a2 √ 1 − z2 z . dz = pdx + qdy Z dz = Z 1 √ 1 + a2 √ 1 − z2 z dx + Z a 1 √ 1 + a2 √ 1 − z2 z dy Z z dz √ 1 − z2 = 1 √ 1 + a2 Z dx + a Z dy + c − p 1 − z2 = 1 √ 1 + a2 (x + ay + c). P. Sam Johnson First Order Partial Differential Equations March 5, 2020 44/63
  • 45. Example Example 29. Solve 2p2 − qz = z2. Solution. The given equation is of the form F(z, p, q) = 0. Let us assume that q = ap. Then p = z a ± √ a2 + 8 2 ! q = a z a ± √ a2 + 8 2 ! . dz = pdx + qdy Z dz = Z z a ± √ a2 + 8 2 ! dx + Z a z a ± √ a2 + 8 2 ! dy Z dz z = a ± √ a2 + 8 2 ! ( Z dx + a Z dy + c) log z = a ± √ a2 + 8 2 ! (x + ay + c). P. Sam Johnson First Order Partial Differential Equations March 5, 2020 45/63
  • 46. Example Example 30. Solve p(1 + q) = q z. Solution. The given equation is of the form F(z, p, q) = 0. Let us assume that q = ap. Then p(1 + ap) = zpa p[(1 + az) + ap] = 0 p = 0, 1 a (az − 1) q = a 1 a (az − 1). dz = pdx + qdy Z dz = 1 a Z (az − 1)dx + a 1 a Z (az − 1)dy Z 1 az − 1 dz = 1 a Z dx + a Z dy + c 1 a log(az − 1) = 1 a (x + ay + c) log(az − 1) = (x + ay + c). P. Sam Johnson First Order Partial Differential Equations March 5, 2020 46/63
  • 47. Type 4 Consider the partial differential equation of the form F1(x, p) = F2(y, q). Let us assume that F1(x, p) = a F2(y, q) = a. Solving for p and q we get p = φ1(x, a) q = φ2(y, a). Since z is a function of x and y dz = ∂z ∂x dx + ∂z ∂y dy = pdx + qdy = φ1(x, a)dx + φ2(y, a)dy z = Z φ1(x, a)dx + Z φ2(y, a)dy + c. P. Sam Johnson First Order Partial Differential Equations March 5, 2020 47/63
  • 48. Examples Example 31. Solve p + q = x + y. Solution. The given equation is of the form F1(x, p) = F2(y, q). p + q = x + y = a(say) p − x = a ⇒ p = x + a q − y = a ⇒ q = y + a dz = pdx + qdy z = Z (x + a)dx + Z (y + a)dy + c z = (x + a)2 2 + (y + a)2 2 + c. P. Sam Johnson First Order Partial Differential Equations March 5, 2020 48/63
  • 49. Examples Example 32. Solve p2 + q2 = x + y. Solution. The given equation is of the form F1(x, p) = F2(y, p). p2 + q2 = x + y = a(say) p2 − x = a ⇒ p = √ x + a y − q2 = a ⇒ q = √ y − a dz = pdx + qdy z = √ x + adx + Z √ y − ady + c z = 2 3 (x + a) 3 2 + 2 3 (y − a) 3 2 + c. P. Sam Johnson First Order Partial Differential Equations March 5, 2020 49/63
  • 50. Example Example 33. Solve p2 + q2 = x2 + y2. Solution. The given equation is of the form F1(x, p) = F2(y, q). From p2 + q2 = x2 + y2 = a2(say), we have, p2 − x2 = a2 ⇒ p = p x2 + a2, y2 − q2 = a2 ⇒ q = p y2 − a2 dz = pdx + qdy z = Z p x2 + a2dx + Z p y2 − a2dy + c z = x 2 p x2 + a2 a2 2 sinh−1 x a + y 2 p y2 − a2 − a2 2 cosh−1 x a + c. P. Sam Johnson First Order Partial Differential Equations March 5, 2020 50/63
  • 51. Example Example 34. Solve q2x(1 + y2) = py2. Solution. The given equation can written in form F1(x, p) = F2(y, q). q2 x(1 + y2 ) = py2 q2 1 + y2 y2 = p x = a2 (say) p x = a2 ⇒ p = a2 x q2 1 + y2 y2 = a2 ⇒ q = ay p 1 + y2 dz = pdx + qdy z = Z a2 x dx + Z ay p 1 + y2 dy + c z = a2 x2 2 + a p 1 + y2 + c. P. Sam Johnson First Order Partial Differential Equations March 5, 2020 51/63
  • 52. Type 5 Consider the partial differential equations of the form F(xm p, yn q) = 0 (1) and F(z, xm p, yn q) = 0, (2) where m and n are constants. Case I. For m 6= 1 and n 6= 1 put X = x1−m ⇒ ∂X ∂x = (1 − m)x−m , Y = y1−n ⇒ ∂Y ∂y = (1 − n)y−n P = ∂z ∂x = ∂z ∂X ∂X ∂x = P(1 − m)x−m ⇒ P(1 − m) = pxm (3) Q = ∂z ∂y = ∂z ∂Y ∂Y ∂y = Q(1 − n)y−n ⇒ Q(1 − n) = qyn (4) where P = ∂x ∂X and Q = ∂z ∂Y . Using expressions (3) and (4), the equations (1) and (2) can be transformed into equations F(P, Q) = 0 and F(z, P, Q) = 0, respectively. These equations are easily solvable by standard method. P. Sam Johnson First Order Partial Differential Equations March 5, 2020 52/63
  • 53. Case 2. For m = 1 and n = 1 the equation is of the form F(xp, yp) = 0. To solve this put X = log x ⇒ ∂X ∂x = 1 x Y = log y ⇒ ∂Y ∂y = 1 y p = ∂z ∂x = ∂z ∂X ∂X ∂x = P 1 x ⇒ P = px (5) q = ∂z ∂y = ∂z ∂Y ∂Y ∂y = Q 1 y ⇒ Q = qy (6) where P = ∂z ∂X and Q = ∂z ∂Y . Substituting the values of (5) and (6) in (1) and (2) we get transformed equatons F(P, Q) = 0 and F(z, P, Q) = 0, respectively. These equations are easily solvable by standard method. P. Sam Johnson First Order Partial Differential Equations March 5, 2020 53/63
  • 54. Examples Example 35. Solve xp + yq = 1. Solution. The given equation is of the form F(xmp, ynq) = 0. Here m = 1 and n = 1. Put X = log x and Y = log y, then xp = P yp = Q. Hence the equation becomes P + Q = 1 which is of the form F(P, Q) = 0. Let z = aX + bY + c be the solution with a + b = 1 ⇒ b = 1 − a z = aX + (1 − a)Y + c z = a log x + (1 − a) log y + c. P. Sam Johnson First Order Partial Differential Equations March 5, 2020 54/63
  • 55. Example Example 36. Find the solution of px2 + qy2 = z2. Solution. The given equation is of the form F(xmp, ynq, z) = 0. Here m = 2 and n = 2. Put x1−2 = X and y1−2 = Y , then x2p = −P, y2q = −Q. Hence the equation becomes P + Q = −z2. Putting Q = aP, we get P + aP = −z2 ⇒ P = − z2 (1+a) and Q = − a z2 (1+a) . dz = PdX + QdY = − z2 (1 + a) dX − a (1 + a) z2 dY − Z dz z2 = 1 (1 + a) Z dX + a Z dY 1 z = 1 (1 + a) (X + aY ) + c 1 z = 1 (1 + a) 1 x + a 1 y + c. P. Sam Johnson First Order Partial Differential Equations March 5, 2020 55/63
  • 56. Example Example 37. Solve x2y2pq = z2. Solution. The given equation is of the form F(xmp, ynq, z) = 0. Here m = 2 and n = 2. Put x1−2 = X and y1−2 = Y , then x2p = −P and y2q = −Q. Hence the equation becomes PQ = z2. Putting Q = a2P, we get P2a2 = z2 ⇒ P = z a and Q = a2 z a = az. dz = PdX + QdY = 1 a zdX + azdY Z dz z = 1 a Z dX + a Z dY log z = 1 a (X + aY ) + c log z = 1 a 1 x + a 1 y + c. P. Sam Johnson First Order Partial Differential Equations March 5, 2020 56/63
  • 57. Example Example 38. Solve p x2 + q y2 = z. Solution. The given equation is of the form F(xmp, ynq, z) = 0. Here m = −2 and n = −2. Put x1−(−2) = x3 = X and y1−(−2) = y3 = Y , then x−2p = 3P and y−2q = 3Q. Hence the equation becomes 3P + 3Q = z. Putting Q = aP, we get 3P + 3aP = z ⇒ P = z 3(1+a) and Q = a z 3(1+a) = a 3(1+a) z. dz = PdX + QdY = 1 3(1 + a) zdX + a 3(1 + a) zdY Z dz z = 1 3(1 + a) Z dX + a Z dY log z = 1 3(1 + a) (X + aY ) + c log z = 1 3(1 + a) (x3 + ay3 ) + c. P. Sam Johnson First Order Partial Differential Equations March 5, 2020 57/63
  • 58. Example Example 39. Solve x4p2 − yzq = z2. Solution. The given equation can be written as (x2p)2 − (yz)q = z2. Putting x1−2 = X and log y = Y , we get x2p = P and yp = Q. Hence the equation becomes P2 − Qz = z2. This is the form F(P, Q, z) = 0. Putting Q = aP, we get P2 − azP = z2 ⇒ P = z(a ± √ a2 + 4) 2 , and Q = a z(a ± √ a2 + 4) 2 . dz = PdX + Qdy = z(a ± √ a2 + 4) 2 dX + a z(a ± √ a2 + 4) 2 dY Z dz z = (a ± √ a2 + 4) 2 Z dX + a Z dY log z = (a ± √ a2 + 4) 2 (X + aY ) + c log z = (a ± √ a2 + 4) 2 1 x + a log y + c. P. Sam Johnson First Order Partial Differential Equations March 5, 2020 58/63
  • 59. Type 6. Consider the partial differential equations of the form F(zm p, zm q) = 0 (1) and F1(x, zm p) = F2(y, zm q) (2) where m is a constant. Case I For m 6= −1 put Z = zm+1, we get ∂Z ∂x = ∂Z ∂z ∂z ∂x ⇒ P = (m + 1)zm p ⇒ P m + 1 = p (3) ∂Z ∂y = ∂Z ∂z ∂z ∂y ⇒ Q = (m + 1)zm q ⇒ Q m + 1 = q, (4) where P = ∂Z ∂x and Z = ∂Z ∂y . Substituting (3) and (4) in (1) and (2) we get transformed equations of the form F(P, Q) = 0 and F1(x, P) = F2(y, Q) and this can be easily solvable by standard method. P. Sam Johnson First Order Partial Differential Equations March 5, 2020 59/63
  • 60. Type 6 Case 2. For m = −1 put Z = log z ∂Z ∂x = ∂Z ∂z ∂z ∂x ⇒ P = 1 z p ⇒ P = z−1 p (5) ∂Z ∂y = ∂Z ∂z ∂z ∂y ⇒ Q = 1 z q ⇒ Q = z−1 q, (6) where P = ∂Z ∂x and Q = ∂Z ∂y. Substituting (5) and (6) in (1) and (2) we have transformed equations of the form F(P, Q) = 0 and F1(x, P) = F2(y, Q) and this can be easily solvable by standard method. P. Sam Johnson First Order Partial Differential Equations March 5, 2020 60/63
  • 61. Examples Example 40. Solve z2(p2 + q2) = x2 + y2. Solution. The given equation is of the form F1(x, zmp) = F2(y, zmq). Putting Z = z2 we get P = 2zp ⇒ P 2 = zp and Q = 2zq ⇒ Q 2 = zq. Hence the equation becomes P 2 2 + Q 2 2 = x2 + y2 ⇒ P2 + Q2 = 4x2 + 4y2 ⇒ P2 − 4x2 = −Q2 + 4y2 = 4a2 (say) P2 − 4x2 = 4a2 and − Q2 + 4y2 = 4a2 dZ = Pdx + Qdy Z dZ = 2 Z p x2 + a2dx + 2 Z p y2 − a2dy Z = 2 x 2 p x2 + a2 + a2 2 sinh−1 x a + 2 y 2 p y2 − a2 − a2 2 cosh−1 y a z2 = x p x2 + a2 + a2 sinh−1 x a + y p y2 − a2 − a2 cosh−1 y a . P. Sam Johnson First Order Partial Differential Equations March 5, 2020 61/63
  • 62. Example Example 41. Solve (z2p + x)2 + (z2q + y)2 = 1. Solution. The given equation is of the form F1(x, zmp) = F2(y, zmq). Putting Z = z3, we get P = 3z2p ⇒ P 3 = z2p and Q = 3z2q ⇒ Q 3 = z2q. Hence the equation becomes P 3 + x 2 + Q 3 + y 2 = 1 ⇒ P = 3(a − x) 1 − Q 3 + y 2 = a2 ⇒ Q = 3( p 1 − a2 − y) dZ = Pdx + Qdy Z dZ = Z 3(a − x)dx + Z 3( p 1 − a2 − y)dy z = 3 ax − x2 2 + 3 ( p 1 − a2)y − y2 2 + c z3 = 3 ax − x2 2 + 3 ( p 1 − a2)y − y2 2 + c. P. Sam Johnson First Order Partial Differential Equations March 5, 2020 62/63
  • 63. References T. Amaranath, An Elementary Course in Partial Differential Equations, Second Edition, Narosa Publishing House, 1997. Ian Sneddon, Elements of Partial Differential Equations, McGraw-Hill Book Company, 1957. P. Sam Johnson First Order Partial Differential Equations March 5, 2020 63/63