1. Analytical Methods for Solving Higher
Order Ordinary Differential Equation.
Rabbiya Ahmed
Reg: CIIT/SP19-BSM-013/ATK
Supervised By :Dr Maimona Rafiq
Department of Mathematics
COMSATS University Islamabad,
Attock Campus
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2. Outlines
• Introduction
• Self-adjoint Method for Solving Higher Order ODEs
• Integrating Factor Technique for Solving Higher Order ODEs
• References
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3. Introduction
Definition 1.
A differential equation containing the derivatives (some finite
number of derivatives ) of one or more dependent variables, with
respect to one or more independent variables, is said to be a
differential equation.
Differential equations are classified into two main parts :
1. Ordinary Differential Equations.
2. Partial Differential Equations.
Definition 2.
If an equation contains only ordinary derivatives of one or more
dependent variables with respect to a single independent variable,
it is said to be an ordinary differential equation (ODE).
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4. Introduction
Example 3.
dy
dx + 5y = ex
Definition 4.
An equation involving partial derivatives of one or more dependent
variables with respect to two or more independent variables is
called a partial differential equation.
Example 5.
∂2u
∂x2 + ∂u
∂y = 0
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5. Introduction
Definition 6.
A solution of a differential equation is a function that satisfies the
equation. When you substitute the function into the differential
equation, you get a true mathematical statement.
Definition 7.
An initial value problem is a problem which seeks to determine a
solution to a differential equation subject to conditions on the
unknown function and its derivative specified at one value of the
independent variable.
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6. Introduction
Definition 8.
A boundary value problem is a problem which seeks to determine a
solution to a differential equation subject to conditions on the
unknown function specified at two or more values of the
independent variable..
Definition 9.
A differential equation Ly = 0 is said to be a self adjoint operator
if the operator L is self adjoint. Thus we have,
Ly = a0yn + a′
0y′ + a2y = (a0y′)′ + a2y
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7. Higher Order ODEs in the Self-Adjoint Form
Second Order ODEs in the Self-Adjoint Form
Consider the second order ODE:
p0(x)u′′
(x) + p1(x)u′
(x) + p2(x)u(x) = 0. (1)
Suppose we have an operator L defined as
L = p0(x)
d2
dx2
+ p1(x)
d
dx
+ p2(x). (2)
Using the substitution p′
0 = p1, we can rewrite Label (1) as
L̃u =
d2
dx2
(p0u) −
d
dx
(p1u) + p2u = 0. (3)
For a given operator L, there exists a corresponding operator L,
known as the adjoint operator associated with L. If the condition 7 / 23
8. Second Order ODE’s in the Self-Adjoint Form
Definition 10.
A second order ODE (1) is said to be in the self-adjoint form if and
only if:
Lu = Lu =
d
dx
p0u′
+ p2u = 0,
where p0(x) 0 on (a, b), p′
0(x), ant p2(x) are continuous
functions on [a, b] and condition and
p′
0 = p1
is satisfied.
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9. Theorem 11.
If a second order self-adjoint ODE
d
dx
(p0(x)u′
) + p2(x)u = 0
verifies the condition
p2(x) =
p0
′′
2
−
p0
′2
4p0
then the solution to equ is:
u(x) =
1
p
(p0(x)
(C1(x) + C0)
where C1 and C2 are arbitrary constants.
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10. Example
Consider the order ODE
d
dx
x
2
y′
+
1
2x
y
′
−
1
8
x
y′
+
1
2x
y
= 0
substitute v = y′ + 1
2x y
Thus
d
dx
hx
2
v′
i
−
1
8x
v = 0
Now Comparing with this equ
d
dx
P0(x)u′
+ P2(x)u
P0(x) =
x
2
and P2(x) = −
1
8x
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11. P2(x) =
P′′
0
2
−
P′
0
4P0
= 0 −
1
4
= −
1
4x
(5)
so the solution is
v(x) = c3
√
x +
c4
√
x
(6)
y′
+
1
2x
y = c3
√
x +
c4
√
x
(7)
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12. Third Order ODE in Self-adjoint Form
Definition 12.
A third order ODE is said to be in the self-adjoint form if and only
if:
Ly = Ly = roy′
′′
+ (q(x)y)′
+ p(x)y = 0.
where r0(x) 0 and conditions
r1(x) = 2r′
0(x), q(x) = r2(x) − r′′
0 (x), p(x) = r3(x) − q′
(x).
are satisfied.
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13. Theorem 13.
If a third order self-adjoint ODE
r(x)y′
′′
+ (q(x)y)′
+ p(x)y = 0
verifies the conditions q = r′′ − 2
3r (r′)2
and
p = −r′′
3 + 2
3r′r′′ − 10
27r2 (r′)3
, then the solution of third order
self-adjoint ODE is
y(x) =
1
3
p
r2(x)
C1x2
+ C2x + C3
where r(x) 0, p(x), q(x) are continuous differentiable functions,
and C1, C2, C3 are arbitrary constants.
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14. Example 14.
Consider a third order ODE
x2
y′
′′
+
−
2
3
y
′
+
−
8
27x
y = 0.
where r(x) = x2. Now, we check the conditions of Theorem 13 to
solve the given ODE :
q = r′′
−
2
3r
r′
2
= −
2
3
,
and p = −
r′′′
3
+
2
3r
r′
r′′
−
10
27r2
r′
3
= −
8
27x
.
We observe that the conditions are satisfied, hence we can obtain
the analytic solution from Theorem 13 as:
y(x) =
1
3
√
x4
C1x2
+ C2x + C3
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15. Integrating Factor techniques for Solving Higher Order
ODEs
Integrating Factor Technique for Third Order ODE
Theorem 15.
Given
y′′′
+ P(x)y′′
+ Q(x)y′
+ R(x)y = f (x),
if we know a solution to the associated integrating factor equation
u′′′
− Pu′′
+ Q − 2P′
u′
+ Q′
− P′′
− R
u = 0
or, alternatively, a solution to
y′′′
+ 2Py′′
+ P′
+ P2
+ Q
y′
+ Q′
− R + QP
y = 0,
then we can find a particular solution to third order ODE.
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16. Example
Consider the third order ODE
u′′′
+ x2
u′′
+ 6xu′
+ 6u = 0. (8)
One solution for above equation is u1 = x2e−x3/3
and another solution is
u2 = x2e−x3/3
R x3/3
52 dx.
Comparing Theorem 15 integrating factor equation by given ode, we
observe that −P = x2, Q − 2P′ = 6x and Q − P′′ − R = 6. Solving
for P, P′, P′, Q, Q′ and R, the corresponding equation is
y′′′
− x2
y′′
+ 2xy′
− 2y = f (x)
Since
u′
u
= p − b from Theorem 15 proof ,
we can solve for b using u = e−x3/3x2. Thus, u′
u = 2
x − x2 and as
u′u = P − b = −x2 − b solving for b we get b = −2
x .
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17. Since a = Q − b′ − b(P − b). replacing b, P, Q and b′, we get
a = 2
x2 . This, replacing u, b and a in [u(y′ + by′ + ay)]′ = uf (x)
and integrating both sides we get:
e−x3/3
x2
y′′
−
2
x′
y′
+
2
x2
y
=
Z
e−x3/3
x2
f (x)dx.
Suppose f (x) = c, then
e−x3/3
x2
y′′
−
2
x
y′
+
2
x2
y
= −ce−x3/3
+ k
and we have that:
y′′
−
2
x
y′
+
2
x2
y = −
c
x2
+ k
ex3/3
x2
To find a solution for this last equation, we use Euler for the howogen-
tous equation y′′ − 2
x y + 2
x2 y = 0 ar
x2
y′′
− 2xy′
+ 2y = 0. (9)
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18. Setting y = xr , we get the fundamental solutions to be y1 = x2 and
y2 = x. Hence, the general solution for the homogeneous part is
yh = c1x2 + c2x. To find a particular solution, we divide the above
equation by x2
y′′
−
2
x
y′
+
2
x2
y = g(x) (10)
with g(x) = − c
x2 + k c3/3
x2 . Using one of the fundamental solutions
for example y = x, since b = −y
y , then, from Equation,
u y′
+ by
′
= ug(x). (11)
We know that u′
u = P − b, where in this case from above equation,
P = −2
x and b = −y′
y = −1
x . Thus, u′
u = −1
x , hence u = x−1.
Thus, replacing u and b into above equation, we have.
x−1
y′
+
−
1
x
y
′
=
1
x
g(x), then
1
x
y′
−
1
x
y
=
Z
1
x
g(x)dx+k
and, for this last equation, we obtain the solution by using the first
degree order integrating factor technique. 18 / 23
19. Integrating Factor techniques for solving fourth order
ODE
Theorem 16.
Given the equation:
y(4)
+ Py′′′
+ Qy′′
+ Ry′
+ Wy = f (x). (12)
If we know one solution of either
u(4)
− Pu′′′
+ u′′
Q − 3P′
+ u′
2Q′
− 3P′′
− R
+ u W + Q′′
− P′′
− R′
= 0
y(4)
+ 3Py′′
+ 3P′
+ 3P2
+ Q
y′′
+ P′′
+ 3P′
P + 2Q′
− R + 2PQ + P3
y′
+ −R′
− PR + P2
Q + P′
Q + Q′′
+ 2Q′
P + W
y = 0
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20. Example
Consider the fourth order ODE
y(4)
− x2
y′′
+ 3xy′′
− 6y′
+
6
x
y = f (x)
In this example, we hate P = −x2, Q = 3x, R = −6 and W = 6
x .
Therefore, −P = x2, Q − 3P′ = 9x, 2Qr − 3Pv − R = 18 and
W + Qn − pw − Rr = 6
x . Hence,
u(4)
−Pu′′′
+u′′
Q − 3P′
+u′
2Q′
− 3P′′
− R
+u W + Q′′
− P′′′
− R′
would be
u(4)
− x2
u′′′
+ u′′
(9x) + u′
(18) + u
6
x
= 0.
One solution of this equation is u = x3e−x3/3. Replacing u′
u = 3
x −x2
and since u′
u = P − b, then u′
u = −x2 − b. Therefore, b = −3
x and
as a = Q − b(P − b) − b, this means a = 6
x2 . In addition, since
c = R − a′ − a(P − b), it implies that c = − 6
x3 . Then, following
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21. Replacing u, a, b and c into the above equation and integrating we
have:
x3
e−x3/3
y′′
−
3
x
y′
+
6
x2
y′
−
6
x3
y
=
Z
x3
e−x3/3
f (x)dx (14)
To find this solution, we first note that y′′′ − 3
x y′′ + 6
x2 y′ − 6
x3 y = 0
can be converted to an Euler equation
x3
y′′′
− 3x2
y′
+ 6xy′
− 6y = 0
Setting y = xr , we obtain y1 = x, y2 = x2 and y3 = x3 fo be the
solutions of the third order ODE
y′′
−
3
x
y′′
+
6
x2
y′
−
6
x3
y.
Thus, we can solve for (14) by using the same methodology as we
did with degree 3.
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