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1. ENGINEERING
MATHEMATICS II
MA-2001
Dr. Umber Sheikh

2. Course Outlines
Week Activity
1 Basic Concepts and Ideas, Differential Equations and their
Classifications
2 Formation of Differential Equations, Initial and Boundary
Conditions, Geometrical Meaning of y’=f(x,y), Separable
Differential Equations
3 Homogeneous Equations, Differential Equations reducible
to Homogeneous Form, Exact Equations, Integrating
Factors
4 Linear Equations, Bernoulli Equations, Orthogonal
Trajectories of Curves, Homogeneous Linear Equations of
Second Order, Differential Operators
5 Euler-Cauchy Equation, Nonhomogeneous Linear
Equations, Reduction of Order, Existance and Uniqueness
Theory,Wronskian
6 Variation of Parameters

3. 7 Higher Order Differential Equations, Higher Order
Nonhomogeneous Equations, Introduction to Vectors,
Matrices, Eigenvalues
8 Homogeneous Systems with Constant Coefficients, Phase
Plane, Critical Points, Criteria for Critical Points
9 Stability, Nonhomogeneous Linear Systems, Power Series
Method,Theory of Power Series Method, Legendre
Equations
10 Legendre Polynomials, Frobenius Method,
11 Laplace Transforms, Properties of Laplace Transforms,
Table of Some Laplace Transforms, Inverse Transform,
Linearity, Shifting
12 Transforms of Derivative and Integral, Differential
Equations Solution of InitialValue Problems

4. Basic Concepts
Equation: Equations describe the relations between the
dependent and independent variables. An equal sign "=" is
required in every equation.
Differential Equations: Equations that involve dependent
variables and their derivatives with respect to the independent
variables are called differential equations.
Example: The following are differential equations involving the
unknown function y.
𝑑𝑦
𝑑𝑥
+ 2 = 0,
𝑑2
𝑦
𝑑𝑥2 +
𝑑𝑦
𝑑𝑥
= 𝑒𝑥,
𝜕2𝑦
𝜕𝑥2
=
𝜕𝑦
𝜕𝑡
.

5. Some Other Examples:
Newton’s Second Law of Motion
Maxwell Equations
Kirchhoff’s Laws
Heat Equation
Wave Equation
Ordinary Differential Equation: Differential equations that
involve only one independent variable are called ordinary
differential equations.
Partial Differential Equation: Differential equations that
involve two or more independent variables are called partial
differential equations.
Examples: Equations 1 and 2 are examples of ordinary
differential equations, since the unknown function y depends
solely on the variable
x. Equation 3 is a partial differential equation, since y depends
on both the independent variables t and x.

6. Order: The order of a differential equation is the highest
derivative that appears in the differential equation.
Degree: The degree of a differential equation is the power of
the highest derivative term.
Linear Equations: A differential equation is called linear if
there are no multiplications among dependent variables and
their derivatives. In other words, all coefficients are functions
of independent variables.
Non-linear: Differential equations that do not satisfy the
definition of linear are non-linear.

7. Equation Indepen-dent
Variable
Depen-dent
Variable
Order Degree
(
𝑑𝑦
𝑑𝑥
)2+2 = 0
𝑥 𝑦 1 2
𝑑2
𝑦
𝑑𝑥2
+
𝑑𝑦
𝑑𝑥
= 𝑒𝑥
𝑥 𝑦 2 1
𝑧(𝑖𝑣) + 𝑧" = −y 𝑦 𝑧 4 1
𝑦′
+ 𝑎𝑦 = 𝑏𝑡 𝑡 𝑦 1 1
𝑓" + 𝑓′ + 𝑔(𝑡)𝑓 = ℎ(𝑡) 𝑡 𝑓 2 1

8. Linear Homogeneous: A differential equation is homogeneous if
every single term contains the dependent variables or their
derivatives.
𝑎𝑛 𝑡 𝑦(𝑛)
𝑡 + 𝑎𝑛−1 𝑡 𝑦(𝑛−1)
𝑡 + ⋯ + 𝑎1 𝑡 𝑦′ 𝑡 + 𝑎0 𝑡 𝑦(𝑡)
= 0.
Linear Non-homogeneous: Differential equations which do not
satisfy the definition of homogeneous are considered to be non-
homogeneous.
𝑎𝑛 𝑡 𝑦(𝑛) 𝑡 + 𝑎𝑛−1 𝑡 𝑦(𝑛−1) 𝑡 + ⋯ + 𝑎1 𝑡 𝑦′ 𝑡 + 𝑎0 𝑡 𝑦(𝑡)
= 𝑔(𝑡).

9. Solutions
General Solution: Solutions obtained from integrating the
differential equations are called general solutions. The general
solution of a order ordinary differential equation contains
arbitrary constants resulting from integrating times.
Particular Solution: Particular solutions are the solutions
obtained by assigning specific values to the arbitrary constants
in the general solutions.
Singular Solutions: Solutions that can not be expressed by the
general solutions are called singular solutions.

10. Explicit Solution: Any solution that is given in the form y = y
(independent variable).
Examples: 𝐴 𝑟 = 𝜋𝑟2
+ 𝑐, 𝑦 𝑡 = 𝑒−𝑡
+ 𝑐 etc.
Implicit Solution: Any solution that isn’t in explicit form.
Examples:𝑦2
𝑥2
= 𝑐, sin 𝑥𝑦 = 𝑐, 𝑥𝑒𝑦
= 𝑐 etc.

11. Conditions
Initial Condition: Constrains that are specified at the initial
point, generally time point, are called initial conditions.
Problems with specified initial conditions are called initial
value problems.
Examples: y’+y=0 y(0)=1,
z’=1, z(1)=-1.
Boundary Condition: Constrains that are specified at the
boundary points, generally space points, are called boundary
conditions. Problems with specified boundary conditions are
called boundary value problems.
Examples: z”+z’+z=1, z(0)=2, z(2)=-1,
f”+f=0, f(0)=0, f(/2)=1.

12. Worksheet 1
Equation Independent
Variable
Dependent
Variable
Order Degree Homo-genity
(
𝑑𝑦
𝑑𝑧
)3+2y = 0
𝑑2𝑦
𝑑𝑥2
+
𝑑𝑦
𝑑𝑥
= 𝑒𝑥
𝑧(𝑖𝑣)
+ 𝑧" = 2
𝑝′
+ 𝑎𝑝 = 𝑏𝑡
𝑓" + 𝑓 + 𝑔(𝑡) = ℎ(𝑡)

13. Worksheet 2
Equation Solution to Check
𝑦′′ + 4𝑦 = 0 𝑦(𝑥) = 𝑐1sin 2𝑥 + 𝑐2cos2𝑥
(𝑦′)4
+ 𝑦2
= −1 𝑦 = 𝑥2
− 1
𝑦′ = 𝑎𝑦, 𝑦(0) = 1 𝑦(𝑡) = 𝑒𝑎𝑡
𝑦′ + 𝑦 = 10 𝑦 𝑡 = 10 − 𝑐𝑒 − 𝑡
𝑤𝑡 + 3𝑤𝑥 = 0 𝑤(𝑥, 𝑡) = 1/(1 + (𝑥 − 3𝑡)2)
𝑥’’ + 4x = 0 𝑥 = cos(2𝑡) + sin(2𝑡) + 𝑐
Check whether the given expression a solution of the corresponding equation?

14. First Order Differential Equations
Standard Form:
Standard form for a first-order differential equation in the
unknown function y(x) is
y’=f(x,y).
Interpretation:
The form shows the slope of the tangent at each point of the
xy-plane. Thus the standard form of the first order differential
equation give the slope field. Solving the equation gives the
original curve family having the slope field expressed by the
differential equation.

15. Separable Equations
Form of Equation: 𝑓 𝑥 𝑑𝑥 = 𝑔 𝑦 𝑑𝑦.
Method of Solution: Integrate both sides will give the solution
of such type of equations.

16. Examples
Exercise: Solve
𝑑𝑝
𝑑𝑡
=
1 − 𝑝2
𝑡
.
Example 1: Solve
𝑑𝑦
𝑑𝑥
=
𝑦
𝑥
Example 2: Solve
𝒅𝒛
𝒅𝒕
=
𝟏
𝒛𝒆𝒕
Solution:
𝑑𝑦
𝑑𝑥
=
𝑦
𝑥
𝑑𝑦
𝑦
=
𝑑𝑥
𝑥
𝑑𝑦
𝑦
=
𝑑𝑥
𝑥
ln 𝑦 = ln 𝑥 + 𝑐
Solution:
𝑑𝑧
𝑑𝑡
=
1
𝑧𝑒𝑡
𝑧𝑑𝑧 =
𝑑𝑡
𝑒𝑡
𝑧𝑑𝑧 =
𝑑𝑡
𝑒𝑡
𝑧2
2
= −𝑒−𝑡
+ 𝑐

17. Homogeneous Equations
(of degree zero)
Form of Equation:
𝑑𝑦
𝑑𝑥
= 𝑓 𝑥, 𝑦
𝑓 𝑥, 𝑦 is a function homogeneous of degree zero, i.e.,
𝑓 𝑡𝑥, 𝑡𝑦 = 𝑡0𝑓 𝑥, 𝑦 = 𝑓 𝑥, 𝑦 .
A function homogeneous of degree n can be defined as
𝑓 𝑡𝑥, 𝑡𝑦 = 𝑡𝑛𝑓 𝑥, 𝑦 .
Method of Solution: Substituting 𝑦 = 𝑣𝑥,
𝑑𝑦
𝑑𝑥
= 𝑣 + 𝑥
𝑑𝑣
𝑑𝑥
,
reduces the equation in form separable in variables v and x.
Integrate both sides and then substituting the value of v will
give the solution.

18. Examples
Example 1: Solve
𝑑𝑦
𝑑𝑥
=
𝒚 − 𝒙
𝒚 + 𝒙
Example 2: Solve
𝑑𝑦
𝑑𝑥
=
𝑦
𝑥
Solution:
𝑓(𝑥, 𝑦) =
𝑦 − 𝑥
𝑦 + 𝑥
𝑓 𝑡𝑥, 𝑡𝑦 =
𝑦 − 𝑥
𝑦 + 𝑥
= 𝑓(𝑥, 𝑦)
𝑑𝑦
𝑑𝑥
=
𝑦 − 𝑥
𝑦 + 𝑥
=
𝑦
𝑥
− 1
𝑦
𝑥
+ 1
Put 𝑦 = 𝑣𝑥,
𝑑𝑦
𝑑𝑥
= 𝑣 + 𝑥
𝑑𝑣
𝑑𝑥
.
𝑣 + 𝑥
𝑑𝑣
𝑑𝑥
=
𝑣 − 1
𝑣 + 1
𝑥
𝑑𝑣
𝑑𝑥
=
−1 − 𝑣2
𝑣 + 1
(𝑣 + 1)𝑑𝑣
1 + 𝑣2
= −
𝑑𝑥
𝑥
𝑡𝑎𝑛−1
𝑣 +
1
2
ln 1 + 𝑣2
= − ln 𝑥 + 𝑐
𝑡𝑎𝑛−1
(
𝑦
𝑥
) +
1
2
ln 1 + (
𝑦
𝑥
)2
= − ln 𝑥 + 𝑐
Solution:
𝑓 𝑥, 𝑦 =
𝑦
𝑥
𝑓 𝑡𝑥, 𝑡𝑦 =
𝑡𝑦
𝑡𝑥
=
𝑦
𝑥
= 𝑓 𝑥, 𝑦
Put 𝑦 = 𝑣𝑥,
𝑑𝑦
𝑑𝑥
= 𝑣 + 𝑥
𝑑𝑣
𝑑𝑥
.
𝑣 + 𝑥
𝑑𝑣
𝑑𝑥
= 𝑣,
𝑥
𝑑𝑣
𝑑𝑥
= 0,
𝑑𝑣
𝑑𝑥
= 0
Integrating we get
𝑣 = 𝑐
or
𝑦
𝑥
= 𝑐 , 𝑦 = 𝑐𝑥.
Exercise: Solve
𝒅𝒚
𝒅𝒙
=
𝒙
𝒚

19. First Order Linear Equations
Form of Equation:
𝑑𝑦
𝑑𝑥
+ 𝑝 𝑥 𝑦 = 𝑞 𝑥 .
Method of Solution: The integrating factor is I.F.=𝑒 𝑝 𝑥 𝑑𝑥
.
Multiplying the equation with this factor gives
𝑒 𝑝 𝑥 𝑑𝑥 𝑑𝑦
𝑑𝑥
+ 𝑒 𝑝 𝑥 𝑑𝑥
𝑝 𝑥 𝑦 = 𝑒 𝑝 𝑥 𝑑𝑥
𝑞 𝑥
𝑑(𝑦𝑒 𝑝 𝑥 𝑑𝑥
)
𝑑𝑥
= 𝑒 𝑝 𝑥 𝑑𝑥
𝑞 𝑥
𝑦𝑒 𝑝 𝑥 𝑑𝑥
= 𝑒 𝑝 𝑥 𝑑𝑥
𝑞 𝑥 𝑑𝑥
Solving this gives the solution y.

20. Examples
Exercise: Solve
𝑑𝑝
𝑑𝑡
+ 𝑝𝑡 = 3.
Example 1: Solve
𝑑𝑦
𝑑𝑥
+ 3𝑦 = 𝑥
Example 2: Solve
𝑑𝑦
𝑑𝑥
=
𝑦
𝑥
Solution: Here 𝑝 𝑥 = 3, 𝑞 𝑥 = 𝑥.
Substituting values in
𝑦𝑒 3𝑑𝑥
= 𝑥𝑒 3𝑑𝑥
𝑑𝑥
𝑦𝑒3𝑥
= 𝑥𝑒3𝑥
𝑑𝑥
𝑦𝑒3𝑥
=
𝑥𝑒3𝑥
3
+
𝑒3𝑥
9
+ 𝑐
Solution:
𝑑𝑦
𝑑𝑥
−
𝑦
𝑥
= 0
𝑝 𝑥 = −
1
𝑥
, 𝑞 𝑥 = 0.
Substituting values in
𝑦𝑒 𝑝 𝑥 𝑑𝑥
= 𝑒 𝑝 𝑥 𝑑𝑥
𝑞 𝑥 𝑑𝑥
𝑦𝑒 −
1
𝑥𝑑𝑥
= 0 𝑑𝑥
𝑦𝑒−ln |𝑥| = 𝑐
𝑦
𝑥
= 𝑐

21. Bernoulli Equations
Form of Equation:
𝑑𝑦
𝑑𝑥
+ 𝑝 𝑥 𝑦 = 𝑞 𝑥 𝑦𝑛
, 𝑛 ∈ ℝ.
Method of Solution: The substitution 𝑧 = 𝑦1−𝑛 transform the
Bernoulli’s equation into a linear equation in z. This linear
equation, after solving and back substituting the value of z
gives the solution of the Bernoulli’s equation.

22. Examples
Example 1: Solve
𝑑𝑦
𝑑𝑥
+ 3𝑦 = 𝑥𝑦2
Exercise: Solve
𝑑𝑦
𝑑𝑥
+ 𝑥𝑦 = 𝑥𝑦2
Solution: Put 𝑧 = 𝑦1−2
=
1
𝑦
⟹ 𝑦 =
1
𝑧
,
𝑑𝑦
𝑑𝑥
= −
1
𝑧2
𝑑𝑧
𝑑𝑥
𝑑𝑧
𝑑𝑥
− 3𝑧 = −𝑥
𝑝 𝑥 = −3 𝑞 𝑥 = −𝑥.
Substituting values in
𝑧𝑒− 3𝑑𝑥
= −𝑥𝑒− 3𝑑𝑥
𝑑𝑥
𝑒−3𝑥
𝑦
= − 𝑥𝑒−3𝑥 𝑑𝑥
𝑦𝑒3𝑥 = 1/(
𝑥𝑒3𝑥
3
+
𝑒3𝑥
9
) + 𝑐
Solution: Put 𝑧 = 𝑦1−2
=
1
𝑦
⟹ 𝑦 =
1
𝑧
,
𝑑𝑦
𝑑𝑥
= −
1
𝑧2
𝑑𝑧
𝑑𝑥
𝑑𝑧
𝑑𝑥
− 𝑥𝑧 = −𝑥
𝑝 𝑥 = −𝑥, 𝑞 𝑥 = −𝑥.
Substituting values in
𝑧𝑒− 𝑥𝑑𝑥
= −𝑥𝑒− 𝑥𝑑𝑥
𝑑𝑥
𝑦𝑒
−𝑥2
2 = −𝑥𝑒
−𝑥2
2 𝑑𝑥
𝑦𝑒
−𝑥2
2 = 𝑒
−𝑥2
2 +c

23. A differential equation
𝑀 𝑥, 𝑦 𝑑𝑥 + 𝑁 𝑥, 𝑦 𝑑𝑦 = 0.
is exact if there exists a function 𝑔(𝑥, 𝑦) such that
𝑑𝑔 𝑥, 𝑦 = 𝑀 𝑥, 𝑦 𝑑𝑥 + 𝑁 𝑥, 𝑦 𝑑𝑦.
Test for exactness: If 𝑀(𝑥, 𝑦) and 𝑁(𝑥, 𝑦) are continuous functions and
have continuous first partial derivatives on some rectangle of the 𝑥𝑦-
plane, then
𝑀 𝑥, 𝑦 𝑑𝑥 + 𝑁 𝑥, 𝑦 𝑑𝑦 = 0
is exact if and only if
𝜕𝑀(𝑥, 𝑦)
𝜕𝑦
=
𝜕𝑁(𝑥, 𝑦)
𝜕𝑥
.
How to obtain an exact equation from a function?
𝑓 𝑥, 𝑦 = 𝑐,
𝜕𝑓(𝑥, 𝑦)
𝜕𝑥
𝑑𝑥 +
𝜕𝑓(𝑥, 𝑦)
𝜕𝑦
𝑑𝑦 = 0,
𝑀 𝑥, 𝑦 𝑑𝑥 + 𝑁 𝑥, 𝑦 𝑑𝑦 = 0.
Exact Equations

24. Method of Solution
To solve an exact equation, first solve the equations
𝜕𝑓(𝑥, 𝑦)
𝜕𝑥
= 𝑀 𝑥, 𝑦 ,
𝜕𝑓(𝑥, 𝑦)
𝜕𝑦
= 𝑁(𝑥, 𝑦)
for 𝑓(𝑥, 𝑦). The solution to the exact equation is then given implicitly
by
𝑓(𝑥, 𝑦) = 𝑐,
where 𝑐 represents an arbitrary constant.

25. Examples
Example 1: Make an exact ODE using 𝑢 𝑥, 𝑦 = 𝑥 + 𝑥2𝑦3
Sol: If 𝑢(𝑥, 𝑦) = 𝑥 + 𝑥2𝑦3 = 𝑐, then
𝑑𝑢 = (1 + 2𝑥𝑦3)𝑑𝑥 + 3𝑥2𝑦2𝑑𝑦 = 0
𝑦′ = 𝑑𝑦/𝑑𝑥 = −(1 + 2𝑥𝑦3)/3𝑥2𝑦2

26. Example 2:
Solve cos 𝑥 + 𝑦 𝑑𝑥 + 3𝑦2 + 2𝑦 + cos 𝑥 + 𝑦 𝑑𝑦 = 0.
Sol: Here M = cos 𝑥 + 𝑦 , 𝑁 = 3𝑦2 + 2𝑦 + cos 𝑥 + 𝑦
Since
𝜕𝑀
𝜕𝑦
= − sin 𝑥 + 𝑦 ,
𝜕𝑁
𝜕𝑥
= − sin 𝑥 + 𝑦 , therefore
equation is exact.
𝑢 𝑥, 𝑦 = cos 𝑥 + 𝑦 𝑑𝑥 + 𝑘 𝑦 = sin 𝑥 + 𝑦 + 𝑘 𝑦
𝜕𝑢
𝜕𝑦
= cos(𝑥 + 𝑦) +
𝑑𝑘(𝑦)
𝑑𝑦
= 3𝑦2 + 2𝑦 + cos 𝑥 + 𝑦
𝑑𝑘(𝑦)
𝑑𝑦
= 3𝑦2
+ 2𝑦 ⇒ 𝑘 𝑦 = 𝑦3
+ 𝑦2
+c
𝑢 𝑥, 𝑦 = cos 𝑥 + 𝑦 𝑑𝑥 + 𝑘 𝑦 = sin 𝑥 + 𝑦 + 𝑦3 + 𝑦2+c

27. Integrating Factors
If an equation is inexact (not exact), it is possible to transform
Such equation into an exact differential equation by a judicious
multiplication.
A function 𝐼(𝑥, 𝑦) is an integrating factor for an inexact equation if the
equation
𝐼 𝑥, 𝑦 𝑀 𝑥, 𝑦 𝑑𝑥 + 𝑁 𝑥, 𝑦 𝑑𝑦 = 0
is an exact equation.
The famous formulas to obtain an integrating factor are as follows:
If
1
𝑁(𝑥,𝑦)
𝜕𝑀(𝑥,𝑦)
𝜕𝑦
−
𝜕𝑁(𝑥,𝑦)
𝜕𝑥
≡ 𝑔 𝑥 , then 𝐼. 𝐹. = 𝐼 𝑥, 𝑦 = 𝑒 𝑔(𝑥)𝑑𝑥
.
If
1
𝑀(𝑥,𝑦)
𝜕𝑀(𝑥,𝑦)
𝜕𝑦
−
𝜕𝑁(𝑥,𝑦)
𝜕𝑥
≡ 𝑓 𝑦 , then 𝐼. 𝐹. = 𝐼 𝑥, 𝑦 = 𝑒− 𝑓(𝑦)𝑑𝑦
.
If M = yf xy , N = xg(xy), then 𝐼. 𝐹. = 𝐼 𝑥, 𝑦 =
1
𝑥𝑀−𝑦𝑁
.

29. Examples and Exercises
Make an exact differential equation from the functions
𝑓 𝑥, 𝑦 = 𝑥2
𝑦 + 6𝑦 − 𝑦3
.
𝑢 𝑥, 𝑦 = 𝑥 + 𝑥2𝑦3.
Check for the exactness and solve
𝑦2 − 2𝑥 𝑑𝑥 + 2𝑥 + 1 𝑑𝑦 = 0,
3𝑥2𝑦 − 1 𝑑𝑥 + 𝑥3 + 6𝑦 − 𝑦2 𝑑𝑦 = 0, 𝑦 0 = 3,
2𝑥𝑦 − 9𝑥2 + 2𝑦 + 𝑥2 + 1
𝑑𝑦
𝑑𝑥
= 0,
2𝑥𝑦 − 9𝑥2
+ 2𝑦 + 𝑥2
+ 1
𝑑𝑦
𝑑𝑥
= 0, 𝑦 0 = −3,
1 + 𝑡2
𝑦′
+ 4𝑡𝑦 = 1 + 𝑡2 −2
, y 0 = 1.

30. OrthogonalTrajectories
Given a one-parameter family of curves 𝐹(𝑥, 𝑦, 𝑐) = 0.
A curve that intersects each member of the family at right angles
(orthogonally) is called an orthogonal trajectory of the family.
The one-parameter families 𝐹(𝑥, 𝑦, 𝑐) = 0 and 𝐺(𝑥, 𝑦, 𝑘) = 0 are
orthogonal trajectories if each member of one family is an orthogonal
trajectory of the other family.
A procedure for finding a family of orthogonal trajectories 𝐺(𝑥, 𝑦, 𝑘) = 0
for a given family of curves 𝐹(𝑥, 𝑦, 𝑐) = 0 is as follows:
Step 1. Determine the differential equation for the given family 𝐹(𝑥, 𝑦, 𝑐) =
0.
Step 2. Replace 𝑦0 in that equation by −1/𝑦0; the resulting equation is the
differential equation for the family of orthogonal trajectories.
Step 3. Find the general solution of the new differential equation. This is
the family of orthogonal trajectories.

31. Example
Find family of curves orthogonal to one parameter family of
quadratic parabolas 𝑦 = 𝑐𝑥^2.
Solution:
𝑦 = 𝑐𝑥2
𝐹 𝑥, 𝑦, 𝑐 = 𝑦 − 𝑐𝑥2 = 0
𝑦
𝑥2 = 𝑐 ⇒
𝑥2
𝑦′
− 2𝑥𝑦
𝑥4 = 0 ⇒ 𝑦′
𝑜𝑙𝑑 =
2𝑦
𝑥
.
Now the slope of the family of new curves is
𝑦′
𝑛𝑒𝑤
=
−1
𝑦′
𝑜𝑙𝑑
=
−1
2𝑦
𝑥
=
−𝑥
2𝑦
.
2𝑦𝑦′ = −𝑥
2 𝑦𝑦′𝑑𝑦 = − 𝑥𝑑𝑥
𝑦2
= −
𝑥2
2
+ 𝑐
is required family of curves.

32. Exercises
1. Find the family of orthogonal trajectories of:
𝑦 = 𝐶𝑥2
+ 2
Answer: 𝑥2
+ 2𝑦2
− 8𝑦 = 𝐶
2. Find the orthogonal trajectories of the family of parabolas with
vertical axis and vertex at the point (−1, 3).
Answer: (𝑥 + 1)2+2(𝑦 − 3)2 = 𝐶

33. Linear Second Order DEs
The most general linear second order differential equation
is in the form.
𝑝 𝑡 𝑦′′ + 𝑞 𝑡 𝑦′ + 𝑟 𝑡 𝑦 = 𝑔 𝑡 .
The constant coefficient linear second order differential
equation is
𝑎𝑦′′
+ 𝑏𝑦′
+ 𝑐𝑦 = 𝑔(𝑡)
where a, b, c are all constants.
Initially we will make our life easier by looking at differential
equations with 𝑔 𝑡 = 0. When 𝑔 𝑡 = 0 we call the
differential equation homogeneous and when 𝑔 𝑡 ≠ 0 we
call the differential equation nonhomogeneous.

34. Solving Second Order, Linear,
Homogeneous ODE with Constant
Coefficients
Consider the solution of the type 𝑦 𝑡 = 𝑒𝑟𝑡
. Substituting in
𝑎𝑦′′
+ 𝑏𝑦′
+ 𝑐𝑦 = 0
We have
𝑎 𝑟2𝑒𝑟𝑡 + 𝑏 𝑟𝑒𝑟𝑡 + 𝑐𝑒𝑟𝑡 = 0
𝑎𝑟2
+ 𝑏𝑟 + 𝑐 = 0
As 𝑒𝑟𝑡 ≠ 0.
This equation is typically called the characteristic equation
This will be a quadratic equation and so we should expect two
roots, r1 and r2. Once we have these two roots we have two
solutions to the differential equation.
𝑦1 𝑡 = 𝑒𝑟1𝑡 and 𝑦2 𝑡 = 𝑒𝑟2𝑡.

35. The roots will have three possible forms. These are
Real, distinct roots, 𝑟1 ≠ 𝑟2 .
Complex root, 𝑟1 = 𝜆 + 𝑖𝜇, 𝑟2 = 𝜆 − 𝑖𝜇 .
Double roots, 𝑟1 = 𝑟2 = 𝑟.
Real Roots
Example: Find two solutions to 𝑦′′ − 9𝑦 = 0.
Solution:The characteristic equation is
𝑟2
− 9 = 0 ⇒ 𝑟 = ±3.
The two roots are 3 and -3.Therefore, two solutions are
𝑦1 𝑡 = 𝑒3𝑡
and 𝑦2 𝑡 = 𝑒−3𝑡
.
The general solution is 𝑦𝑐 𝑡 = 𝑐1𝑒3𝑡 + 𝑐2𝑒−3𝑡

36. Example: Solve the following IVP 𝑦′′ + 11𝑦′ + 24𝑦 =
0, 𝑦 0 = 0, 𝑦′ 0 = −7.
Solution:The characteristic equation is 𝑟2 + 11𝑟 + 24 =
0 ⇒ 𝑟 + 8 𝑟 + 3 = 0 ⇒ 𝑟 = −8, −3.
The general solution and its derivative is
𝑦𝑐 𝑡 = 𝑐1𝑒−3𝑡 + 𝑐2𝑒−8𝑡
𝑦′𝑐 𝑡 = −3𝑐1𝑒−3𝑡 − 8𝑐2𝑒−8𝑡
Putting the initial conditions, we have the following system
of equations
0 = 𝑦𝑐 0 = 𝑐1 + 𝑐2
−7 = 𝑦′𝑐 0 = −3𝑐1 − 8𝑐2
Solving we get 𝑐1 =
−7
5
and 𝑐2 =
7
5
.Thus the solution is
𝑦𝑐 𝑡 =
−7
5
𝑒−3𝑡 +
7
5
𝑒−8𝑡

37. Complex Roots
Example: Solve the following IVP 𝑦′′
− 4𝑦′
+ 9𝑦 = 0, 𝑦 0 =
0, 𝑦′
0 = −8.
Solution:The characteristic equation is 𝑟2
− 4𝑟 + 9 = 0 ⇒ 𝑟1,2 =
2 ± 5𝑖.
The general solution and its derivative is
𝑦𝑐 𝑡 = 𝑐1𝑒2𝑡
cos 5𝑡 + 𝑐2𝑒2𝑡
sin 5𝑡
Putting the initial conditions, we have the following system of
equations
0 = 𝑦𝑐 0 = 𝑐1
so
𝑦𝑐 𝑡 = 𝑐2𝑒2𝑡
sin 5𝑡
𝑦′𝑐 𝑡 = 2𝑐2𝑒2𝑡
sin 5𝑡 + 5𝑐2𝑒2𝑡
cos 5𝑡
𝑦′𝑐 0 = 5𝑐2 ⇒ −8 = 5𝑐2 ⇒ 𝑐2 =
−8
5
Thus the solution is 𝑦 𝑡 =
−8
5
𝑒2𝑡 sin 5𝑡

38. Repeated Roots
Example: Solve the following IVP 𝑦′′
− 4𝑦′
+ 4𝑦 = 0, 𝑦 0 =
12, 𝑦′
0 = −3.
Solution:The characteristic equation is 𝑟2
− 4𝑟 + 4 = 0 ⇒
𝑟 − 2 2 = 0 ⇒ 𝑟 = 2,2.
The general solution and its derivative is
𝑦𝑐 𝑡 = 𝑐1𝑒2𝑡 + 𝑐2𝑡𝑒2𝑡
𝑦′𝑐 𝑡 = 2𝑐1𝑒2𝑡 + 𝑐2𝑒2𝑡 + 2𝑐2𝑡𝑒2𝑡
Putting the initial conditions, we have the following system of
equations
12 = 𝑦𝑐 0 = 𝑐1
−3 = 𝑦′
𝑐
0 = 2𝑐1 + 𝑐2
Solving we get 𝑐1 = 12 and 𝑐2 = −27.Thus the solution is 𝑦 𝑡 =
12𝑒2𝑡
− 27𝑡𝑒2𝑡

39. Exercises: Solve the following IVPs
𝑦′′ + 3𝑦′ − 10𝑦 = 0, 𝑦 0 = 4, 𝑦′ 0 = −2
3𝑦′′
− 2𝑦′
− 8𝑦 = 0, 𝑦 0 = −6, 𝑦′
0 = −18
4𝑦′′ − 5𝑦′ = 0, 𝑦 −2 = 0, 𝑦′ −2 = 7
𝑦′′ − 6𝑦′ − 2𝑦 = 0
𝑦′′
− 8𝑦′
+ 17𝑦 = 0, 𝑦 0 = −4, 𝑦′
0 = −1
4𝑦′′
− 24𝑦′
+ 37𝑦 = 0, 𝑦 𝜋 = 1, 𝑦′
𝜋 = 0
𝑦′′ + 16𝑦 = 0, 𝑦
𝜋
2
= −10, 𝑦′
𝜋
2
= 3
16𝑦′′
− 40𝑦′
+ 25𝑦 = 0, 𝑦 0 = 3, 𝑦′
0 = −
9
4
𝑦′′ + 14𝑦′ + 49𝑦 = 0, 𝑦 −4 = −1, 𝑦′ −4 = 5

40. Reduction of Order
Let 𝑦(𝑥) = 𝑦1(𝑥) be the known solution of second order DE
𝑎2 (𝑥)𝑦” (𝑥) + 𝑎1 (𝑥)𝑦’ (𝑥) + 𝑎0 (𝑥)𝑦(𝑥) = 0.
Assume 𝑦2 = 𝑢(𝑥)𝑦1(𝑥) is the other solution.Then
𝑎2 (𝑥)𝑦2”(𝑥) + 𝑎1(𝑥)𝑦2’(𝑥) + 𝑎0(𝑥)𝑦2(𝑥) = 0
𝑎2 (𝑥){𝑢(𝑥)𝑦1(𝑥)}” + 𝑎1 (𝑥){𝑢(𝑥)𝑦1(𝑥)}’ + 𝑎0(𝑥)𝑢(𝑥)𝑦1(𝑥) = 0
𝑎2 (𝑥){𝑢’’(𝑥)𝑦1(𝑥) + 2𝑢’(𝑥)𝑦’1(𝑥) + 𝑢(𝑥)𝑦”1(𝑥)} + 𝑎1(𝑥) {𝑢’(𝑥)𝑦1(𝑥)
+ 𝑢(𝑥)𝑦’1(𝑥)} + 𝑎0(𝑥)𝑢(𝑥)𝑦1(𝑥) = 0
𝑢’’(𝑥)𝑎2(𝑥)𝑦1(𝑥) + 𝑢’(𝑥){2𝑎2(𝑥)𝑦’1(𝑥) + 𝑎1(𝑥)𝑦1(𝑥)} + 𝑢(𝑥){𝑎2(𝑥)𝑦”1(𝑥)
+ 𝑎1(𝑥)𝑦’1(𝑥) + 𝑎0 (𝑥)𝑦1(𝑥)} = 0
Since 𝑎2 (𝑥)𝑦”1(𝑥) + 𝑎1(𝑥)𝑦’1(𝑥) + 𝑎0(𝑥)𝑦1(𝑥) = 0,
therefore
𝑢’’(𝑥)𝑎2(𝑥)𝑦1(𝑥) + 𝑢’(𝑥){2𝑎2 (𝑥)𝑦’1(𝑥) + 𝑎1(𝑥)𝑦1(𝑥)} = 0
𝑢’’(𝑥)𝑎2(𝑥)𝑦1(𝑥) = −𝑢’(𝑥){2𝑎2 (𝑥)𝑦’1(𝑥) + 𝑎1(𝑥)𝑦1(𝑥)}
𝑢’’(𝑥) = − 2𝑦’1(𝑥) − 𝑎1 (𝑥)
𝑢’(𝑥) 𝑦1(𝑥) 𝑎2(𝑥)
or 𝑢(𝑥) = 𝑐1 𝑒
−
𝑎1(𝑥)
𝑎2(𝑥)
𝑑𝑥
𝑑𝑥 .

41. Examples and Exercises
Solve
𝑦"(𝑥) − 𝑦(𝑥) = 0, 𝑦1(𝑥) = 𝑒𝑥
,
𝑥2𝑦"(𝑥) − 3𝑥𝑦′(𝑥) + 4𝑦(𝑥) = 0, 𝑦1(𝑥) = 𝑥2,
𝑦"(𝑥) + 36𝑦(𝑥) = 0, 𝑦1(𝑥) = 𝑒6𝑖𝑥,
𝑦"(𝑥) + 36𝑦(𝑥) = 0, 𝑦1(𝑥) = 𝑠𝑖𝑛6𝑥,
𝑦"(𝑥) − 𝑦(𝑥) = 0, 𝑦1(𝑥) = 𝑐𝑜𝑠𝑥.

42. Solutions of Linear Homogeneous
Equations; theWronskian
Theorem . (Existence and UniquenessTheorem)
Consider the initial value problem
𝑦′′ + 𝑝(𝑡)𝑦′ + 𝑞(𝑡)𝑦 = 𝑔(𝑡), 𝑦(𝑡0) = 𝑦0, 𝑦′(𝑡0) = 𝑦′0
If 𝑝(𝑡), 𝑞(𝑡) and 𝑔(𝑡) are continuous and bounded on an open
interval 𝐼 containing 𝑡0, then there exists exactly one solution
𝑦(𝑡) of this equation, valid on 𝐼.

43. Wronskian: Given two functions 𝑓(𝑡), 𝑔(𝑡), the Wronskian is
defined as
𝑊(𝑓, 𝑔)(𝑡) = 𝑓𝑔′ − 𝑓′𝑔
Remark: One way to remember this definition could be using the
determinant,
𝑊(𝑓, 𝑔)(𝑡) =
𝑓 𝑔
𝑓′ 𝑔′
Main property of the Wronskian:
• If 𝑊(𝑓, 𝑔) ≡ 0, then 𝑓 anf 𝑔 are linearly dependent.
• Otherwise, they are linearly independent.

44. Example: Check if the given pair of functions are linearly
dependent or not.
(a) 𝑓(𝑡) = 𝑒𝑡, 𝑔(𝑡) = 𝑒−𝑡
(b) 𝑓(𝑡) = sin𝜔𝑡, 𝑔(𝑡) = cos𝜔𝑡
(c) 𝑓(𝑡) = 𝑡 + 1, 𝑔(𝑡) = 4𝑡 + 4
(d) 𝑓(𝑡) = 2𝑡, 𝑔(𝑡) = |𝑡|
Suppose𝑦_1 (t),𝑦_2 (t) are two solutions of
𝑦′′ + 𝑝(𝑡)𝑦′ + 𝑞(𝑡)𝑦 = 0
Then
(I)We have either W(𝑦_1,𝑦_2)≡0 orW(𝑦_1,𝑦_2) never zero;
(II) IfW(𝑦_1,𝑦_2)≠0, then y =𝑐_1 𝑦_1+𝑐_2 𝑦_2is the general
solution.They are also called to form a fundamental set of
solutions. As a consequence, for any Ics 𝑦(𝑡_0) = 𝑦_0, 𝑦′(𝑡_0)=〖
𝑦′〗_0, there is a unique set of (𝑐_1,𝑐_2) that give a unique
solution.

45. Theorem (Abel’sTheorem):Let y1, y2 be two (linearly independent)
solutions to 𝑦′′ + 𝑝(𝑡)𝑦′ + 𝑞(𝑡)𝑦 = 0 on an open interval I.Then, the
Wronskian 𝑊(𝑦1, 𝑦2) on I is given by
𝑊(𝑦1, 𝑦2)(𝑡) = 𝐶 · exp( −𝑝(𝑡) 𝑑𝑡),
for some constant 𝐶 depending on 𝑦1, 𝑦2, but independent on 𝑡 in 𝐼.

46. Example: Given
𝑡2𝑦′′ − 𝑡(𝑡 + 2)𝑦′ + (𝑡 + 2)𝑦 = 0.
Find 𝑊(𝑦1, 𝑦2) without solving the equation.
Answer.We first find the 𝑝(𝑡)
p(t) = −(t + 2)t
which is
which is valid for 𝑡 ≠ 0. By Abel’sTheorem, we have
𝑊(𝑦1, 𝑦2)(𝑡) = 𝐶 · exp( −𝑝(𝑡) 𝑑𝑡)=𝐶 · exp 𝑡 𝑡 + 2 𝑑𝑡 =
C. 𝑒𝑡+2 ln 𝑡 = 𝑡2𝐶𝑒𝑡
Example: If y1, y2 are two solutions of
ty′′ + 2y′ + tety = 0,
andW(y1, y2)(1) = 2, findW(y1, y2)(5).
Example 5. IfW(f, g) = 3e4t, and f = e2t, find g.

47. Example: For the equation
2𝑡2𝑦′′ + 3𝑡𝑦′ − 𝑦 = 0, 𝑡 > 0
given one solution 𝑦1 =
1
𝑡
, find a second linearly independent solution.
Solution: Use Abel’sTheorem and Wronskian. By Abel’s
Theorem, and choose C = 1, we have
𝑊(𝑦1, 𝑦2)(𝑡) = 𝐶 · exp( −𝑝(𝑡) 𝑑𝑡)=𝐶 · exp −
3𝑡
2𝑡2 𝑑𝑡 = 𝑡−3/2.
By definition of theWronskian,
𝑊 𝑦1, 𝑦2 = 𝑦1𝑦2
′ − 𝑦1
′𝑦2 =
𝑦2
′
𝑡
+
𝑦2
𝑡2
= 𝑡−3/2
Solve this for 𝑦2 (taking c=0): 𝑦2 =
2
3
𝑡
Example: Consider the equation
𝑡2
𝑦′′ − 𝑡(𝑡 + 2)𝑦′ + (𝑡 + 2)𝑦 = 0, 𝑡 > 0
Given 𝑦1 = 𝑡, find the general solution.
Example: Given the equation 𝑡2
𝑦′′
− (𝑡 −
3
16
) 𝑦 = 0, 𝑡 > 0
and 𝑦1 = 𝑡1/4
𝑒2 𝑡
, find 𝑦2.

48. Linear Differential Equations
An nth-order linear differential equation has the form
(4.1)
where g(x) and the coefficients bj(x) ( j = 0,1,2,..., n) depend
solely on the variable x. In other words, they do
not depend on y or any derivative of y.
If g(x) = 0, then above Equation is homogeneous; if not, 4.1 is
nonhomogeneous.
A linear differential equation has constant coefficients if all
the coefficients bj(x) in above equation are constants; if one or more
of these coefficients
is not constant, the above equation has variable coefficients.

49. Theorem 4.1. Consider the initial-value problem given by the
linear differential
equation 4.1 and the n initial conditions
y(x0)=c0, y’(x0)=c1, y’’(x0)=c2,…, y(n-1)(x0)=cn-1

50. The general solution to the linear differential equation 𝐿(𝑦) =
𝑓(𝑥), 𝑦 = 𝑦ℎ + 𝑦𝑝 where 𝑦𝑝 denotes one solution to the
differential equation and 𝑦ℎ is the general solution to the
associated homogeneous equation, 𝐿(𝑦) = 0 . Methods for
obtaining 𝑦ℎ when the differential equation has constant
coefficients are given in previous lectures. In this lecture, we give
methods for obtaining a particular solution 𝑦𝑝 once 𝑦ℎ is known.
Second Order Non-homgeneous
Linear Differential Equations

51. Method of UndeterminedCoefficients
This method is applicable only if 𝑓(𝑥) and all of its derivatives can be
written in terms of the same finite set of linearly independent functions,
which we denote by {𝑦1(𝑥), 𝑦2(𝑥), … , 𝑦𝑛(𝑥)}. The method is initiated by
assuming a particular solution of the form
𝑦𝑝(𝑥) = 𝐴1𝑦1(𝑥) + 𝐴2𝑦2(𝑥) + ⋯ + 𝐴𝑛𝑦𝑛(𝑥),
where 𝐴1, 𝐴2, … , 𝐴𝑛 denote arbitrary multiplicative constants. These
arbitrary constants are then evaluated by substituting the proposed
solution into the given differential equation and equating the
coefficients of like
terms.

52. 𝒇(𝒙) = 𝒑𝒏(𝒙), an nth-degree polynomial in 𝒙.
Assume a solution of the form
𝑦𝑝 = 𝑒𝑎𝑥(𝐴𝑛𝑥𝑛 + 𝐴𝑛 − 1𝑥𝑛 − 1 + ⋯ +
𝐴2𝑥2 + 𝐴1𝑥 + 𝐴0)
where 𝐴𝑗 (𝑗 = 0,1,2, … , 𝑛) is a constant to be determined.
𝒇(𝒙) = 𝒌𝒆𝒂𝒙 where 𝒌 and 𝒂 are known constants.
Assume a solution of the form
𝑦𝑝 = 𝐴𝑒𝑎𝑥
where 𝐴 is a constant to be determined.
𝒇(𝒙) = 𝒌𝟏𝒔𝒊𝒏𝒃𝒙 + 𝒌𝟐𝒄𝒐𝒔𝒃𝒙 where 𝒌𝟏, 𝒌𝟐, and 𝒃 are known
constants.
Assume a solution of the form
𝑦𝑝 = 𝐴 sin 𝑏𝑥 + 𝐵 cos 𝑏𝑥,
where 𝐴 and 𝐵 are constants to be determined.

53. Generalizations
If 𝑓(𝑥) is the product of terms considered in all the cases given above,
take 𝑦𝑝 to be the product of the corresponding assumed solutions and
algebraically
combine arbitrary constants where possible. In particular, if 𝑓(𝑥) =
𝑒𝑎𝑥𝑝𝑛(𝑥) is the product of a polynomial with an exponential, assume
𝑦𝑝 = 𝑒𝑎𝑥(𝐴𝑛𝑥𝑛 + 𝐴𝑛 − 1𝑥𝑛 − 1 + ⋯ +
𝐴2𝑥2 + 𝐴1𝑥 + 𝐴0)
where Aj is as in 1st case. If, instead, 𝑓(𝑥) = 𝑒𝑎𝑥𝑝𝑛(𝑥)𝑠𝑖𝑛𝑏𝑥 is the
product of a polynomial, exponential, and sine term, or if 𝑓(𝑥) =
𝑒𝑎𝑥𝑝𝑛(𝑥)𝑐𝑜𝑠𝑏𝑥 is the product of a polynomial, exponential, and cosine
term, then assume
𝑦𝑝
= 𝑒𝑎𝑥(𝐴𝑛𝑥𝑛 + 𝐴𝑛 − 1𝑥𝑛 − 1 + ⋯ +
𝐴2𝑥2 + 𝐴1𝑥 + 𝐴0)𝑠𝑖𝑛𝑏𝑥 + 𝑒𝑎𝑥(𝐵𝑛𝑥𝑛
+ 𝐵𝑛 − 1𝑥𝑛 − 1 + ⋯ +
𝐵2𝑥2 + 𝐵1𝑥 + 𝐵0)𝑐𝑜𝑠𝑏𝑥
where 𝐴𝑗 and𝐵𝑗 (𝑗 = 0,1,2, … , 𝑛) are constants which still must be
determined.
If 𝑓(𝑥) is the sum (or difference) of terms already considered, then we
take 𝑦𝑝 to be the sum (or difference) of the corresponding assumed
solutions and algebraically combine arbitrary constants where possible.

54. Modifications
If any term of the assumed solution, disregarding multiplicative
constants, is also a term of
(the homogeneous solution), then the assumed solution must be
modified by multiplying it by 𝑥𝑚, where 𝑚 is the smallest positive
integer such that the product of 𝑥𝑚 with the assumed solution has no
terms in common with 𝑦ℎ.
Limitations of the Method
In general, if 𝑓(𝑥) is not one of the types of functions considered
above, or if the differential equation does not have constant
coefficients, then the method of variations of parameters is preferred.

55. Variation of parameters is another method for finding a particular solution of
the nth-order linear differential equation 𝑳(𝑦) = 𝜑(𝑥), once the solution of the
associated homogeneous equation 𝑳(𝑦) = 0 is known. If 𝑦1 𝑥 , 𝑦2 𝑥 , … , 𝑦𝑛 𝑥
are 𝑛 linearly independent solutions of 𝑳(𝑦) = 0, then the general solution of
𝑳(𝑦) = 0 is
𝑦ℎ = 𝑐1𝑦1(𝑥) + 𝑐2𝑦2(𝑥) + ⋯ + 𝑐𝑛𝑦𝑛(𝑥)
Methodology:
A particular solution of 𝑳(𝑦) = 𝑓(𝑥) has the form
𝑦𝑝 = 𝑣1𝑦1(𝑥) + 𝑣2𝑦2(𝑥) + ⋯ + 𝑣𝑛𝑦𝑛(𝑥)
where 𝑦𝑖 = 𝑦𝑖 𝑥 (𝑖 = 1,2, … , 𝑛) is given in above Equation and 𝑣𝑖 (𝑖 =
1,2, … 𝑛) is an unknown function of 𝑥 which still must be determined.
Variation of Parameters

56. To find 𝑣𝑖, first solve the following linear equations simultaneously for 𝑣′𝑖:
𝑣′1𝑦1 + 𝑣′2𝑦2 + ⋯ + 𝑣′
𝑛𝑦𝑛 = 0
𝑣′1𝑦′1 + 𝑣′2𝑦′2 + ⋯ + 𝑣′
𝑛𝑦′𝑛 = 0
⋮
𝑣′1𝑦1
(𝑛−2)
+ 𝑣′2𝑦2
(𝑛−2)
+ ⋯ + 𝑣′
𝑛𝑦𝑛
(𝑛−2)
= 0
𝑣′1𝑦1
(𝑛−1)
+ 𝑣′2𝑦2
(𝑛−1)
+ ⋯ + 𝑣′
𝑛𝑦𝑛
𝑛−1
= 𝜑 𝑥 .
Then integrate each to obtain 𝑣𝑖, disregarding all constants of integration.
This is permissible because we are seeking only one particular solution.
Example 6.1: For the special case 𝑛 = 3, Equations reduce to
𝑣′1𝑦1 + 𝑣′2𝑦2 + 𝑣′
3𝑦3 = 0
𝑣′1𝑦′1 + 𝑣′2𝑦′2 + 𝑣′
3𝑦′3 = 0
𝑣′1𝑦1′′ + 𝑣′2𝑦2′′ + 𝑣′
3𝑦3′′ = 𝜑 𝑥 .
For the case 𝑛 = 2, Equations become
𝑣′1𝑦1 + 𝑣′2𝑦2 = 0
𝑣′1𝑦′1 + 𝑣′2𝑦′2 = 𝜑 𝑥 .
and for the case 𝑛 = 1, we obtain the single equation
𝑣′1𝑦1 = 𝜑 𝑥 .

57. Since 𝑦1 𝑥 , 𝑦2 𝑥 , … , 𝑦𝑛 𝑥 are 𝑛 linearly independent solutions of the same
equation L(y)=0, theirWronskian is not zero.This means that the system 6.9
has a nonzero determinant and can be solved uniquely for 𝑣′𝑖.
Scope of the Method
The method of variation of parameters can be applied to all linear differential
equations. It is therefore more powerful than the method of undetermined
coefficients, which is restricted to linear differential equations with constant
coefficients and particular forms of 𝜑 𝑥 .
Nonetheless, in those cases where both methods are applicable, the method
of undetermined coefficients is usually the more efficient and, hence,
preferable.
As a practical matter, the integration of may be impossible to perform. In
such an event other methods (in particular, numerical techniques) must be
employed.

58. Initial-Value Problems
Initial-value problems are solved by applying the initial conditions
to the general solution of the differential equation. It must be
emphasized that the initial conditions are applied only to the
general solution and not to the homogeneous solution 𝑦ℎ that
possesses all the arbitrary constants that must be evaluated.The
one exception is when the general solution is the homogeneous
solution; that is, when the differential equation under consideration
is itself homogeneous.

59. Example: Solve 𝑦′′ − 𝑦′ − 2𝑦 = 4𝑥2.
Solution: 𝑦ℎ = 𝑐1𝑒2𝑥 + 𝑐2𝑒−𝑥. Here 𝜑 𝑥 = 4𝑥2, a second degree polynomial.We
assume that
𝑦𝑝 = 𝐴2𝑥2 + 𝐴1𝑥 + 𝐴0
Thus 𝑦′𝑝 = 2𝐴2𝑥 + 𝐴1 and 𝑦′′𝑝 = 2𝐴2. Substituting these results into the
differential equation, we have
2𝐴2 − (2𝐴2𝑥 + 𝐴1) − 2(𝐴2𝑥2
+ 𝐴1𝑥 + 𝐴0) = 4𝑥2
.
Equating the coefficients of like powers of 𝑥, we obtain
−2𝐴2 = 4, −2𝐴2 − 2𝐴1 = 0,2𝐴2 − 𝐴1 − 2𝐴0 = 0
Solving this system, we find that 𝐴2 = −2, 𝐴1 = 2, 𝐴0 = −3.
Hence 𝑦𝑝 = −2𝑥2
+ 2𝑥 − 3 and the general solution is
𝑦𝑔 = 𝑦ℎ + 𝑦𝑝 = 𝑐1𝑒2𝑥 + 𝑐2𝑒−𝑥 − 2𝑥2 + 2𝑥 − 3

60. Example: Solve 𝑦′′
− 𝑦′
− 2𝑦 = sin(2𝑥) .
Solution:Again 𝑦ℎ = 𝑐1𝑒2𝑥
+ 𝑐2𝑒−𝑥
.
Assume that 𝑦𝑝 = 𝐴 sin 2𝑥 + 𝐵 cos 2𝑥 .
Thus, 𝑦′
𝑝 = 2𝐴 cos 2𝑥 − 2𝐵 sin 2𝑥 and 𝑦′′
𝑝 = −4𝐴 sin 2𝑥 −
4𝐵 cos 2𝑥 Substituting these results into the differential equation, we
have (−6𝐴 + 2𝐵) sin 2𝑥 + (−2𝐴 − 6𝐵) cos 2𝑥 = sin(2𝑥)
Equating coefficients of like terms, we obtain
−6𝐴 + 2𝐵 = 1, −2𝐴 − 6𝐵 = 0
Solving this system, we find that 𝐴 = −3/20 and 𝐵 = 1/20.Then
𝑦𝑝 = −
3
20
sin 2𝑥 +
1
20
cos 2𝑥 .
and the general solution is
𝑦𝑔 = 𝑦ℎ + 𝑦𝑝 = 𝑐1𝑒2𝑥 + 𝑐2𝑒−𝑥 −
3
20
sin 2𝑥 +
1
20
cos 2𝑥 .

61. Example: Solve 𝑦′′′
+ 𝑦′
= sec 𝑥 .
This is a third-order equation with 𝑦ℎ = 𝑐1 + 𝑐2 cos 𝑥 + 𝑐3 sin 𝑥 .
It follows that
𝑦𝑝 = 𝑣1 + 𝑣2 cos 𝑥 + 𝑣3 sin 𝑥 .
So the set of Equations becomes
𝑣′1 + 𝑣′2 cos 𝑥 + 𝑣′
3 sin 𝑥 = 0
𝑣′2 sin 𝑥 + 𝑣′
3 cos 𝑥 = 0
−𝑣′
2 cos 𝑥 − 𝑣′
3 sin 𝑥 = sec 𝑥 .
Solving this set of equations simultaneously, we obtain
𝑣′1 = sec 𝑥 , 𝑣′2 = −1, 𝑣′
3 = − tan 𝑥 .
Thus 𝑣1 = ln sec 𝑥 + tan 𝑥 , 𝑣2 = −𝑥, 𝑣3 = ln cos 𝑥 ,
Gives 𝑦𝑝 = ln sec 𝑥 + tan 𝑥 − 𝑥 cos 𝑥 + ln cos 𝑥 sin 𝑥 .
The general solution is therefore
𝑦𝑔 = 𝑦ℎ + 𝑦𝑝
= 𝑐1 + 𝑐2 cos 𝑥 + 𝑐3 sin 𝑥 + ln sec 𝑥 + tan 𝑥 − 𝑥 cos 𝑥 + ln cos 𝑥 sin 𝑥 .

62. LaplaceTransform
Let 𝑓(𝑥) be defined for 0 ≤ 𝑥 < ∞ and let 𝑠 denote an arbitrary real variable.
The Laplace transform of 𝑓(𝑥), designated by either ℒ{𝑓(𝑥)} or 𝐹(𝑠),is
ℒ 𝑓 𝑥 = 𝐹 𝑠 =
0
∞
𝑒−𝑠𝑥
𝑓(𝑥)𝑑𝑥
for all values of 𝑠 for which the improper integral converges. Convergence
occurs when the limit
lim
𝑅→∞ 0
𝑅
𝑒−𝑠𝑥
𝑓(𝑥)𝑑𝑥
exists. If this limit does not exist, the improper integral diverges and 𝑓(𝑥) has no
Laplace transform. When evaluating the integral, the variable 𝑠 is treated as a
constant because the integration is with respect to 𝑥.
The Laplace transforms for a number of elementary functions can be found in
Annexure of the Book.

63. ☺ (Linearity). If ℒ 𝑓 𝑥 = 𝐹 𝑠 and ℒ 𝑔 𝑥 = 𝐺 𝑠 , then
for any two constants 𝑐1 and 𝑐2
ℒ 𝑐1𝑓 𝑥 + 𝑐2𝑔 𝑥 = 𝑐1ℒ 𝑓 𝑥 + 𝑐2ℒ 𝑔 𝑥 = 𝑐1𝐹(𝑠) + 𝑐2𝐺(𝑠)
☺ If ℒ 𝑓 𝑥 = 𝐹(𝑠), then for any constant 𝑎, ℒ 𝑒𝑎𝑥𝑓 𝑥 = 𝐹(𝑠 − 𝑎)
☺ If ℒ 𝑓 𝑥 = 𝐹(𝑠), then for any positive integer 𝑛, ℒ 𝑥𝑛
𝑓 𝑥 = −1 𝑛 𝑑𝑛
𝑑𝑠𝑛 [𝐹 𝑠 ]
☺ If ℒ 𝑓 𝑥 = 𝐹(𝑠) and if lim
𝑥→0
𝑥>0
𝑓(𝑥)
𝑥
exists, then ℒ
1
𝑥
𝑓(𝑥) = 𝑠
∞
𝐹(𝑡) 𝑑𝑡
☺ If ℒ 𝑓 𝑥 = 𝐹(𝑠), then ℒ 0
𝑥
𝑓(𝑡) 𝑑𝑡 =
1
𝑠
𝐹(𝑠)
☺ If 𝑓(𝑥) is periodic with period 𝑤, that is, 𝑓 𝑥 + 𝑤 = 𝑓(𝑥), then
ℒ 𝑓 𝑥 = 0
𝑤
𝑒−𝑠𝑥
𝑓(𝑥) 𝑑𝑥
1 − 𝑒−𝑤𝑠
Properties of LaplaceTransforms

64. Inverse LaplaceTransform
An inverse Laplace transform of 𝐹(𝑠) designated by ℒ−1{𝐹(𝑠)}, is another
function 𝑓(𝑥) having the property that ℒ 𝑓 𝑥 = 𝐹(𝑠).
Methodology
The simplest technique for identifying inverse Laplace transforms is to
recognize them, either from memory or from a table.
If 𝐹(𝑠) is not in a recognizable form, then occasionally it can be transformed
into such a form by algebraic manipulation.

65. Manipulating Denominators
-The method of completing the square deals with quadratic polynomials
-The method of partial fractions
Manipulating Numerators
A factor 𝑠 − 𝑎 in the numerator may be written in terms of the factor 𝑠 − 𝑏,
where both 𝑎 and 𝑏 are constants, through the identity 𝑠 − 𝑎 = 𝑠 − 𝑏 +
(𝑏 − 𝑎).
♥ (Linearity). If the inverse Laplace transforms of two functions 𝐹(𝑠) and
𝐺(𝑠) exist, then for any constants 𝑐1 and𝑐2,
ℒ−1 𝑐1𝐹 𝑠 + 𝑐2𝐺 𝑠 = 𝑐1ℒ−1{𝐹 𝑠 } + 𝑐2ℒ−1{𝐺 𝑠 }

66. Convolution
The convolution of two functions 𝑓(𝑥) and 𝑔(𝑥) is
𝑓 𝑥 ∗ 𝑔 𝑥 =
0
𝑥
𝑓(𝑡)𝑔(𝑥 − 𝑡)𝑑𝑡
Theorem 𝑓 𝑥 ∗ 𝑔 𝑥 = 𝑔(𝑥) ∗ 𝑓(𝑥).
Theorem. (Convolution Theorem). If ℒ 𝑓 𝑥 = 𝐹(𝑠) and ℒ 𝑔 𝑥 =
𝐺(𝑠), then
ℒ 𝑓 𝑥 ∗ 𝑔 𝑥 = ℒ 𝑓 𝑥 ℒ 𝑔 𝑥 = 𝐹(𝑠)𝐺(𝑠)
The inverse Laplace transform of a product is computed using a
convolution.
ℒ−1 𝐹(𝑠)𝐺(𝑠) = 𝑓 𝑥 ∗ 𝑔 𝑥 = 𝑔 𝑥 ∗ 𝑓(𝑥)
If one of the two convolutions in above Equation is simpler to calculate,
then that convolution is chosen when determining the inverse Laplace
transform of a product.