1. The document discusses solving linear differential equations, both homogeneous and non-homogeneous equations. It provides examples of solving equations of varying orders using techniques like finding the auxiliary equation, using inverse operators, and determining the complementary function and particular integrals. 2. Key concepts covered include the auxiliary equation, complementary function, particular integral, and using inverse operators to find particular integrals when solving non-homogeneous equations. 3. Various types of linear differential equations are solved as examples, including first, second, and third order equations with exponential, trigonometric, and polynomial terms. Complete solutions are presented combining the complementary function and particular integrals.

1. LINEAR DIFFERENTIAL EQUATIONS
SEBASTIAN VATTAMATTAM
1. Homogeneous Equations
Example 1.1. Solve
d2 y dy
+ a + by = 0, a, b ∈ R....(1)
dx2 dx
Suppose y = emx ....(2) is a solution of (1).
Then,
m2 emx + amemx + bemx = 0 ⇔ m2 + am + b = 0....(3)
So, (2) is a solution of (1) iff m is a root of (3).
Let m1 , m2 be the roots of (3).
Case 1: m1 , m2 are real and distinct.
y = c1 em1 x + c2 em2 x
Case 2: m1 = p + iq, m2 = p − iq
y = epx (c1 cos qx + c2 sin qx)
Case 3: m1 = m2 = m
(c1 + c2 x)emx
Example 1.2. Solve
y + y − 2y = 0
The auxiliary equation is
m2 + m − 2 = 0 ⇒ m1 = −1, m2 = 2
Date: 15 June 2010.
Key words and phrases. Complementary Function, Particular Integral, Complete Solution.
1

2. 2 SEBASTIAN VATTAMATTAM
General Solution is
y = c1 e−x + c2 e2x
Example 1.3. Solve
y − 2y + 10y = 0
The auxiliary equation is
m2 − 2m + 10 = 0 ⇒ m1 = 1 + 3i, m2 = 1 − 3i
General Solution is
y = ex (c1 cos 3x + c2 sin 3x
Example 1.4. Solve
y − 2y + 10y = 0 given that y(0) = 4, y (0) = 1
The auxiliary equation is
m2 − 2m + 10 = 0 ⇒ m1 = 1 + 3i, m2 = 1 − 3i
General Solution is
y(x) = ex (c1 cos 3x + c2 sin 3x)
y(0) = c1 ⇒ c1 = 4....(1)
y (x) = ex (c1 cos 3x + c2 sin 3x − 3c1 sin 3x + 3c2 cos 3x)
Therefore
y (0) = c1 + 3c2 ⇒ c1 + 3c2 = 1....(2)
From (1) and (2), c1 = 4, c2 = −1 Therefore, the solution
is
y(x) = ex (4 cos 3x − sin 3x)
Remember
d x
e f (x) = ex [f (x) + f (x)]
dx
1
Example 1.5. Solve
y + n2 y = 0
1Similar to Ex 4, p.190 in [1]

3. ORDINARY DIFFERENTIAL EQUATIONS 3
The auxiliary equation is
m2 + n2 = 0 ⇒ m1 = ni, m2 = −ni
General Solution is
y = c1 cos nx + c2 sin nx
Example 1.6. Solve
y + 8y + 16y = 0
The auxiliary equation is
m2 + 8m + 16 = 0 ⇒ m1 = −4, −4
General Solution is
y(x) = (c1 + c2 x)e−4x
Example 1.7. Solve
y − 4y + 4y = 0, y(0) = 3, y (0) = 1
General Solution is
y(x) = (c1 + c2 x)e2x
y(0) = c1 ⇒ c1 = 3....(1)
y (x) = c2 e2x + 2(c1 + c2 x)e2x
Therefore
y (0) = c2 + 2c1 ⇒ c2 + 2c1 = 1....(2)
From (1) and (2), c1 = 3, c2 = −5 Therefore, the solution
is
y(x) = (3 − 5x)e2x
2
Example 1.8. Solve
2y − 5y − 4y + 3y = 0
The auxiliary equation is
2m3 − 5m2 − 4m + 3 = 0
2Ex 1, p.88 of [1]

4. 4 SEBASTIAN VATTAMATTAM
Solution by synthetic division:
1 2 -5 -4 3 -1 2 -5 -4 3
Try m = 1 2 -3 -7 Try m = −1 -2 7 -3
2 -3 -7 [-4] 2 -7 3 [0]
2m3 −5m2 −4m+3 = (m+1)(2m2 −7m+3) = (m+1)(2m−1)(m−3) = 0
⇒ m = 1/2, −1, 3
General Solution is
y = Aex/2 + Be−x + Ce3x
3
Example 1.9. Solve
y + 3y + 3y + 2y = 0
The auxiliary equation is
m3 + 3m2 + 3m + 2 = 0
Solution by synthetic division:
-1 1 3 3 2 -2 1 3 3 2
Try m = −1 -1 -2 -1 Try m = −2 -2 -2 -2
1 2 1 [1] 1 1 1 [0]
m3 + 3m2 + 3m + 2 = 0 = (m + 2)(m2 + m + 1) = 0
⇒ m = −2, −1/2 ± i1/2
General Solution is
√ √
3x 3x
y = Ae−2x + e−x/2 (C cos + D sin )
2 2
2. Non-homogeneous Equations
Inverse Operators
(1)
1
φ(x) = φ(x)dx
D
3Ex 3, p.190 in [1]

5. ORDINARY DIFFERENTIAL EQUATIONS 5
(2)
1
φ(x) = eax φ(x)e−ax dx
D−a
(3)
1 ax 1 ax
e = e , if f (a) = 0
f (D) f (a)
(4) If f (a) = 0, then
1 ax 1 ax
e =x e , if f (a) = 0
f (D) f (a)
(5) If f (a) = 0, then
1 ax 1 ax
e = x2 e , if f (a) = 0
f (D) f (a)
(6)
1 1
2)
sin(ax + b) = 2)
sin(ax + b), if f (−a2 ) = 0
f (D f (−a
(7) If f (−a2 ) = 0, then
1 1
2)
sin(ax + b) = x sin(ax + b), if f (−a2 ) = 0
f (D f (−a2 )
(8)
1 ax 1
e φ(x) = eax φ(x)
f (D) f (D + a)
Example 2.1. Solve
(D − 2)2 y = 8(e2x + sin 2x + x2 )
(1) Find the Complementary Function (CF)
(m − 2)2 = 0 ⇒ m = 2, 2
CF = (c1 + c2 x)e2x
(2) Find Particular Integrals(PI)
(a)
1 2x 1 2x 2 1 2x x2 2x
P I1 = e =x e =x e = e
(D − 2)2 2(D − 2) 2 2
Note that f (2) = f (2) = 0

6. 6 SEBASTIAN VATTAMATTAM
(b)
1 1 1 11
P I2 = sin 2x = 2 sin 2x = sin 2x = − sin 2
(D − 2)2 D − 4D + 4 −4 − 4D + 4 4D
(c)
1 1 1 1 1 2D
P I3 = x2 = x2 = x2 = (1−D/2)−2 x2 = [1+
(D − 2)2 (2 − D)2 4(1 − D/2)2 4 4 2
(3) Complete solution is
y = CF + P I1 + P I2 + P I3
References
[1] P. Kandasamy etc,Engineering Mathematics- For First Year ,S. Chand & Company Ltd
Mary Bhavan, Vallikkad Road, Ettumanoor - 686631
E-mail address: vattamattam@gmail.com