Differential Equations: Decomposition Solutions

Question 1
Solve
y ′′ + 6y ′ + 13y = 4e2x

using the method of decomposition.

Step 1: Rewrite the second-order DE as a system
Let
z = y′.

Then
z ′ = y ′′ .

From the differential equation,
y ′′ + 6y ′ + 13y = 4e2x ,

we obtain
y ′′ = −6y ′ − 13y + 4e2x .

Since
z = y′,

the corresponding system is
y ′ = z,
z ′ = −13y − 6z + 4e2x .

This is a non-homogeneous system of first-order differential equations.

Step 2: Consider the homogeneous system
First solve
y ′ = z,
z ′ = −13y − 6z.

The coefficient matrix is
0 1
A=( ).
−13 −6


Step 3: Determine the eigenvalues
Solve
∣ A − λI ∣= 0.

Hence
∣∣ −λ 1 ∣
∣−13 −6 − λ∣∣ = 0.

Expanding,
(−λ)(−6 − λ) − 1(−13) = 0.

Therefore
λ(6 + λ) + 13 = 0,

, or
λ2 + 6λ + 13 = 0.

Using the quadratic formula,
−6 ± √36 − 52 −6 ± √−16
λ= = = −3 ± 2i.
2 2

Hence
λ1 = −3 + 2i, λ2 = −3 − 2i.


Step 4: Determine the eigenvectors

Eigenvector corresponding to
λ1 = −3 + 2i

Solve
(A − λI)v = 0.

Thus
3 − 2i 1 v1
( ) (v ) = 0.
−13 −3 − 2i 2

The first row gives
(3 − 2i)v1 + v2 = 0,

so
v2 = −(3 − 2i)v1 .

Choose
v1 = 1.

Then
v2 = −3 + 2i.

Hence
1
v1 = ( ).
−3 + 2i


Eigenvector corresponding to
λ2 = −3 − 2i

Similarly,
1
v2 = ( ).
−3 − 2i


Step 5: Form the complementary solution
The complex solution is
1 1
X = c1 ( ) e(−3+2i)x + c2 ( ) e(−3−2i)x .
−3 + 2i −3 − 2i