Question

y" — 9y = 0. + e.

Answer 1

$\frac{d^2y(x)}{dx^2}-9y(x)=e$

The general solution will be the sum of the complimentary solution and particular solution.

The complimentary solution:

$\frac{d^2y(x)}{dx^2}-9y(x)=0$

Solution will be proportional for e^kx for some constant k.

Substitute y(x)= e^kx into equalation.

k²e^kx-9e^kx=0

e^kx(k²-9)=0

As e^kx can not be equal 0, so k²-9=0

k₁=3, k₂=-3

The complimentary solution is

y(x)=C₁e^3x+C₂e^-3x

Particular solution:

y(x)=Be^3x

y''=9Be^3x

-9Be^3x=e

$B=\frac{-e}{9e^{3x}}$

y(x)=-e/9

Answer:y(x)=C₁e^3x+C₂e^-3x-e/9