Question

Find the general solution to y"-y'-2y=2e^3x

Answer 1

Given differential equation is

y"-y'-2y=2e^3x

i.e. $\frac{d²y}{dx²}-\frac{dy}{dx}-2y = 2e^{3x}$

This is a second order linear differential equation.

General solution of it comprises of two parts, complementary function and particular integral.

Complementary function:

Auxiliary equation is m²-m-2=0

=> (m-2)(m+1) = 0

=> m = 2, -1

So complementary function is

$Ae^{-x}+Be^{2x}$ , where A, B are constants

Particular integral:

Particular integral of the given differential equation is

$\frac{1}{D²-D-2}(2e^{3x})$

= $e^{3x}\frac{1}{(D+3)²-(D+3)-2}(2)$ [since $\frac{1}{f(D)}ke^{ax} = e^{ax}\frac{1}{f(D+a)}k$ ]

= $2e^{3x}\frac{1}{D²+5D+4}(1)$

= $\frac{2e^{3x}}{4}\frac{1}{1+\frac{D²+5D}{4}}(1)$

= $\frac{e^{3x}}{2}[1+\frac{D²+5D}{4}]^{-1}(1)$

= $\frac{e^{3x}}{2}[1-\frac{D²+5D}{4}+(\frac{D²+5D}{4})^{2}-...∞](1)$

= $\frac{e^{3x}}{2}$ , Since Dⁿ(1) = 0 for every natural number n.

So the general solution is

y = $Ae^{-x}+Be^{2x}$ + $\frac{e^{3x}}{2}$ , where A, B are arbitrary constants.