# Non-order DE

#### idontknow

Solve equation : $$\displaystyle y''=y+e^{2x}$$ .

#### DarnItJimImAnEngineer

$y = \Sigma c_n e^{\lambda_n x}$
$\lambda = 2$ or $\lambda^2 - 1 = 0$
$\rightarrow y=\frac{1}{3} e^{2x} + c_1 e^x + c_2 e^{-x}$
$c_1, ~c_2$ from initial/boundary conditions

2 people

#### skipjack

Forum Staff
$e^{-x}y'' - e^{-x}y = e^x \\ e^{-x}y' + e^{-x}y = e^x + 2\text{A} \\ e^xy' + e^xy = e^{3x} + 2\text{A}e^{2x} \\ e^xy = \frac13e^{3x} + \text{A}e^{2x} + \text{B} \\ y = \frac13e^{2x} + \text{A}e^x + \text{B}e^{-x}$

4 people

#### idontknow

Looks similiar to Wronskian . $$\displaystyle dW(y,e^{-x} )=e^{x}dx$$.

1 person

#### v8archie

Math Team
$e^{-x}y'' - e^{-x}y = e^x \\ e^{-x}y' + e^{-x}y = e^x + 2\text{A} \\ e^xy' + e^xy = e^{3x} + 2\text{A}e^{2x} \\ e^xy = \frac13e^{3x} + \text{A}e^{2x} + \text{B} \\ y = \frac13e^{2x} + \text{A}e^x + \text{B}e^{-x}$
Very nice.

The more standard approach would be to solve the characteristic polynomial for $y''-y=0$ ($r=\pm1$) and them find a particular solution for the original equation, probably using the method of undetermined coefficients with $y_p=Ae^{2x}$.

1 person