Question

y''(x) - y'(x) + 2y(x) = 2e^x

Answer 1

Let's solve this problem

$y''\:-y'\:+2y=2e^x:$

Second-order linear non-homogeneous differential equation with constant coefficients

A second order linear, non-homogeneous ODE has formed of $ay''+by'+cy=g\left(x\right)$

The General solution to $a\left(x\right)y''+b\left(x\right)y'+c\left(x\right)y=g\left(x\right)$ can be written as

$y=y_h+y_p$

$y_h$ is the solution to the homogeneous ODE $a\left(x\right)y''+b\left(x\right)y'+c\left(x\right)y=0$

$y_p$ the particular solution, is any function That satisfies the non-homogenous equation

Find $y_h$ by solving $y''\:-y'\:+2y = 0$; $y=e^{\frac{x}{2}}\left(c_1\cos \left(\frac{\sqrt{7}x}{2}\right)+c_2\sin \left(\frac{\sqrt{7}x}{2}\right)\right)$

Find $y_p$ by solving $y''\:-y'\:+2y=2e^x:$ $y = e^x$

The general solution $y=y_h+y_p$

$y=e^{\frac{x}{2}}\left(c_1\cos \left(\frac{\sqrt{7}x}{2}\right)+c_2\sin \left(\frac{\sqrt{7}x}{2}\right)\right)+e^x$

$\mathrm{Plotting:}\:e^{\frac{x}{2}}\left(c_1\cos \left(\frac{\sqrt{7}x}{2}\right)+c_2\sin \left(\frac{\sqrt{7}x}{2}\right)\right)+e^x\quad \mathrm{assuming}\quad \:c_1=1\quad \:c_2=1$