Question

Solve this please :) : y" + 8y' +15y =4x^2

Answer 1

The characteristic equation for this ODE is

$r^2+8r+15=(r+5)(r+3)=0$

and has roots at $r=-5,r=-3$, which admits the characteristic solution

$y_c=C_1e^{-5x}+C_2e^{-3x}$

For the particular solution, we can try finding a quadratic polynomial

$y_p=ax^2+bx+c$
${y_p}'=2ax+b$
${y_p}''=2a$

and substituting into the ODE gives

$2a+8(2ax+b)+15(ax^2+bx+c)=4x^2$
$15ax^2+(16a+15b)x+(2a+8b+15c)=4x^2$
$\implies\begin{cases}15a=4\\16a+15b=0\\2a+8b+15c=0\end{cases}\implies a=\dfrac4{15},b=-\dfrac{64}{225},c=\dfrac{392}{3375}$

so that the particular solution is

$y_p=\dfrac4{15}x^2-\dfrac{64}{225}x+\dfrac{392}{3375}$

and the general solution is

$y=y_c+y_p$
$y=C_1e^{-5x}+C_2e^{-3x}+\dfrac4{15}x^2-\dfrac{64}{225}x+\dfrac{392}{3375}$