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=\frac4{15},b=-(64)/(225),c=(392)/(3375)`

so that the particular solution is


`y_p=\frac4{15}x^2-(64)/(225)x+(392)/(3375)`

and the general solution is


`y=y_c+y_p`

`y=C_1e^(-5x)+C_2e^(-3x)+\frac4{15}x^2-(64)/(225)x+(392)/(3375)`