Question

8) (10) Solve the equation y''=x^2-y+3y' y(0)=1 and y'(0) =-1

Answer 1

`y''-3y'+y=x^2`

The corresponding homogeneous equation,

`r^2-3r+1=0`

has two roots at
`r=\frac{3\pm\sqrt5}2`, so that the characteristic solution is

`y_c=C_1e^((3+\sqrt5)/2\,x)+C_2e^((3-\sqrt5)/2\,x)`

For the particular solution, assume one of the form

`y_p=ax^2+bx+c`

`\implies{y_p}'=2ax+b`

`\implies{y_p}''=2a`

Substituting
`y_p` and its derivatives into the non-homogeneous ODE gives

`2a-3(2ax+b)+(ax^2+bx+c)=x^2`

`ax^2+(-6a+b)x+(2a-3b+c)=x^2`

`\implies\begin{cases}a=1\\-6a+b=0\\2a-3b+c=0\end{cases}\implies a=1,b=6,c=16`

Then the general solution to the ODE is

`\boxed{y(x)=C_1e^((3+\sqrt5)/2\,x)+C_2e^((3-\sqrt5)/2\,x)+x^2+6x+16}`