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 $r=\dfrac{3\pm\sqrt5}2$, 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

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