Exercise
Solve $$y''-5y'+3y = 2-10x+3x^2$$. Is solution stable? The initial conditions are $y(0)= 0 $ $y'(0)=0$
My solution:
So first we solve homogeneous part:
$y''-5y'+3y = 0 \Rightarrow \lambda^2-5\lambda+3\lambda = 0$
$$\lambda_{1,2} = \frac{5+/- \sqrt{13}}{2}$$
Hence $y_H = \alpha_1e^{\frac{5+ \sqrt{13}}{2}}+\alpha_2e^\frac{5- \sqrt{13}}{2}$
I will guess the particular solution :
$$y_P = Ax^2+Bx+C$$
After calculations i got that $y_P=x^2$
Hence $y= \alpha_1e^{\frac{5+ \sqrt{13}}{2}x}+\alpha_2e^{\frac{5- \sqrt{13}}{2}x} + x^2$
Now we put the initial conditions in:
And this is where i am stuc. I cannot find values of $\alpha_1 , \alpha_2$.
My question
Did i have some mistakes until this point? If not, can someone please help me to find $\alpha_1 , \alpha_2$?
And solution is not stable since $Re(\lambda_1,\lambda_2) > 0$, right?