# Find the general solution for differential equation $y''-6y'+9y=e^x((2x+1)\cos x+(x+3)\sin x)$ [closed]

Solution for homogeneous equation $y''-6y'+9y=0$ is $y_h=c_1e^{3x}+c_2(x)xe^{3x}$.

How to use the method of variation of parameters and the method of undetermined coefficients on this equation?

## closed as off-topic by choco_addicted, Shahab, user91500, Stefan Mesken, Claude LeiboviciMar 18 '16 at 7:38

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – choco_addicted, Shahab, user91500, Stefan Mesken, Claude Leibovici
If this question can be reworded to fit the rules in the help center, please edit the question.

Consider one of the solution of the homogeneous equation and replace the constant by a function to be determined : $$y=f(x)e^{3x}$$ $y'=(f'+3f)e^{3x}$
$y''=(f''+6f'+9f)e^{3x}$
that we put into the ODE : $$y''-6y'+9y=(f''+6f'+9f-6(f'+3f)+9f)e^{3x}=f''e^{3x}=e^x\left((2x+1)\cos(x)+(x+3)\sin(x) \right)$$ $$f''=e^{-2x}\left((2x+1)\cos(x)+(x+3)\sin(x) \right)$$ Two successive integrations leads to : $$f=c_1x+c_2+\frac{1}{25}e^{-2x}\left( (10x+21)\cos(x)-(5x+3)\sin(x) \right)$$ $$y=e^{3x}(c_1x+c_2)+\frac{1}{25}e^{x}\left( (10x+21)\cos(x)-(5x+3)\sin(x) \right)$$
Note : A simpler way consists in looking for the particular solution on the form $y_p=e^x\left( (ax+b)\cos(x)+(cx+d)\sin(x) \right)$ which is suggested by the form of the right term of the non-homogeneous ODE.