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 Leibovici Mar 18 '16 at 7:38

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Consider one of the solution of the homogeneous equation and replace the constant by a function to be determined : $$y=f(x)e^{3x}$$ $y'=(f'+3f)e^{3x}$


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.


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