Answer to Question #105508 in Differential Equations for khushi

Question

Using the method of undermined coefficients, find the general solution of the
differential equation
y^iv-2y'''+2y''=3e^-x+2e^-xx+e^-x sin x

Answer 1

1)Solve homogeneous equation $y^{(4)}-2y'''+2y''=0$.

Its characteristic equation is $\lambda^4-2\lambda^3+2\lambda^2=0$, solutions of which are $\lambda_1=\lambda_2=0, \lambda_3=1+i, \lambda_4=1-i$.

So general solution of this equation is $y=Ax+B+e^x(C\cos x+D\sin x)$

2)Particular soltuion of $y^{(4)}-2y'''+2y''=3e^{-x}+2xe^{-x}+e^{-x}\sin x$

Since $-1$ and $-1+i$ are not roots of the characteristic equation, pariclular solution of $y^{(4)}-2y'''+2y''=3e^{-x}+2xe^{-x}+e^{-x}\sin x$ is $y=(U+Vx)e^{-x}+e^{-x}(W\cos x+T\sin x)$.

Calculate derivative of $z=(\alpha+\beta x)e^{-x}+e^{-x}(\gamma\cos x+\delta\sin x)$ :

$z'=\beta e^{-x}-(\alpha+\beta x)e^{-x}+e^{-x}(-\gamma\sin x+\delta\cos x)-$

$-e^{-x}(\gamma\cos x+\delta\sin x)=$

$=(-\beta x+\beta-\alpha)e^{-x}+e^{-x}((\delta-\gamma)\cos x+(-\gamma-\delta)\sin x)$

Then $y'=(-Vx+V-U)e^{-x}+$

$+e^{-x}((T-W)\cos x+(-W-T)\sin x)$

$y''=(Vx-2V+U)e^{-x}+e^{-x}(-2T\cos x+2W\sin x)$

$y'''=(-Vx+3V-U)e^{-x}+$

$+e^{-x}((2W+2T)\cos x+(2T-2W)\sin x)$

$y^{(4)}=(Vx-4V+U)e^{-x}+e^{-x}(-4W\cos x-4T\sin x)$

So $y^{(4)}-2y'''+2y''=(5Vx-14V+5U)e^{-x}+$

$+e^{-x}((-8W-8T)\cos x+(8W-8T)\sin x)$ and we obtain

$\begin{cases} -14V+5U=3\\ 5V=2\\ -8W-8T=0\\ 8W-8T=1 \end{cases}$

That is $V=\frac{2}{5}, U=\frac{43}{25}, T=-\frac{1}{16}, W=\frac{1}{16}$

3)General solution of $y^{(4)}-2y'''+2y''=3e^{-x}+2xe^{-x}+e^{-x}\sin x$ is $y=Ax+B+e^x(C\cos x+D\sin x)+$

$+\frac{1}{25}(43+10x)e^{-x}+\frac{1}{16}e^{-x}(\cos x-\sin x)$

Answer:

$Ax+B+e^x(C\cos x+D\sin x)+$

$+\frac{1}{25}(43+10x)e^{-x}+\frac{1}{16}e^{-x}(\cos x-\sin x)$