Answer to Question #110616 in Differential Equations for ABHISHEK TOMAR

Question

Using the method of undetermined coefficients, find the general solution of the differential equation,
y'''' - 2y''' + 2y'' = 3e^(-x) + 2e^(-x)x + e^(-x)sinx.

Answer 1

$y^{IV}-2y^{III}+2y^{II}=(3+2x)e^{-x}+e^{-x} \sin x$

Solution find in form

$y=y_0+y_1+y_2$

1) $y_0$ find from equation

$y^{IV}-2y^{III}+2y^{II}=0\\ \lambda ^4-2\lambda^3+2\lambda^2=0\\ \lambda_1=\lambda_2=0, \lambda_3=1-i, \lambda_4=1+i\\ y_0=C_1+C_2x+C_3e^x\cos x+C_4e^x \sin x$

2) $y_1$ find in form

$y_1=(ax+b)e^{-x}$

for equation

$y^{IV}-2y^{III}+2y^{II}=(3+2x)e^{-x}\\ y_1'=(a-ax-b)e^{-x}\\ y_1''=(ax-2a+b)e^{-x}\\ y_1'''=(3a-ax-b)e^{-x}\\ y_1''''=(ax-4a+b)e^{-x}\\$

input in equation

$(ax-4a+b)e^{-x}-2(3a-ax-b)e^{-x}+\\ +2(ax-2a+b)e^{-x}=(3+2x)e^{-x}\\ x:a+2a+2a=2\\ x^0:b-4a-6a+2b-4a+2b=3\\ a=\frac{2}{5}\\ b=\frac{43}{25}\\ y_1=(\frac{2}{5}x+\frac{43}{25})e^{-x}$

3) $y_2$ find in form

$y_2=ke^{-x}\cos x+me^{-x}\sin x$

for equation

$y^{IV}-2y'''+2y''=e^{-x}\sin x\\ y'_2=(m-k)e^{-x}\cos x-(k+m)e^{-x}\sin x\\ y''_2=-2me^{-x}\cos x+2ke^{-x}\sin x\\ y'''_2=(2m+2k)e^{-x}\cos x+(2m-2k)e^{-x}\sin x\\ y''''_2=-4ke^{-x}\cos x-4me^{-x}\sin x\\ -4ke^{-x}\cos x-4me^{-x}\sin x-\\ -2(2m+2k)e^{-x}\cos x-2(2m-2k)e^{-x}\sin x-\\ -4me^{-x}\cos x+4ke^{-x}\sin x=e^{-x}\sin x\\ \cos x: -4k-4m-4k-4m=0\\ \sin x:-4m-4m+4k+4k=1\\ m=-\frac{1}{16}\\ k=\frac{1}{16}\\ y_2=\frac{1}{16}e^{-x}\cos x-\frac{1}{16}e^{-x}\sin x$

than

$y=C_1+C_2x+C_3e^x\cos x+C_4e^x\sin x+\\ +(\frac{2}{5}x+\frac{43}{25})e^{-x}+\frac{1}{16}e^{-x}\cos x-\frac{1}{16}e^{-x}\sin x$