Answer to Question #211150 in Differential Equations for Daniel

Question

Solve the following equations by the use of the operator D

1. D²y -2Dy + y= sinx +x²

2.D²y -6Dy +6y= e^3x + e^-3x

Answer 1

1. Let us solve the equation $D^2y -2Dy + y=\sin x +x^2.$ Its characteristic equation $k^2-2k+1=0$ is equivalent to $(k-1)^2=0,$ and hence has the solutions $k_1=k_2=1.$ It follows that the general solution is of the form $y=(C_1+C_2x)e^x+y_p,$ where $y_p=a\sin x+b\cos x+cx^2+dx+e.$ It follows that $Dy_p=a\cos x-b\sin x+2cx+d$ and $D^2y_p=-a\sin x-b\cos x+2c.$ We have the equality $-a\sin x-b\cos x+2c-2(a\cos x-b\sin x+2cx+d)+a\sin x+b\cos x+cx^2+dx+e=\sin x +x^2.$It is equivalent to $-2a\cos x+2b\sin x+cx^2+(d-4c)x+e+2c-2d=\sin x +x^2.$ It follows that $-2a=0,\ 2b=1,\ c=1,\ d-2c=0,\ e+2c-2d=0.$ Therefore, $a=0,\ b=\frac{1}{2},\ c=1,\ d=2,\ e=2.$ We conclude that the general solution is the following:

$y=(C_1+C_2x)e^x+\frac{1}{2}\cos x+x^2+2x+2.$

2. Let us solve the equation $D^2y -6Dy +6y= e^{3x} + e^{-3x}.$ Its characteristic equation $k^2-6k+6=0$ is equivalent to $(k-3)^2=3,$ and hence has the solutions $k_1=3+\sqrt{3}$

and $k_2=3-\sqrt{3}.$ It follows that the general solution is of the form $y=C_1e^{(3+\sqrt{3})x}+C_2e^{(3-\sqrt{3})x}+y_p,$ where $y_p=a e^{3x} + be^{-3x}.$ Therefore, $Dy_p=3a e^{3x} -3be^{-3x}$ and $D^2y_p=9a e^{3x} + 9be^{-3x}.$ We get the equality $9a e^{3x} + 9be^{-3x}-6(3a e^{3x} -3be^{-3x})+6(a e^{3x} +be^{-3x})=e^{3x} +e^{-3x}.$ It is equivalent to $-3a e^{3x} + 33be^{-3x}=e^{3x} +e^{-3x}.$ It follows that $-3a=33b=1,$ and thus $a=-\frac{1}{3},\ b=\frac{1}{33}.$ We conclude that the general solution is the following:

$y=C_1e^{(3+\sqrt{3})x}+C_2e^{(3-\sqrt{3})x}-\frac{1}{3}e^{3x}+\frac{1}{33}e^{-3x}.$