Toronto Math Forum

$$y''-6y'+8y=48sinh(2x)$$
$$y(0)=0 \ y'(0)=0 \ sinhx=\frac{{e^x-e^{-x}}}{2}$$
$r^2-6r+8=0$\\
(r-2)(r-4)=0\\
r=2 \ r=4 \\
$y(x)=c_1e^{2x}+c_2e^{4x}$\\
$48sinh(2x)=48(\frac{e^x-e^{-x}}{2})=48(\frac{e^{2x}}{2}-\frac{e^{-2x}}{2})=24e^{2x}-24e^{-2x}$\\
$y"-6y'+8y=24e^{2x}$\\
$y_1(x)=Axe^{2x}$\\
$y_1'(x)=2Axe^{2x}+Ae^{2x}$\\
$y_{1}''(x)=4Axe^{2x}+4Ae^{2x}$\\
Substitute \  them\  back \ to \ the \ original \ equation\\
$4Axe^{2x}+4Ae^{2x}-12Axe^{2x}-6Ae^{2x}+8Axe^{2x}=24e^{2x}$\\
$-2Ae^{2x}=24Ae^{2x}$\\
A=-12\\
$y_1(x)=-12xe^{2x}$\\
$y''-6y'+8y=-24e^{-2x}$\\
$y_2(x)=Ae^{-2x}$\\
$y_2(x)=-2Ae^{-2x}$\\
$y_2(x)=4Ae^{-2x}$\\
Substitute \  them\  back \ to \ the \ original \ equation\\
$4Ae^{-2x}+12-2Ae^{-2x}+8Ae^{-2x}=-24Ae^{-2x}$\\
$24Ae^{-2x}=-24e^{-2x}$\\
A=-1\\
$y_2(x)=-e^{-2x}$\\
$y(x)=c_1e^{2x}+c_2e^{4x}-12xe^{2x}-e^{-2x}$\\
$y(0)=c_1+c_2-1=0$\\
$y'(x)=2c_1e^{2x}+4c_2e^{4x}-24xe^{2x}-12e^{2x}+2e^{-2x}$\\
$y'(0)=2c_1+4c_2-12+2=0$\\
$c_1=-3 \ c_2=4$\\
$y(x)=-3e^{2x}+4e^{4x}-12xe^{2x}-e^{-2x}$