Question

d^y/dx^-4y=xsinhx

Answer 1

Solution;

$\frac{d^2y}{dx^2}-4y=xsinhx$

For the complementary solution:

The characteristic polynomial is;

$m^2-4=0$

$m^2=4$

$m=\displaystyle _-^+2$

Hence the homogeneous solution is;

$y_h=c_1e^{2x}+c_2e^{-2x}$

For the particular Integral;

$P.I=\frac{1}{D^2-4}xsinhx$

$=\frac{1}{(D+2)(D-2)}x\frac{e^x-e^{-x}}{2}$

$=\frac{e^x}{2(D+2)(D-2)}x-\frac{e^{-x}}{2(D+2)(D-2)}$

$=\frac{e^x}{2(D+1+2)(D+1-2)}x-\frac{e^{-x}}{2(D-1+2)(D-1-2)}x$

$=\frac{e^x}{2(D^2+2D-3)}x-\frac{e^{-x}}{2(D^2-2D-3)}x$

$=\frac{e^x}{2}[\frac{1}{-3(1-\frac{D^2+2D}{3})}]x-\frac{e^{-x}}{2}[\frac{1}{-3(1-\frac{(2D-D^2)}{3})}]x$

$=\frac{e^x}{-6}[1+\frac{D^2+2D}{3}+...]x+\frac{e^{-x}}{6}[1-\frac{2D-D^2}{3}+...]x$

$=\frac{-e^x}{6}[x+\frac23]+\frac{e^{-x}}{6}[x+\frac23]$

$=\frac{-x}{3}(\frac{e^x-e^{-x}}{2})-\frac29(\frac{e^x-e^{-x}}{2})$

Hence;

$P.I=\frac{-x}{3}sinhx-\frac29sinhx$ =$-\frac19(3x+2)sinhx$

Complete solution is ;

$y=C.F+P.I$

$y=c_1e^{2x}+c_2e^{-2x}-\frac 19(3x+2)sinhx$