Answer to Question #114191 in Differential Equations for Neha

Question

solve the following initial value problem d^2y/dx^2 + dy/dx -2y=-6 sin2x-18cos2x where y(0)=2, y'(0)=2

Answer 1

The given differential equation is

$\frac{d^2y}{dx^2}+\frac{dy}{dx}-2y=-6 \ sin2x-18 \ cos2x$ .

The auxiliary equation is $m^2+m-2=0$

$\implies m^2+2m-m-2=0$

$\implies m(m+2)-1(m+2)=0$

$\implies (m+2)(m-1)=0$

$\implies m=1,-2$

$\therefore C.F.=A\ e^x+B\ e^{-2x}$

Now , $P.I.=\frac{1}{D^2+D-2}(-6\ sin2x-18\ cos2x)$

$=\frac{1}{D^2+D-2}(-6\ sin2x)+\frac{1}{D^2+D-2}(-18\ cos2x)$

$= \frac{-6}{-4+D-2}(sin2x)+\frac{-18}{-4+D-2}(cos2x)$

$=\frac{-6}{D-6}(sin2x)-\frac{18}{D-6}(cos2x)$

$=-6×\frac{D+6}{D^2-36}(sin2x)-18×\frac{D+6}{D^2-36}(cos2x)$

$=\frac{6}{40}×(2\ cos2x+6\ sin2x)+\frac{18}{40}×(-2\ sin2x +6\ cos2x)$

$=(\frac{18}{20}+\frac{-18}{20})sin2x+(\frac{6}{20}+\frac{54}{20})cos2x$

$=3 \ cos2x$ .

Therefore the complete solution is

$y=A\ e^x+B \ e^{-2x}+3\ cos2x$ ...................(1)

Differentiating (1) ,we get

$y'=A\ e^x-2B\ e^{-2x}-6\ sin2x$ .............(2)

But given that $y(0)=2$ and $y'(0)=2$ .

Therefore from equation (1) and (2) ,we get

$A+B+3=2 \ ,\implies A+B=-1$ .............(3)

$A-2B=2$ .........(4)

Subtracting (4) from (3) ,we get

$3B=-3 \ , \implies B=-1$

Putting the value of B in equation (3) ,we get

A=0.

$\therefore y=-e^{-2x}+3 \ cos2x$