Answer to Question #199957 in Differential Equations for sania khan

Question

solve undetermined coefficient y'' - y' = sinx

Answer 1

Given equation is -${y^{''}}-y'=sinx$ ......A)

$\dfrac{d^{2}y}{dx^{2}}-\dfrac{dy}{dx}= sinx$

D = $\dfrac{d}{dx}$

writing the equation in normal form -

$(D^{2}-D)y=sinx$

writing in auxiliary form we get ,

auxiliary equation is -

$m^{2}-m =0$

m=0,1

CF can be written as $y=C_1e^{0x}+C_2e^{x}$

$y=C_1+C_2e^{x}$

now , we have to this by undetermined coefficient , for $sinx$ method let the particular solution is $y=Acosx+Bsinx$

$y^{'}=-Asinx+Bcosx$

$y{''}=-Acosx-Bsinx$

now putting the value of $y^{'},y{''}$ in equation A) , we get -

$=-Acosx-Bsinx +Asinx-Bcosx=sinx$

$=cosx(-A-B)$ $+sinx(-B+A)$ =$sinx$

comparing the coefficient of $sinx\ and \ cosx \ o$n both side of equation -

$=-A-B=0$

$=A-B=1$

on solving we get ,

$=$ $B=\dfrac{-1}{2}$

$=A=\dfrac{1}{2}$

So our particular solution will be -

$y=\dfrac{1}{2}cos x+\dfrac{1}{2}sinx$

now complete solution is given by -

$y=CF+PI$

$y=$ $C_1+C_2e^{x}+\dfrac{1}{2}(cosx+sinx)$