Question

How to find particular integral for y''+3y'+2y=e^e^x

Answer 1

′′+3y′+2y=eexy″+3y′+2y=eex

Complimentary function

Auxiliaryequationr2+3r+2=0⟹(r+2)(r+1)=0⟹r=−2,−1Auxiliaryequationr2+3r+2=0⟹(r+2)(r+1)=0⟹r=−2,−1

yc=Ae−x+Be−2xyc=Ae−x+Be−2x

Particular Integral

(D+1)(D+2)y=eex;Let(D+2)y=z(D+1)(D+2)y=eex;Let(D+2)y=z

(D+1)z=eex(D+1)z=eex

dzdx+y=eexdzdx+y=eex

I.F.=e∫dx=exI.F.=e∫dx=ex

Multiplying by I.F. exz=∫exeexdxexz=∫exeexdx

Put ex=tex=t ; exdx=dtexdx=dt

exz=∫etdt=et=eexexz=∫etdt=et=eex

z=e−xeexz=e−xeex

(D+2)y=e−xeex(D+2)y=e−xeex

dydx+2y=e−xeexdydx+2y=e−xeex

I.F=e∫2dx=e2xI.F=e∫2dx=e2x

Multiplying by I.F. and integrating

e2xy=∫e2xe−xeexdx=∫exeexdx=eexe2xy=∫e2xe−xeexdx=∫exeexdx=eexas above.

So yp=e−2xeexyp=e−2xeex

The total solution is

y=Ae−x+Be−2x+e−2xeexy=Ae−x+Be−2x+e−2xeex