Question

Find the particular integral of
(D^2+1)y=x^2sin2x

Answer 1

Answer:

(D2+1)y=x2sinx

Let,yh=emx, be a trial solution of the corresponding homogeneous equation (D2+1)y=0 ,for some real or complex values of m.

that is (D2+1)emx=0

⟹(m2+1)emx=0

⟹m2+1=0(since,emx≠0 for any m)

⟹m=±i

So,the solution yh of the homogeneous equation is eix or e−ix

Thus yh is some linear combination of eix and e−ix:

yh=C1eix+C2e−ix

C1,C2∈R being arbitrary.

Now,let yp be the particular solution of the given differential equation.

Then,(D2+1)yp=x2sinx

⟹yp=1(D2+1)x2sinx=I1(D2+1)x2eix=Ieix1((D+i)2+1)x2=Ieix1(D2+2Di)x2=Ieix12Di⋅1(1−i2D)x2=Ieix12Di⋅(1−i2D)−1x2=Ieix12Di⋅(1+i2D−14D2+⋯)x2=Ieix12Di⋅(x2+ix−12)=Ieix12i∫(x2+ix−12)dx=Ieix12i(x33+ix22−x2)=Ieix(−ix36+x24+ix4)=I(cosx+isinx)(−ix36+x24+ix4)=−x36cosx+x4cosx+x24sinx

Hence the general solution y to the given differential equation is given by :

y=yh+yp=C1eix+C2e−ix−x36cosx+x4cosx+x24sinx;

C1,C2∈R being arbitrary