Solve (D²+1)y=x²sin2x

1.	Solve (D²+1)y=x²sin2x
Answer» $\large { \underline{ \bf \red{Question:- }}}$ SOLVE (D²+1)y=x²sin2x $\large { \underline{ \bf \purple{<klux>SOLUTION</klux>:- }}}$ (D2+1)y=x2sinx(D2+1)y=x2sin⁡x Let,yh=emxyh=emx, be a trial solution of the corresponding homogeneous equation (D2+1)y=0(D2+1)y=0 ,for some real or complex values of mm. $that \: is (D2+1)emx=0(D2+1)emx=0 \\ ⟹(m2+1)emx=0⟹(m2+1)emx=0 \\ ⟹m2+1=0(⟹m2+1=0 \\ (since,emx≠0emx≠0 for any m)\\ ⟹m=±i⟹m=±i \\$ So,the solution yhyh of the homogeneous equation is eixeix or e−ixe−ix Thus yhyh is some linear combination of eixeix and e−ix:e−ix: $yh=C1eix+C2e−ixyh=C1eix+C2e−ix$ C1,C2∈RC1,C2∈R being arbitrary. Now,let ypyp be the particular solution of the given DIFFERENTIAL equation. $Then(D2+1)yp= \\ x2sinx(D2+1)yp= \\ 2sin⁡x$ $yp=1(D2+1)x2sinx \\ =I1(D2+1)x2eix \\ =Ieix1((D+i)2+1)x2 \\ =Ieix1(D2+2Di)x2 \\ =Ieix12Di⋅1(1−i2D)x2 \\ =Ieix12D \\ (1−i2D)−1x2=Ieix12Di⋅ \\ (1+i2D−14)x2 \\ =Ieix12Di⋅(x2+ix−12) \\ =Ieix12i∫(x2+ix−12)dx \\ =Ieix12i(x33+ix22−x2) \\ =Ieix(−ix36+x24+ix4) \\ =I(cosx+isinx)(−ix36+$ Hence the general solution YY to the given differential equation is given by : $y=yh+yp \\ =C1eix+C2e−ix−x36cosx+x4cosx+x24sinx;y=yh+yp \\ =C1eix+C2e−ix−x36cos⁡x+x4cos⁡x+x24sin⁡x;$ $\large { \underline{ \bf \pink{Therefore:- }}}$ $C1,C2∈RC1,C2∈r being \: \: arbitrary.$

Answer»

$\large { \underline{ \bf \red{Question:- }}}$

SOLVE (D²+1)y=x²sin2x

$\large { \underline{ \bf \purple{<klux>SOLUTION</klux>:- }}}$

(D2+1)y=x2sinx(D2+1)y=x2sin⁡x

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

$that \: is (D2+1)emx=0(D2+1)emx=0 \\ ⟹(m2+1)emx=0⟹(m2+1)emx=0 \\ ⟹m2+1=0(⟹m2+1=0 \\ (since,emx≠0emx≠0 for any m)\\ ⟹m=±i⟹m=±i \\$

So,the solution yhyh of the homogeneous equation is eixeix or e−ixe−ix

Thus yhyh is some linear combination of eixeix and e−ix:e−ix:

$yh=C1eix+C2e−ixyh=C1eix+C2e−ix$

C1,C2∈RC1,C2∈R being arbitrary.

Now,let ypyp be the particular solution of the given DIFFERENTIAL equation.

$Then(D2+1)yp= \\ x2sinx(D2+1)yp= \\ 2sin⁡x$

$yp=1(D2+1)x2sinx \\ =I1(D2+1)x2eix \\ =Ieix1((D+i)2+1)x2 \\ =Ieix1(D2+2Di)x2 \\ =Ieix12Di⋅1(1−i2D)x2 \\ =Ieix12D \\ (1−i2D)−1x2=Ieix12Di⋅ \\ (1+i2D−14)x2 \\ =Ieix12Di⋅(x2+ix−12) \\ =Ieix12i∫(x2+ix−12)dx \\ =Ieix12i(x33+ix22−x2) \\ =Ieix(−ix36+x24+ix4) \\ =I(cosx+isinx)(−ix36+$