Question

The general solution of the differential equation x^2D^2+4xD+2y = x^2+1/x^2​

Answer 1

Answer:

Given question is

x2d2ydx2+4xdydx+2y=ex

Replacing

d2ydz2−dydz+4dydz+2y=eez

d2ydz2+3dydz+2y=eez

(D2+3D+2)y=eez where D=ddz

(D+1)(D+2)y=eez

To find complementary function

Auxiliary equation is

(D+1)(D+2)=0

D=−1,−2

y=Ae−z+Be−2z

=Ae−ln(x)+Be−2ln(x)

y=Ax+Bx2

To find particular integral

y=1(D+1)(D+2)⋅eez

1D−af(x)=eax∫e−axf(x)dx

So

y=1D+2(1D+1eez)

=1D+2(e−z∫ez⋅eezdz)

To solve integral put u=ez and du=ezdz

Solving

y=1D+2(e−zeez)

=e−2z∫e2z⋅eezdz

put u=ez

du=ezdz

Integral become ∫ueudu=eu(u−1)

Therefore

y=e−2z⋅eez(ez−1)

y=e−2lnxeelnx(elnx−1)

y=1x2ex(x−1)

y=ex(x−1)x2

Complete solution is

y=Ax+Bx2+ex(x−1)x2

y=Ax+B+ex(x−1)x2