Answer to Question #137294 in Differential Equations for Nikhil

Question

[(D-1)^2.(D^2+1)^2]y=sin^2x/2+e^x

Answer 1

The Auxiliary equation is given by,

$[(m-1)^2(m+1)^2]=0$

The roots of the above equation are,

1,1,-1,-1

Therefore The complimentary function,

C.F.=$(c_1+c_2x)e^x+(c_3+c_4x)e^{-x}$ ......(1)

Particular Integral, P.I.= $\frac{sin^2x}{2(D^2-1)^2}+\frac{e^x}{(D^2-1)^2}$

= $\frac{sin^2x}{2(-1-1)^2}+\frac{xe^x}{2(D^2-1)\times 2D}$

=$\frac{sin^2x}{2(-2)^2}+\frac{x^2e^x}{4(3D^2-1)}$

=$\frac{sin^2x}{8}+\frac{e^x}{4(3.1-1)}$

=$\frac{sin^2x}{8}+\frac{e^x}{8}$

So complete solution=C.F+P.I.

y =$(c_1+c_2x)e^x+(c_3+c_4x)e^{-x}+\frac{sin^2x+e^x}{8}$