Question

Solve (d^2-2D=1)y=x^2+1+sin2x

Answer 1

Answer:

kindly tell me

by which method?

Answer 2

Given:

(D²-2D+1)y = x²+1+sin(2x)

To find:

We are asked to find the general solution of given equation

Solution:

The given equation is of the form $f(D)y=Q(x)$

General solution of $f(D)y=Q(x)$  is y=C.F(complimentary function) + P.I(particular integral)

Step-1: Calculating C.F

From the given equation,$f(D) = D^{2}-2D+1$

So Auxilary equation is ,$f(m) = m^{2}-2m+1 = 0$

Solving above auxilary equation we get  $m=1,1$

Here roots of auxilary equation are real and equal

C.F = $(c_1+c_{2}x)e^{x}$

Step-2: Calculating P.I

For the purpose of simplicity let us calculate P.I individually i.e.,

P.I = (P.I)₁+(P.I)₂+(P.I)₃

In general P.I = $\frac{1}{f(D)} (Q(x))$ = $\frac{1}{D^{2}-2D+1 }(x^{2} +1+sin2x)$

Now  (P.I)₁ = $\frac{1}{D^{2}-2D+1 }x^{2}$

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

                 = $[1+(D^{2}-2D) ]^{-1} x^{2}$

                 = $[1-D^{2}+2D+ D^{4} +4D^{2} -4D^{3}..............]x^{2}$

        (P.I)₁  = $x^{2} -2+2(2x)+0+4(2)-4(0)$

         (P.I)₁ = $x^{2} +4x+6$

Now (P.I)₂ =  $\frac{1}{D^{2}-2D+1 }(1)$

                =  $\frac{1}{D^{2}-2D+1 }(e^{0x} )$

                =   $\frac{1}{0-0+1 }(e^{0x} )$

        (P.I)₂ = 1

Now (P.I)₃ =  $\frac{1}{D^{2}-2D+1 }(sin2x)$

here substitute D²= -b²  where b= coefficient of x in  sin function  

so put D²= -4

         (P.I)₃ =  $\frac{1}{-4-2D+1 }(sin2x)$

                  =  $\frac{1}{-3-2D }(sin2x)$

                  =  $\frac{-3+2D}{(-3-2D)(-3+2D)}(sin2x)$

                  =   $\frac{-3+2D}{(9+16)}(sin2x)$

                  =  $\frac{-3+2D}{25}(sin2x)$

                  = $\frac{1}{25}[-3sin2x + 2Dsin2x]$

        (P.I)₃  = $\frac{1}{25}[-3sin2x +4cos2x]$

Now P.I = $x^{2} +4x+7+\frac{1}{25}[4cos2x-3sin2x]$

The general solution isy = C.F+P.I

                                        y=  $(c_1+c_{2}x)e^{x}+ x^{2} +4x+7+\frac{1}{25}[4cos2x-3sin2x]$