# Answer in Calculus for Kaustav Borah #205033

(D − 1)

2

(D

2 + 1)

2y = sin2

(

x

2

) + e

x + x

Answer:-

“(D-1)^2(D^2+1)^2y=sin^2(x/2)+e^x+x”

substitute “y=e^{kx}”

then

“Dy-frac{dy}{dx}-ke^{kx}”

“D^2y=frac{d^2y}{dx^2}=k^2e^{kx}”

“D^4y=frac{d^4y}{dx^4}=k^4e^{kx}”

“(y”-2y’+y)(y^{4}+2y”+y)=0\n(k^2e^{kx}-2ke^{kx}+e^{kx})(k^4e^{kx}+2k^2e^{kx}+e^{kx})=0\ne^{kx}(k^2-2k+1)(k^4+2k^2+1)=0\nk=1;k=i;k=-i”

So, we have three fundamental solutions:

“e^x;e^{ix};e^{-ix}”

Each of those three fundamental solutions satisfies the homogeneous equation and also any linear combination of those. Each of roots is double and to obtain three more fundamental solutions we need to multiply the corresponding fundamental solution by :

“xe^x;xsinx;xcosx\”

So the general solution to the homogeneous equation is

“Y=Ae^x+Bxe^x+Csinx+Dxsinx+Ecosx+Fxcosx”

where , , , , , are arbitrary constants.

Since

“sin^2frac{x}{2}=frac{1}{2}-frac{cosx}{2}”

“(y”-2y’+y)(y^{(4)}+2y”+y)=frac{1}{2}-frac{cosx}{2}+e^x+x”

To get ^{x} a particular solution

“widetilde{y_1}=ax^2e^x”

now do second, third , fourth order differentiation of “widetilde{y_1}” we get

“a=frac{1}{32}\n widetilde{y_1}=frac{x^2e^x}{32}”

similarly

to get “(-frac{cosx}{2})” a particular solution

“widetilde{y_2}=bx^2cosx”

now do second, third , fourth order differentiation of “widetilde{y_2}” we get

“b=0\n widetilde{y_2}=0”

to get “(x+frac{1}{2})” a particular solution

“widetilde{y_3}=cx+d”

“widetilde{y_3}’=c\n widetilde{y_3}”= widetilde{y_3}^{(4)}=0″

we get “(-2c+cx+d)(cx+d)=x+frac{1}{2}”

c=0

“widetilde{y_3}=0”

“widetilde{y}= widetilde{y_1}+ widetilde{y_2}+ widetilde{y_3}=frac{x^2e^x}{32}”

answer

“boxed{y=Y+ widetilde{y}=Ae^x+Bxe^x+Csinx+Dxsinx+Ecosx+Fxcosx+frac{x^2e^x}{32}}”