Answer to Question #205033 in Calculus for Kaustav Borah

Question

(D − 1)

2

(D

2 + 1)

2y = sin2

(

x

2

) + e

x + x

Answer 1

Answer:-

$(D-1)^2(D^2+1)^2y=sin^2(x/2)+e^x+x$

substitute $y=e^{kx}$

then

$Dy-\frac{dy}{dx}-ke^{kx}$

$D^2y=\frac{d^2y}{dx^2}=k^2e^{kx}$

$D^4y=\frac{d^4y}{dx^4}=k^4e^{kx}$

$(y''-2y'+y)(y^{4}+2y''+y)=0\\ (k^2e^{kx}-2ke^{kx}+e^{kx})(k^4e^{kx}+2k^2e^{kx}+e^{kx})=0\\ e^{kx}(k^2-2k+1)(k^4+2k^2+1)=0\\ k=1;k=i;k=-i$

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 𝑘 is double and to obtain three more fundamental solutions we need to multiply the corresponding fundamental solution by 𝑥 :

$xe^x;xsinx;xcosx\\$

So the general solution to the homogeneous equation is

$Y=Ae^x+Bxe^x+Csinx+Dxsinx+Ecosx+Fxcosx$

where 𝐴, 𝐵, 𝐶,𝐷, 𝐸, 𝐹 are arbitrary constants.

Since 

$sin^2\frac{x}{2}=\frac{1}{2}-\frac{cosx}{2}$

$(y''-2y'+y)(y^{(4)}+2y''+y)=\frac{1}{2}-\frac{cosx}{2}+e^x+x$

To get 𝑒^x a particular solution

$\widetilde{y_1}=ax^2e^x$

now do second, third , fourth order differentiation of $\widetilde{y_1}$ we get

$a=\frac{1}{32}\\ \widetilde{y_1}=\frac{x^2e^x}{32}$

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

$b=0\\ \widetilde{y_2}=0$

to get $(x+\frac{1}{2})$ a particular solution

$\widetilde{y_3}=cx+d$

$\widetilde{y_3}'=c\\ \widetilde{y_3}''= \widetilde{y_3}^{(4)}=0$

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

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