Answer to Question #199568 in Differential Equations for rajkumar

Question

(D-1)²(D²+1)² y= sin ²(x/2)+e^x+x

Answer 1

$y(x)=y_c+y_p$

General solution:

$y_c=c_1e^x+c_2xe^x+c_3cosx+c_4sinx+x(c_5cosx+c_6sinx)$

For particular solution:

$sin^2(x/2)=(1-cosx)/2$

Then:

Particular solution:

$PI=\frac{1}{f(D)}(sin^ 2(x/2)+e^x+x)$

$PI_1=\frac{1}{f(D)}e^x=\frac{x^2e^x}{2!(1+1)^2}=\frac{x^2e^x}{8}$

$PI_2=\frac{1}{f(D)}(-\frac{1}{2}cosx)=-\frac{1}{2}(\frac{1}{-2D}cosx+\frac{x}{4}\int xsinxdx)=$

$=\frac{sinx}{4}-\frac{x(sinx-xcosx)}{4}$

$PI_3=\frac{1}{f(D)}(x+1/2)$

$(D-1)^{-2}=1+2D+...$

$(D^2+1)^{-2}=1-2D+...$

$PI_3=(1-2D)(x+2.5)=x+2.5-2=x+1/2$

$y_p=PI_1+PI_2+PI_3$