Answer to Question #304879 in Differential Equations for adhi

Question

y′′+ 4y= sin 5x+x−1

Answer 1

The characteristic equation $k^2+4=0$ has roots $k_1=2i$ and $k_2=-2i.$ Therefore, the general solution of the differential equation is of the form:

$y=C_1\cos 2x+C_2\sin 2x+y_p,$

where $y_p=a\cos 5x+b\sin 5x+cx+d.$ It follows that

$y_p'=-5a\sin 5x+5b\cos 5x+c, y_p''=-25a\cos 5x-25b\sin 5x$

Then

$-25a\cos 5x-25b\sin 5x+4(a\cos 5x+b\sin 5x+cx+d)=\sin 5x +x-1.$

We get the following equation

$-21a\cos 5x-21b\sin 5x+4cx+4d=\sin 5x +x-1,$

and hence

$-21a=0,\ -21b=1,\ 4c=1,\ 4d=-1.$

It follows that $a=0,\ b=-\frac{1}{21},\ c=\frac{1}4,\ d=-\frac{1}4.$

We conclude that the general solution of the differential equation is of the form

$y=C_1\cos 2x+C_2\sin 2x-\frac{1}{21}\sin 5x+\frac{1}4x-\frac{1}4.$