Answer to Question #304879 in Differential Equations for adhi

Question #304879

y′′+ 4y= sin 5x+x−1

1
Expert's answer
2022-03-02T16:58:24-0500

The characteristic equation k2+4=0k^2+4=0 has roots k1=2ik_1=2i and k2=2i.k_2=-2i. Therefore, the general solution of the differential equation is of the form:

y=C1cos2x+C2sin2x+yp,y=C_1\cos 2x+C_2\sin 2x+y_p,

where yp=acos5x+bsin5x+cx+d.y_p=a\cos 5x+b\sin 5x+cx+d. It follows that

yp=5asin5x+5bcos5x+c,yp=25acos5x25bsin5xy_p'=-5a\sin 5x+5b\cos 5x+c, y_p''=-25a\cos 5x-25b\sin 5x

Then

25acos5x25bsin5x+4(acos5x+bsin5x+cx+d)=sin5x+x1.-25a\cos 5x-25b\sin 5x+4(a\cos 5x+b\sin 5x+cx+d)=\sin 5x +x-1.

We get the following equation

21acos5x21bsin5x+4cx+4d=sin5x+x1,-21a\cos 5x-21b\sin 5x+4cx+4d=\sin 5x +x-1,

and hence

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

It follows that a=0, b=121, c=14, d=14.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=C1cos2x+C2sin2x121sin5x+14x14.y=C_1\cos 2x+C_2\sin 2x-\frac{1}{21}\sin 5x+\frac{1}4x-\frac{1}4.




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