solve D.E. dy/dx +y/x =y^2sin(x)

Question

solution of question

Answer 1

dy/dx+y/x=y²sin(x).

Let v=1/y, then dv/dx=-(1/y²)(dy/dx), dy/dx=-y²(dv/dx)=-(1/v²)(dv/dx).

The DE becomes:

-(dv/dx)/v²+1/(vx)=sin(x)/v², -(dv/dx)/v+1/x=sin(x)/v.

Multiply through by -v:

dv/dx-v/x=-sin(x).

Multiply through by 1/x:

(1/x)(dv/dx)-v/x²=-sin(x)/x.

(d/dx)(v/x)=-sin(x)/x.

v/x=-∫(sin(x)/x)dx, 1/(xy)=-∫(sin(x)/x)dx.

There is no standard solution for the indefinite integral, but sin(x)=∑(-1)ⁿx²ⁿ⁺¹/(2n+1)! as a series (for integer n≥0).

Therefore, 1/xy=-∫∑(-1)ⁿx²ⁿ/(2n+1)!=∑(-1)ⁿ⁺¹x²ⁿ⁺¹/(2n+1)[(2n+1)!]+C.

y=1/(Cx+∑(-1)ⁿ⁺¹x²⁽ⁿ⁺¹⁾/(2n+1)[(2n+1)!].

I suspect that the original DE should have been dy/dx+y/x=xy²sin(x). If so we get to:

(d/dx)(v/x)=-sin(x). 1/(xy)=cos(x)+C, y=1/(Cx+xcos(x)).