Help with calc please?

Use partial fractions to solve the separable equation dy/dx=-2y/x^2-1

1 Answer
Jul 4, 2018

I’ll interpret the equation as #dy/dx = -(2y)/x^2 -1#

Explanation:

Let’s start with a small rearrangement.

#dy/dx + (2y)/x^2 =-1#

We now can see the equation is in the form #dy/dx + f(x)y = g(x)#, a linear differential equation. The first step here is to find the integrating factor which will always be given by #e^(intf(x) dx)#.

It follows that #I= e^(int 2/x^2dx) = e^(-2/x)#

Multiply both sides of the equation by #e^(-2/x)#

#dy/dxe^(-2/x) + e^(-2/x)(2y)/x^2 = -e^(-2/x)#

#(ye^(-2/x))’ = e^(-2/x)#

#ye^(-2/x) = int e^(-2/x) dx + C#

However, #int e^(-2/x) dx# can’t be integrated in terms of elementary functions so it is completely allowable for us to leave it as this in the final answer.

#y = (int e^(-2/x)dx+ C)/e^(-2/x)#

Hopefully this helps!