Question

Help with calc please?

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

Answer 1

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!