Question

Separable Differential Equation with Initial Value?

`x^2(dy/dx)=y-xy`

Initial condition : y(-1)=-1

This is what I separated it to

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

After integration, I have:

`ln(y)=(-xln(x)-1)/(x)`

I know that I will have to introduce "e," but I am getting tripped up here.

Any help would be appreciated.

Thanks!

Answer 1

`|y| = (e^(-1/x-1))/(|x|)`

Explanation:

We have:

`x^2 \ dy/dx = y-xy` with Initial condition `y(-1)=-1`

Which we can write as:

`x^2 \ dy/dx = y(1-x) => 1/y \ dy/dx = (1-x)/x^2`

So, as indicated, the ODE is separable, so we can "separate the variables" to get:

`int \ 1/y \ dy = int \ 1/x^2 - 1/x \ dx`

Both integrals are of standard functions, so we can directly integrate, to get:

`ln |y| = -1/x - ln |x| + A`

Using the initial condition `y(-1)=-1` we have:

`ln |-1| = -1/(-1) - ln |-1| + A => A = -1`

Thus we have:

`ln |y| = -1/x - ln |x| + 1`

Noting that we can write `lne=1` we get:

`ln |y| = -1/x - lne - ln |x|`

`:. ln |y| + ln |x| + lne = -1/x`

`:. ln |xye| = -1/x`

`:. |xye| = e^(-1/x)`

`:. |y| = (e^(-1/x))/(e|x|)`

`:. |y| = (e^(-1/x-1))/(|x|)`