Linear Differential equation problem

[erisedk]

Homework Statement

Solution of the differential equation
$(\cos x )dy = y (\sin x - y) dx , 0 < x < \dfrac{\pi}{2}$ is

Homework Equations

The Attempt at a Solution

Only separation of variables, homogenous and linear DEs are in the syllabus, therefore it must be one of those. It obviously isn't the former two, so it must be a linear DE. I have no idea how to convert this into a linear form, especially because of the $y^2$ term. Please help.

 

[Mark44]

Homework Statement

Solution of the differential equation
$(\cos x )dy = y (\sin x - y) dx , 0 < x < \dfrac{\pi}{2}$ is

Homework Equations

The Attempt at a Solution

Only separation of variables, homogenous and linear DEs are in the syllabus, therefore it must be one of those. It obviously isn't the former two, so it must be a linear DE. I have no idea how to convert this into a linear form, especially because of the $y^2$ term. Please help.

It's not linear, due to the y² term. It might be homogeneous, which is an ambiguous term that can mean two different kinds of diff. equations.

 

[erisedk]

I'm not sure what the other homogenous is, I'm talking about the one in which we get y/x terms, and substitute that as V.
Okay, so how do I proceed? It probably involves substitutions, but I'm not sure what to substitute.

 

[vela]

You can't express y' as a function of y/x, so the differential equation isn't homogeneous. You need a different approach.

You can rearrange the terms slightly to get $(y\sin x)dx - \cos x\,dy = y^2\,dx$. Why would you want to do this? It's because the lefthand side is exact—that is, if $v = -y\cos x$, you have
$$dv = \frac{\partial v}{\partial x}\,dx + \frac{\partial v}{\partial y}\,dy = (y\sin x)dx - \cos x\,dy.$$ So try out the substitution $v = -y\cos x$ and see what you get after you rewrite the original differential equation in terms of $v$ and $x$.