## khalissa 2 years ago how to differentiate methods used in solving first-order differential equations?like separable,homogeneous equation,exact,etc.

1. vf321

It's been a while since I took DE, but since separable is, well, separable, then it must ultimately be expressed as$F(y)dy=G(x)dx$which implies that it was originally$\frac{dy}{dx}=\frac{G(x)}{F(y)}$Thus, if you ever see a problem with $$\frac{dy}{dx}$$ on one side and something which can group into two distinct functions for x and y that are all one term on the other, you know it's separable.

2. vf321

A homogenous function $$f(x, y)$$ of degree $$n$$ can be expressed as $f(\alpha x, \alpha y)=\alpha^nf(x, y)$When you encounter a DE, if it can be simplified into the following form:$\frac{dy}{dx}=f(x, y)$And then you check that $$f$$ is homogenous. If it is, use the substitution of $$v=y/x$$ and replace every instance of $$y$$ with $$x v$$. You'll notice the LHS becomes $$x\frac{dv}{dx}$$. The DE will now be separable for v and x.

3. vf321

Also of note, a homogenous LINEAR DE implies that $f(x)\frac{dy}{dx}+g(x)y=0$The solution to this is just a special case of the solution to ALL first-order linear DEs, which can be re-written in the form $\frac{dy}{dx}+f(x)y=g(x)$for different functions $$f$$ and $$g$$. Wikipedia has a splendid, concise, understandable proof for this. I encourage you to go through the steps of the proof whenever you solve these instead of plugging into the equation Wikipedia gives you at the end for y(x). http://en.wikipedia.org/wiki/Linear_differential_equation#First_order_equation

4. khalissa

thanks!it do helps a lot =)

5. vf321

However, there are also exact first-order DEs, which solve the slightly more complicated $a(x, y)y'(x)+b(x, y)=0$which is often written as$Mdy+Ndx=0$where $$M=a(x, y)$$ and $$N=b(x, y)$$. The solution to this is exact if, using subscript notation to denote partials, $$M_x=N_y$$. This is because of Green's Theorem, actually. If you'd like to to find out why, then make a new question regarding this and then type @vf321 so that I can answer it. But given that condition, the solution will be what you get when you find the integral:$\int Mdx\cup\int N dy=c$

6. vf321

Whoops! That equation should be $Ndy+Mdx=0$with $$M=b(x, y)$$ and $$N=a(x, y)$$. The integral is right, however.