# Solving a differential equation

1. Feb 2, 2015

### MarcL

1. The problem statement, all variables and given/known data
Solve (xy+y2+x2) dx -( x2 )dy = 0

2. Relevant equations

3. The attempt at a solution
So I can see it isn't separable and linear, so I thought of solving it through substitution

i did y=ux and dy= u dx + x du
and I substituted them in my first equation giving me
(xu+u2+x^2) dx - x2(udx + xdu) = 0]

I'm stuck here... the answer key factors out x^2 from the first equation leaving it with x^2(u+u^2+1), which doesn't make sense to me.. because if I factored out x^2 I would be left with (x-1u+x-2u2+1)
So any help on how I am seeing this wrong?

Last edited by a moderator: May 7, 2017
2. Feb 2, 2015

### ShayanJ

$\partial_y M=\partial_y(xy+y^2+x^2)=x+2y$
$\partial_x N=\partial_x(-x^2)=-2x$
$\partial_y M \neq \partial_x N!!!$

3. Feb 2, 2015

### MarcL

I know how to verify if exact, hence why I chose substitution, but I am kinda stuck at how to factor my xα. Unless I didn't understand your answer correctly, to me it seems as if you're solving to see whether or not if exact.

4. Feb 2, 2015

### ShayanJ

That means your differential isn't exact which means it can't be integrated in this form, so you should turn it into an exact differential. There is such a method in your toolbox which you can't integrate the differential without it. You remember it?

5. Feb 2, 2015

### MarcL

Well I know I can use substitution to reach a separable equation. If not, I know I can use υ(x,y) and multiply it to my equation. My question was more towards the substitution method because I don't completely grasp the subject.

6. Feb 2, 2015

### ShayanJ

By some manipulation, you can get $y'=\frac{y}{x}+\frac{y^2}{x^2}+1$. Now try your substitution!

7. Feb 2, 2015

### MarcL

I'm sorry, I'm not understanding. why you took the derivative of y. I was taught ( and my book explains) a different way. For instance, we just have to find a coeffiecient of xα to then create a separable DE. anyway, I'll check again later, maybe this will help getting a clearer answer? :/

8. Feb 2, 2015

### ShayanJ

$M dx+N dy=0 \Rightarrow N dy=-M dx \Rightarrow \frac{dy}{dx}=- \frac{M}{N}$

9. Feb 2, 2015

### Ray Vickson

If you substitute $y = x u$ into the DE
$$x^2 \frac{dy}{dx} = x^2 + xy + y^2 \; \Rightarrow \frac{dy}{dx} = 1 + \frac{y}{x} + \left(\frac{y}{x}\right)^2$$
the resulting DE for $u$ is very simple and is straightforward to solve.

Last edited by a moderator: May 7, 2017
10. Feb 2, 2015

### MarcL

Yeah it was a stupid mistake, I was going fast and stressing out over an exam. I should've noticed that replacing y = ux into the equation can allow me to factor out x^2 because the function is homogeneous.