Solving a differential equation

[MarcL]

Homework Statement

Solve (xy+y²+x²) dx -( x² )dy = 0

Homework Equations

to verify if exact [PLAIN]http://upload.wikimedia.org/math/5/2/c/52cc749bb1c32abf1dccf613bd847a6e.pngM/[B][PLAIN]http://upload.wikimedia.org/math/5/2/c/52cc749bb1c32abf1dccf613bd847a6e.pngy = [PLAIN]http://upload.wikimedia.org/math/5/2/c/52cc749bb1c32abf1dccf613bd847a6e.pngN/[B][PLAIN]http://upload.wikimedia.org/math/5/2/c/52cc749bb1c32abf1dccf613bd847a6e.pngx[/B][/B][/B][/B]

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?

 

[ShayanJ]

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

 

[MarcL]

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

 

[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?

 

[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.

 

[ShayanJ]

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

 

[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 x^α to then create a separable DE. anyway, I'll check again later, maybe this will help getting a clearer answer? :/

 

[ShayanJ]

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

 

[Ray Vickson]

Homework Statement

Solve (xy+y²+x²) dx -( x² )dy = 0

Homework Equations

to verify if exact [PLAIN]http://upload.wikimedia.org/math/5/2/c/52cc749bb1c32abf1dccf613bd847a6e.pngM/[B][PLAIN]http://upload.wikimedia.org/math/5/2/c/52cc749bb1c32abf1dccf613bd847a6e.pngy = [PLAIN]http://upload.wikimedia.org/math/5/2/c/52cc749bb1c32abf1dccf613bd847a6e.pngN/[B][PLAIN]http://upload.wikimedia.org/math/5/2/c/52cc749bb1c32abf1dccf613bd847a6e.pngx[/B][/B][/B][/B]

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?

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.

 

[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.