How Do I Solve This Differential Equation Involving y/x?

[boris90]

Hi all,

this is my first post here. I just registered today. I have a little problem: solving differential equations generally. Our professor at my college sent us some math problems for practice, because we're having a final exam at the end of January. Anyway, the first problem is solving a differential equation and I can't seem to remember how to solve it (step by step). I know the first one should be putting all y's on one side and all x's on the other side of the equation, then we should do the integration and so on.

My problem looks like this: y' = $$\frac{y}{x}$$ + ($$\frac{y}{x}$$)² (whole fraction y/x is squared)

How do I solve this problem? I keep getting to a point where it is obvious I'm going the wrong way. If anyone can help me, that person would make me happy, because I've been struggling with this all day now, and still haven't found anything similar that could help me.

Thanks!

 

[RadioactivMan]

I do not know too much about this but it looks like a Bernoulli differential equation
http://en.wikipedia.org/wiki/Bernoulli_differential_equation
p(x)=-1/x
q(x)=1/x^2

I hope this helps.

 

[boris90]

Thanks ;). I would like to see some step by step help, too. Wouldn't hurt ;)

 

[AlephZero]

The terms in (y/x) suggest the substitution

y = vx
y' = v + v'x

That gives an equation in v and x which is separable.

 

[RadioactivMan]

using the sustitution z=y^(-1) the equation get the form of
z`+ (1-2)(-1/x)z=(1-2)(1/x^2)
and you can solve

As AlephZero said:

u=y/x them y=ux them
y`= u + xu` them the equation get the form of
u+xu`=g(u) them is
x(du/dx)= g(u) - u
du(g(u)-u)=dx/x
int(du(g(u)-u)=int(dx/x)for your equation would be
u=y/x and y`=u+xu`
them the equation get the form
xu`=u^2
u`/u^2 = 1/x
them
du/u^2 = dx/x
-1/u = lnx + C
u(x) = -1/[lnx + C]
them
y(x)= - x/[lnx +C]

PD: I need to learn to write in latex.