Nonlinear Differential Equation solving help please

[thejakeisalie]

Homework Statement

Consider the following differential equation:
[itex]x^{2}\frac{dy}{dx}=x^{2}-xy+y^{2}[/itex]
State whether this equation is linear or nonlinear and find all it's solutions

Homework Equations

I think that the Bernoulli differential equation is relevant, but I'm not sure:
[itex]y'+P(x)y=Q(x)y^{n}[/itex]

The Attempt at a Solution

Ok, so this is really a past exam question, and I've been struggling to remember the method & can't find it anywhere in my notes.

First, I tried to rearrange into something similar to the Bernoulli equation, so I could solve using the method from the wikipedia article (wikipedia dot org slash)wiki/Bernoulli_differential_equation.

The rearrangement I got is

[itex]\frac{y'}{y^{2}}-\frac{1}{y^{2}}+\frac{1}{xy}=\frac{1}{x^{2}}[/itex]

and then I'd use the substitution w=1/y, w'=(-1/y^2)y' but I'm not sure what to do about that pesky -1/y² in the middle. If someone could point me towards the right method to sovle this, I'd be very greatful.

N.B. The course I'm on doesn't cover nonlinear DE's very much, and the only ones that we do have exact solutions. So this should have an exact solution.

 

[Dustinsfl]

Edit. Yes you can use Bernoulli's Equation since you have the y^2

$$
y'+\frac{1}{x}y=1+\frac{y^2}{x^2}
$$

 

[DeIdeal]

Hmm...

I might be neglecting something that makes this more difficult, but wouldn't a basic "z(x)=y/x" substitution work?

I mean, if you divide the original equation with x², you get

[itex]\frac{dy}{dx} = 1 - \frac{y}{x} + \frac{y^2}{x^2}[/itex]

I haven't studied DEs that much and have never had to use anything fancy like the Bernoulli differential equation, but from what I can tell, z=y/x seems obvious here. The DE that follows after the substitution is easy to solve, unless I made a mistake somewhere.

 

[thejakeisalie]

Edit. Yes you can use Bernoulli's Equation since you have the y^2

$$
y'+\frac{1}{x}y=1+\frac{y^2}{x^2}
$$

That's what I thought, but dividing

[itex]1+\frac{y^{2}}{x^{2}}[/itex]

by y² doesn't yield something of the form Q(x), it leaves

[itex]\frac{1}{y^{2}}+\frac{1}{x^{2}}[/itex]

Unless I'm approaching that method entirely wrong, which I may well be, I don't see how I can use the [itex]w'=\frac{-1}{y^{2}}y'[/itex] substitution to then solve it.

 

[Dustinsfl]

I am sorry you have a Riccatis Equation.

Take a look at how to solve this equation and then give it a try.