Differential equation x^2*y'=y^2

1. Apr 22, 2014

Mr. Fest

Hello,

I have trouble solving the following differential equation.

I am trying to learn how to solve that form of DEs.

The DE is:

x2*dy/dx = y2

There are no initial-value problem, but the solution should be given such that y is defined for all x.

The most important for me is to learn how to solve this type of DE.

I tried solving it using the separable equation way.

x2*dy/dx = y2 --> dy/y2 = dx/x2 and then integrate both sides to get -1/y = -1/x + C --> -y = -x + 1/C --> y = x - 1/C but apparently this doesn't solve the equation...

Hope you guys can help me.

Sincerely,
Mr. Fest

2. Apr 22, 2014

hilbert2

How do you solve the equation $-1/y=-1/x+C$ for $y$? The answer is NOT $-y=-x+1/C$ !

3. Apr 22, 2014

Mr. Fest

Hi,

You are right. The time was around 2am over here when I wrote that. My brain must have been fried after 10 hours of math work.

The answer is ofc:

-1/y = -1/x + C --> -1 = y(-1/x + C) --> -1/(-1/x + C) = y --> -1/((-1+Cx)/x) --> -x/(-1 + Cx) = y

And, the specific solution, that, y is defined for all real x gives us that C = 0, for if not, then for some x giving Cx = 1, y would not be defined.

So, one could say that the initial-value problem is that y must be defined for all real x giving rise to C = 0 and ultimately y = -x/-1 --> y = x

This satisfies: x^2*dy/dx = y^2 as this means that x^2*1 = x^2 and also, y is defined for all real x.

If there is anything wrong with my answer, please correct me.

4. Apr 25, 2014

vanhees71

Your implicit solution is correct, i.e., we indeed have
$$\frac{1}{y}=\frac{1}{x}-C=\frac{1-C x}{x}.$$
This gives
$$y(x)=\frac{x}{1-C x}.$$
Now it's easy to check that this indeed satisfies the given differential equation for any value of $C$.