Solving Difficult Differential Equations

[ShayanJ]

Could somebody help me to solve this really tough differential equations?

$$\left( {x}{y} \sqrt{{x}^{2} - {y}^{2}}+{x} \right){y'}={y}-{x}^{2} \sqrt{{x}^{2}-{y}^{2}}$$

Another tough one.I tried MATLAB which gave me sth about lambertW function which I haven't heard about before.Is there an easier answer?

$$\left( {x}^{2}-{y} \right){y'}+{x}=0$$

thanks

 

[homayoun]

really it's tough enough to confuse my mind!
somebody answer these 2 questions please
Thanks

 

[ShayanJ]

I didn't think these two are this hard that I don't get the answer after writing this thread.
These are just two exersises of a textbook on O.D.E.s
You don't have even a little tip?

 

[coelho]

Ok... first of all, one should not post textbook exercises in this section. Anyway, i will show how the lambertW funtion appears from the second DE:

$$(x^2-y)y'+x=0$$

Using the substituition:

$$u=x^2-y$$

we get:

$$y=x^2-u$$ and $$y'=2x-u'$$.

So, the DE becomes:

$$u(2x-u')+x=0$$

Isolating u', we get:

$$u'=\frac{(2u+1)x}{u}$$

Writing u' as du/dx and separating the variables:

$$\left[ \frac{u}{2u+1} \right] du = x dx$$

Integrating both sides gives:

$$\int \frac{u du}{2u+1} = \frac{x^2}{2}+c$$

The substituition v=2u+1 is useful, and gives:

$$\int \frac{1}{4} \left[ 1-\frac{1}{v} \right] dv = \frac{x^2}{2}+c$$

Which, integrated, leads to:

$$v - ln(v)=2x^2+4c$$

Ok... now we have to isolate v in this equation. And this is where the lambertW function appears. The http://en.wikipedia.org/wiki/Lambert_W_function" function W(x) is defined as the function that satisfies this equation:

$$x=W(x)e^{W(x)}$$

Taking logarithms, we get:

$$ln(x) = ln[W(x)] + W(x)$$

And this resembles very much the function v (which is actually v(x)) we got when solving the DE. I can't say that v(x)=W(x) because we have the - sign instead the + sign. Anyway, I am pretty sure the function v(x) is related to the lambertW function, as you mentioned that MATLAB gave it as a possible solution.

I didnt try to solve the first equation. But probably there is some substituition like the one i did that can make it a bit easier to solve.