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.

 

[blue_raver22]

I understand the frustration that comes with solving difficult differential equations. These types of equations can be challenging and may require advanced mathematical techniques. I would recommend seeking help from a mathematician or a colleague who specializes in solving differential equations. They may be able to provide insights or offer alternative methods for solving these equations. Additionally, there are numerous resources available online, such as forums and tutorials, that can provide step-by-step guidance on solving differential equations. It is also worth considering using specialized software or programming languages, such as MATLAB, to tackle these equations. However, it is important to note that some equations may not have an easy or straightforward solution, and it may require a combination of approaches to solve them. Don't give up and keep exploring different methods until you find a solution.