Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Solve $(2x^2 + y^2)\,dx - xy \, dy = 0$

Attempted :

The equation is not exact because $ M_y \ne N_x $ for $ M = 2x^2 + y^2 $ and $ N = xy$

Or is it exact?

The equation is also not separable.

The equation is also not homogenous, I don't think.

So.. what do I do?

Thanks.

share|improve this question
    
It's not exact. But, multiply the equation by $x^{-3}$. You will then have an exact equation. See here for the method used to find the integrating factor $x^{-3}$. –  David Mitra Feb 17 '13 at 21:53

2 Answers 2

up vote 2 down vote accepted

Hint: Divide by $xy$ and put $u = y/x$. The equation will then become separable.


To elaborate, divide by $xy$ to get:

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

Put $u = y/x$, $y' = xu' + u$:

$$ xu' + u = \frac{2}{u} + u $$

Rearrange to get:

$$ uu' = \frac{2}{x} $$

Integrate both sides, solve for $u$ and put back $u = y/x$ to get the solution in terms of $y$.

share|improve this answer
    
Now I have,$$y' = \frac{2x^2 + y^2}{xy} $$ $$ yy' = 2x + \frac{y^2}{x}$$ –  40Plot Feb 17 '13 at 21:51
    
@40Plot You're on the write track. See my edit. –  Ayman Hourieh Feb 17 '13 at 21:54
    
In most problem (that I've seen so far) that are homogeneous, you set $ u =\frac{y}{x}$ , if I set $ u =\frac{x}{y}$ would that be a problem or would it just give you the same thing? –  40Plot Feb 17 '13 at 22:05
    
@40Plot $u = x/y$ would work if you differentiate it with respect to $y$ and replace $x$, $x'$ in the original equation instead. I use $u = y/x$ in my solution as is commonly done. –  Ayman Hourieh Feb 17 '13 at 22:11

An alternative approach showing how to arrive at the right integrating factor. Split the terms in two groups as follows

$$2x^2dx+(y^2dx-xydy)=0$$

Now for $2x^2dx$ the integrating factor is trivial $1$ leading to solution $x^3=C$. Then $\mu_1=\phi(x^3)$ where $\phi$ is an arbitrary function will be the most general integrating factor for this part.

For the second part it is easy to see that $\frac{1}{xy^2}$ separates variables giving solution $\frac{x}{y}=C$. Therefore, the most general integrating factor will be $\mu_2=\frac{1}{xy^2}\psi\left(\frac{x}{y}\right)$.

Now we want to make $\mu_1\equiv\mu_2$. Set $\psi(t)=\frac{1}{t^2}$ to make $\mu_2$ independent of $x$. Hence, $\mu_2=\frac{1}{x^3}$. This implies $\phi(t)=\frac{1}{t}$. So $\mu_1\equiv\mu_2=\frac{1}{x^3}$

$$2\frac{dx}{x}+\left(\frac{y}{x}\right)^2d\left(\frac{x}{y}\right)=0$$ $$2\frac{dx}{x}+\frac{d\left(\frac{x}{y}\right)}{\left(\frac{x}{y}\right)^2}=0$$ $$d\left[\ln\left(x^2\right)\right]-d\left[\left(\frac{x}{y}\right)^{-1}\right]=0$$ $$d\left[\ln\left(x^2\right)-\left(\frac{y}{x}\right)\right]=0$$ $$x^2=Ce^{\frac{y}{x}}$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.