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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?


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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
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 $y$.

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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


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 $\mu_1\equiv\mu_2=\frac{1}{x^3}$

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

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