First-order nonlinear ODE (approach is not clear to me)

**Rafael Almeida** · October 18th 2010, 06:58 PM

I've been given the following ODE to be solved:

$(yx^2)y' + (2y^2x + 3x^3) = 0$

But I don't really know how to start. A direct integration is (obviously enough) not possible and it's not in the form to be solved by an integrating factor (or so I understand).

Then it could be solved by separation of variables (though I can't see what the separation itself would be), or by using the exact equation technique (though I'm not even sure how to check if it's in the appropriate form).

How do I get started?

**Prove It** · October 18th 2010, 07:15 PM

At the moment this is unsolvable exactly.

Are you sure you haven't got a "square" in the wrong spot. If it was

$(yx^2)\frac{dy}{dx} + (2yx^2 + 3x^3) = 0$ this would be separable...

**Krizalid** · October 18th 2010, 07:26 PM

Originally Posted by Rafael Almeida

I've been given the following ODE to be solved:

$(yx^2)y' + (2y^2x + 3x^3) = 0$

$\displaystyle\frac{dy}{dx}=-\frac{2{{y}^{2}}x+3{{x}^{3}}}{{{x}^{2}}y}=-2\cdot \frac{y}{x}-3\cdot \frac{x}{y}.$

homogeneous ODE.

**Rafael Almeida** · October 18th 2010, 07:29 PM

I just double, no, triple-checked it here, and the square is indeed at the "y", as I posted it. What might have happened is that the professor mistyped it when writing the exam (I'm doing a post-mortem), but that would be at least unexpected since I don't remember anybody complaining about it during the exam nor did he make any corrections.

So, the equation as I originally posted is not solvable at all? What does "unsolvable exactly" mean? Because, according to Wolfram Alpha (solve - Wolfram|Alpha[(y+x^2)dy/dx+%2B+2y^2+x+%2B+3x^3+%3D%3D+0]), it does give me a solution.

PS: I forgot to mention, the original problem says to solve the IVP given by the equation I posted and the condition $y(2) = 2$

**Ackbeet** · October 19th 2010, 02:38 AM

The LHS is almost a product rule. Try this:

$yy'x^{2}+2y^{2}x+3x^{3}=0,\;\quad y(2)=2$ is the original DE.

Now $(x^{2}y^{2})'=2xy^{2}+2x^{2}yy'.$ Hence,

$2yy'x^{2}+2y^{2}x+3x^{3}=yy'x^{2},\;\quad y(2)=2.$

Let $u=x^{2}y^{2}.$ Then

$u'=2xy^{2}+2x^{2}yy'=2\dfrac{u}{x}+2x^{2}yy'$ , from which we may obtain

$\dfrac{u'}{2}-\dfrac{u}{x}=x^{2}yy'.$ Substitute this into the DE as we re-wrote it, and you obtain the following:

$u'+3x^{3}=\dfrac{u'}{2}-\dfrac{u}{x},\;\quad u(2)=16.$

This equation is first-order linear.

**Danny** · October 19th 2010, 05:01 AM

Another approach. For $(yx^2)y' + (2y^2x + 3x^3) = 0$ try letting $u = y^2$ . It should make it linear in $u$ .

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