Where did I go wrong in solving this ODE?

[twoflower]

Hi all,

I've been just solving this one:

$$y' + \frac{y}{x} = 3\sqrt[3]{\left(xy\right)^2}\arctan x$$

The problem is, one of the solutions I got doesn't pass the original equation and I can't find the mistake. Here it is:

After substituting

$$z = \sqrt[3]{y}$$

and thus getting

$$3z^2z' + \frac{z^3}{x} = 3\sqrt[3]{x^2}z^2\arctan x$$

dividing with $z^2$ (and so getting the condition for y not to be particular solution $y \equiv 0$)

$$3z' + \frac{z}{x} = 3x^{\frac{2}{3}}\arctan x$$

To solve this, I first solved the homogenous equation

$$3z' + \frac{z}{x} = 0$$

and few steps I won't write here I got

$$\log |z|^3 = \log |x| + C$$

$$|z|^3 = e^{C}|x|$$

$$z_1 = C\sqrt[3]{x}$$

$$z_2 = -C\sqrt[3]{x}$$

Well, I think this is the problematic step although I think it's ok.

To get the proper $C = C(x)$ for the non-homogenous equation I expressed $z$ in terms of

$C(x)$ and put it to the equation. It involved another ODE at the end of which I got

$$\log |C| = \log e^{C}\frac{1}{\sqrt[3]{x^2}}$$

$$C = Q\frac{1}{\sqrt[3]{x^2}}$$

I know that I should actually also write that

$$C_2 = -Q\frac{1}{\sqrt[3]{x^2}}$$

but this won't give me anything new since changing the sign in front of $C$ in expression

$$z = \pm C\sqrt[3]{x}$$

will give all possibilites.

Concerning $Q$, I got

$$Q = \left(\frac{x^2 + 1}{2}\arctan x - \frac{x}{2} + R\right)$$

and so

$$C = \frac{1}{\sqrt[3]{x^2}}\left(\frac{x^2 + 1}{2}\arctan x - \frac{x}{2} + R\right)$$

$$z = \pm\frac{1}{\sqrt[3]{x}}\left(\frac{x^2 + 1}{2}\arctan x - \frac{x}{2} + R\right)$$

and finally

$$y = z^3 = \pm\frac{1}{x}\left(\frac{x^2 + 1}{2}\arctan x - \frac{x}{2} + R\right)$$
Well, with the plus-signed solution, it satisfies the original equation while with the minus sign it doesn't. Which is
something that I think can be seen already from the original ODE since the left side will get negative sign while the right
side doesn't depend on the sign of $y$Anyway, can you see where I did a mistake?Thank you very much!

 

[saltydog]

Hey Twoflower. What up? Me, when I got to:

$$3z^{'}+\frac{z}{x}=3x^{2/3}\text{ArcTan[x]}$$

I'd treat it like a regular first order ODE and solve for the integrating factor. You know:

$$\sigma=x^{1/3}$$

so that:

$$d\left[x^{1/3}z\right]=x\text{Arctan[x]}$$

leaving:

$$x^{1/3}z=-\frac{x}{2}+\frac{1}{2}\text{Arctan[x]}+<br /> \frac{x^2}{2}\text{Arctan[x]}+c$$

 

[twoflower]

Hey Twoflower. What up? Me, when I got to:

$$3z^{'}+\frac{z}{x}=3x^{2/3}\text{ArcTan[x]}$$

I'd treat it like a regular first order ODE and solve for the integrating factor. You know:

$$\sigma=x^{1/3}$$

so that:

$$d\left[x^{1/3}z\right]=x\text{Arctan[x]}$$

leaving:

$$x^{1/3}z=-\frac{x}{2}+\frac{1}{2}\text{Arctan[x]}+<br /> \frac{x^2}{2}\text{Arctan[x]}+c$$

Thank you Saltydog. Your solution is basically the same I got excepting I also have the same solution also with the minus sign.

We had been told integrating factor as an alternative to method of variation of parameters I used here since I like it more.

You know, there has to be some flaw in my approach since the minus-sign solution doesn't satisfy the ODE while the same solution, only with positive sign, does. And I can't see, why should I exclude the minus-sign solution...