Where did I go wrong in solving this ODE?

[twoflower]

Hi all,

I've been just solving this one:

<br /> y&#039; + \frac{y}{x} = 3\sqrt[3]{\left(xy\right)^2}\arctan x<br />

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

<br /> z = \sqrt[3]{y}<br />

and thus getting

<br /> 3z^2z&#039; + \frac{z^3}{x} = 3\sqrt[3]{x^2}z^2\arctan x<br />

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

<br /> 3z&#039; + \frac{z}{x} = 3x^{\frac{2}{3}}\arctan x<br />

To solve this, I first solved the homogenous equation

<br /> 3z&#039; + \frac{z}{x} = 0<br />

and few steps I won't write here I got

<br /> \log |z|^3 = \log |x| + C<br />

<br /> |z|^3 = e^{C}|x|<br />

<br /> z_1 = C\sqrt[3]{x}<br />

<br /> z_2 = -C\sqrt[3]{x}<br />

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

<br /> \log |C| = \log e^{C}\frac{1}{\sqrt[3]{x^2}}<br />

<br /> C = Q\frac{1}{\sqrt[3]{x^2}}<br />

I know that I should actually also write that

<br /> C_2 = -Q\frac{1}{\sqrt[3]{x^2}}<br />

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

<br /> z = \pm C\sqrt[3]{x}<br />

will give all possibilites.

Concerning Q, I got

<br /> Q = \left(\frac{x^2 + 1}{2}\arctan x - \frac{x}{2} + R\right)<br />

and so

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

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

and finally

<br /> y = z^3 = \pm\frac{1}{x}\left(\frac{x^2 + 1}{2}\arctan x - \frac{x}{2} + R\right)<br />
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 yAnyway, can you see where I did a mistake?Thank you very much!

 

[saltydog]

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

3z^{&#039;}+\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^{&#039;}+\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...