Hi all,(adsbygoogle = window.adsbygoogle || []).push({});

I've been just solving this one:

[tex]

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

[/tex]

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

[tex]

z = \sqrt[3]{y}

[/tex]

and thus getting

[tex]

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

[/tex]

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

[tex]

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

[/tex]

To solve this, I first solved the homogenous equation

[tex]

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

[/tex]

and few steps I won't write here I got

[tex]

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

[/tex]

[tex]

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

[/tex]

[tex]

z_1 = C\sqrt[3]{x}

[/tex]

[tex]

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

[/tex]

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

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

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

[tex]

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

[/tex]

[tex]

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

[/tex]

I know that I should actually also write that

[tex]

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

[/tex]

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

[tex]

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

[/tex]

will give all possibilites.

Concerning [itex]Q[/itex], I got

[tex]

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

[/tex]

and so

[tex]

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

[/tex]

[tex]

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

[/tex]

and finally

[tex]

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

[/tex]

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 [itex]y[/itex]

Anyway, can you see where I did a mistake?

Thank you very much!

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# Homework Help: ODE: Can't find mistake

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