IVP, what am I doing wrong?

**matt3D** · February 9th 2010, 06:31 PM

Hi, I am trying to solve a differential ivp and I can't seem to get the correct answer. The homogeneous differential equation is: $3x^2y+7y^3-3x^3y'=0, y(1)=1$ Here is my work:

Thanks for your time!
-Matt

**Random Variable** · February 9th 2010, 07:43 PM

It looks like you just made one silly error.

$y' = \frac{y}{x} + \frac{7}{3} \frac{y^{3}}{x^{3}}$

**matt3D** · February 9th 2010, 08:37 PM

Ah, thank you! But I am still not getting the right answer.
After the integration:
$-\frac{3}{14v^2}=ln|x|+c$
$v^{-2}=-\frac{14}{3}ln|x|-\frac{14}{3}c$
$v=\frac{1}{\sqrt{\frac{14}{3}ln|x|-\frac{14}{3}c}}$
$\frac{y}{x}=\frac{1}{\sqrt{\frac{14}{3}ln|x|-\frac{14}{3}c}}$
$y=\frac{x}{\sqrt{\frac{14}{3}ln|x|-\frac{14}{3}c}}$
After the substitution for $y(1)=1$ I get $c=-\frac{3}{14}$

**Random Variable** · February 9th 2010, 09:12 PM

It looks correct to me (except you're missing a negative sign inside of the radical). What answer is given in the book?

You can multiply the top and bottom by 3 and get the equivalent solution

$y(x) = \frac{3x}{\sqrt{9-42 \ln x}}$

**matt3D** · February 9th 2010, 11:17 PM

Thank you! Yea, I forgot the negative... I am using webwork online so it tells me if the answer I put in is correct.

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