MHB -2.2.2 Separable eq y'=(x^2)/y(1+x^3)

[karush]

1000
use Separable Equations to solve

$$y'= \frac{x^2}{y(1+x^3)}$$
Multiply both sides by the denominator
$$y(1+x^3)y'=x^2$$
Subtract $x^2$ from both sides
$$-x^2 +y(1+x^3)y'=0$$

ok was trying to follow an example but ?
View attachment 8242

 

[MarkFL]

We are given:

$$y'=\frac{x^2}{y(1+x^3)}$$

I would separate the variables:

$$y\,dy=\frac{x^2}{x^3+1}\,dx$$

Now integrate...

 

[karush]

use Separable Equations to solve

\begin{align*}\displaystyle
y'&= \frac{x^2}{y(1+x^3)}\\
yy'&=\frac{x^2}{(1+x^3)}\\
\int y \, dy&=\int \frac{x^2}{(1+x^3)} \, dx\\
\frac{y^2}{2}&=\frac{ \ln(x^3 + 1)}{3} + c\\
3y^2−2\ln|1+x^3|&=c\\
x&\ne−1, \\
y&\ne 0
\end{align*}

kinda winged it...

 

[tkhunny]

You do need to make up your mind about the absolute value. Yes, in this case.

 

[MarkFL]

Please don't edit you post after someone has replied based on that post. I have reverted the OP so my subsequent post makes sense, but now the attachment is bad.

 

[karush]

there was no reply yet when i edited the OP