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[SOLVED] 2.2.2 y'=(x^2)/y(1+x^3) Separable Equations

karush

Well-known member
Jan 31, 2012
2,928
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
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
We are given:

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

I would separate the variables:

\(\displaystyle y\,dy=\frac{x^2}{x^3+1}\,dx\)

Now integrate...
 

karush

Well-known member
Jan 31, 2012
2,928
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

Well-known member
MHB Math Helper
Jan 27, 2012
267
You do need to make up your mind about the absolute value. Yes, in this case.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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

Well-known member
Jan 31, 2012
2,928
there was no reply yet when i edited the OP