-b.2.2.1 separate variables y'=\dfrac{x^2}{y}

[karush]

$\tiny{b.2.2.1 \quad 48}$
320
solve $\quad y'=\dfrac{x^2}{y}$

$\begin{array}{lll}
\textit{Rewrite as}
&y\dfrac{dy}{dx}=x^2\implies y\ dy=x^2\ dx \\ \\
\textit{Integrate Thru}
&\displaystyle\int y \, dy = \displaystyle\int x^2 \, dx\\ \\
&\dfrac{y^2}{2} = \dfrac{x^3}{3}+c\\ \\
\textit{thus}
&3y^2-2x^3+c; \quad y\ne 0
\end{array}$

I think this is OK but always get confused about the c

 

[topsquark]

Well done.

Two things...
1) \(\displaystyle 3y^2 - 2x^3 + c\) is not an equation.

2) \(\displaystyle \dfrac{y^2}{2} = \dfrac{x^3}{3} + c\) and \(\displaystyle 3y^2 - 2x^3 + c = 0\) have two different values for c. You can keep it like this and use c for both (almost everyone does) but you really should mention that you are redefining the value of c.

-Dan

 

[karush]

ok looks like left out the =$3y^2-2x^3=c; \quad y\ne 0$
this was the book answer

 

[karush]

I made a pdf of 22 differential equation problems with replies from MHB
it is still in draft mode with markup corrections to be made but so far these have attracted 7K+ views

hope it opens!

https://drive.google.com/file/d/1ARMDHlIAnGE6DEpUfDL6D4lNndt0OFeq/view?usp=sharing

 

[HOI]

Yes, integration of $\int y dy= \int x^2dx$ gives $\frac{y^2}{2}= \frac{x^3}{3}+ c$
where "c" is the "constant of integration" and can be any number.

Now clear the fractions by multiplying by 6:
$3y^2= 2x^3+ 6c$.

Since c can be any constant, so can 6c. Sometimes people just write
$3y^2= 2x^3+ c$ again, understanding that this "c" is not the same number as the first "c" but is still just an undetermined constant. Some people prefer to write "c'" or "C" to make it clear that this is not the same number.

We could just as well have written the original integration as $\frac{y^2}{2}= \frac{x^3}{3}+ \frac{c}{6}$ since "c/6" is just as much an undermined constant as "c" is! Then multiplying by 6 give $3y^2= 2x^3+ c$.