# Thread: Help w/ General Solution

1. ## Help w/ General Solution

Need assistance arriving at this general solution for a particular ODE:

$e^{y^{3}}=csc(x)y^{2}y'$
$=> y' = \frac{e^{y^{3}}}{csc(x)y^{2}}$
$\frac{dy}{dx}= \frac{e^{y^{3}}}{csc(x)y^{2}}$
...??
$=>3cos(x)+C = e^{-y^{3}}$

Ty

2. Originally Posted by mmattson07
Need assistance arriving at this general solution for a particular ODE:

$e^{y^{3}}=csc(x)y^{2}y'$
$=> y' = \frac{e^{y^{3}}}{csc(x)y^{2}}$
$\frac{dy}{dx}= \frac{e^{y^{3}}}{csc(x)y^{2}}$
...??
$=>3cos(x)+C = e^{-y^{3}}$

Ty
That solution looks fine. But if you want to make y the subject, start by taking the log of both sides.

3. Could you maybe show a few steps as to hint how to arrive at that solution?

Thanks.

4. Originally Posted by mmattson07
Could you maybe show a few steps as to hint how to arrive at that solution?

Thanks.
Did you try the first step I suggested? Please show what you've tried.

5. Oh I thought you meant take the log of both sides if i wanted the soltuon in the form y=
So take the log of both sides of the original DE? Wouldn't that just make things uglier? I tried solving using Separation of Variables but I got stuck where indicated...

6. Originally Posted by mmattson07
Oh I thought you meant take the log of both sides if i wanted the soltuon in the form y=
So take the log of both sides of the original DE? Wouldn't that just make things uglier? I tried solving using Separation of Variables but I got stuck where indicated...
NO!
Originally Posted by mmattson07
Need assistance arriving at this general solution for a particular ODE:

$e^{y^{3}}=csc(x)y^{2}y'$
$=> y' = \frac{e^{y^{3}}}{csc(x)y^{2}}$
$\frac{dy}{dx}= \frac{e^{y^{3}}}{csc(x)y^{2}}$
...??
$=>3cos(x)+C = e^{-y^{3}}$ Mr F says: Take log to the base e of both sides of this line!

Ty
Look, if you're studying DE's this should not be a difficult thing to do. Do what I've said to do, post what you do, say where you're stuck.