Help w/ General Solution

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February 16th 2010, 11:08 PM
mmattson07

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
February 17th 2010, 01:21 AM
mr fantastic

Quote:

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.
February 17th 2010, 06:04 PM
mmattson07

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

Thanks.
February 18th 2010, 01:25 AM
mr fantastic

Quote:

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.
February 18th 2010, 08:21 PM
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...
February 19th 2010, 12:58 AM
mr fantastic

Quote:

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!

Quote:

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.