# [SOLVED] integral of perfect derivative

• Feb 15th 2010, 08:48 PM
snaes
[SOLVED] integral of perfect derivative
The Differential Equation: $\displaystyle 3y^2 y''(x) + 2e^y y'(x)+2y y'(x)+6y(y'(x))^2=0$ is a perfect derivative. Find the first integral.
$\displaystyle 3y^2 \dfrac{d^2y}{dx^2} + 2e^y \dfrac{dy}{dx} + 2y \dfrac{dy}{dx}+6y(\dfrac{dy}{dx})^2=0$

not sure which notation is best so i put 2 ways of the same problem.

This is the product rule split up but i just cannot see how to do this backwards. Any help would be great! Thanks.
• Feb 15th 2010, 09:42 PM
mr fantastic
Quote:

Originally Posted by snaes
The Differential Equation: $\displaystyle 3y^2 y''(x) + 2e^y y'(x)+2y y'(x)+6y(y'(x))^2=0$ is a perfect derivative. Find the first integral.
$\displaystyle 3y^2 \dfrac{d^2y}{dx^2} + 2e^y \dfrac{dy}{dx} + 2y \dfrac{dy}{dx}+6y(\dfrac{dy}{dx})^2=0$

not sure which notation is best so i put 2 ways of the same problem.

This is the product rule split up but i just cannot see how to do this backwards. Any help would be great! Thanks.

I suspect there's a typo in the question. I think it should be
$\displaystyle 3y^2 \dfrac{d^2y}{dx^2} + 2e^y \dfrac{dy}{dx} + 2y {\color{red} e^y} \dfrac{dy}{dx}+6y(\dfrac{dy}{dx})^2=0$.
• Feb 16th 2010, 09:49 AM
snaes
That would definately help. I think its a typo as well. Thanks.
• Feb 16th 2010, 09:57 AM
Jester
There might be a typo but for your original problem, you could have

$\displaystyle \frac{d}{dx}\left( 3 y^2 y' + 2 e^y + y^2\right) = 0.$
• Feb 16th 2010, 12:43 PM
snaes
Yeah, the problem had a typo, but its fixed now thanks.