# Thread: [SOLVED] integral of perfect derivative

1. ## [SOLVED] integral of perfect derivative

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

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.

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

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
$3y^2 \dfrac{d^2y}{dx^2} + 2e^y \dfrac{dy}{dx} + 2y {\color{red} e^y} \dfrac{dy}{dx}+6y(\dfrac{dy}{dx})^2=0$.

3. That would definately help. I think its a typo as well. Thanks.

4. There might be a typo but for your original problem, you could have

$
\frac{d}{dx}\left( 3 y^2 y' + 2 e^y + y^2\right) = 0.
$

5. Yeah, the problem had a typo, but its fixed now thanks.