1. ## Implicit Differentiation

Hi again. I was having some trouble with this implicit differentiation problem and was wondering if I could get some help.

$\displaystyle \frac{dy}{dx} - (3 + 2x)(1 + y^2) = 0$

Thanks

2. Originally Posted by Danog
Hi again. I was having some trouble with this implicit differentiation problem and was wondering if I could get some help.

$\displaystyle \frac{dy}{dx} - (3 + 2x)(1 + y^2) = 0$

Thanks
What is your aim? To find the second derivative of y?

3. Agh, my bad. The questions asks to find the general solution to the equation, and to express y as a function of x.

4. Originally Posted by Danog
Hi again. I was having some trouble with this implicit differentiation problem and was wondering if I could get some help.

$\displaystyle \frac{dy}{dx} - (3 + 2x)(1 + y^2) = 0$

Thanks
$\displaystyle \frac{dy}{dx} - (3 + 2x)(1 + y^2) = 0$
$\displaystyle \frac{dy}{dx} = (3 + 2x)(1 + y^2)$
$\displaystyle \frac{dy}{(1 + y^2)} =(3 + 2x) \cdot dx$
just integrate both side....

5. Originally Posted by Danog
Agh, my bad. The questions asks to find the general solution to the equation, and to express y as a function of x.
This isn't really 'implicit differentiation', then. It is the solution of Ordinary Differential Equations!

6. Originally Posted by ramiee2010
$\displaystyle \frac{dy}{dx} - (3 + 2x)(1 + y^2) = 0$
$\displaystyle \frac{dy}{dx} = (3 + 2x)(1 + y^2)$
$\displaystyle \frac{dy}{(1 + y^2)} =(3 + 2x) \cdot dx$
just integrate both side....
You use the lazy notation :P.

To be COMPLETELY correct...

$\displaystyle \frac{dy}{dx} = (3 + 2x)(1 + y^2)$

$\displaystyle \frac{1}{1 + y^2}\,\frac{dy}{dx} = 3 + 2x$

$\displaystyle \int{\frac{1}{1 + y^2}\,\frac{dy}{dx}\,dx} = \int{3 + 2x\,dx}$

$\displaystyle \int{\frac{1}{1 + y^2}\,dy} = 3x + x^2 + C_1$

$\displaystyle \arctan{y} + C_2 = 3x + x^2 + C_1$

$\displaystyle \arctan{y} = 3x + x^2 + C$, where $\displaystyle C = C_1 - C_2$

$\displaystyle y = \tan{(3x + x^2 + C)}$.

7. Originally Posted by Prove It
You use the lazy notation :P.

To be COMPLETELY correct...

it was just a hint not a complete solution .

8. Originally Posted by ramiee2010
it was just a hint not a complete solution .
I'm aware of that. And it was correct. I'm just saying you were using lazy shortcuts that many mathematicians would vomit on...