# Thread: Derivative and differential equation problem?

1. ## Derivative and differential equation problem?

Is my answer for this: right? I'm having a lot of trouble with this because of the "yy'" part. I know obviously what y' means but yy' confuses things. Any help? Could someone walk me through this? :/

2. ## Re: Derivative and differential equation problem?

$y y' - e^x=0$

$y \dfrac{dy}{dx} = e^x$

$y dy = e^x dx$

3. ## Re: Derivative and differential equation problem?

Originally Posted by romsek
$y y' - e^x=0$

$y \dfrac{dy}{dx} = e^x$

$y dy = e^x dx$

Oh okay. So when I plug in (0,4) I get c=7? But I don't know how that relates to any of the options :/ I'm feeling really stupid right now. This question is just going over my head for some reason..

4. ## Re: Derivative and differential equation problem?

From romsek's $\displaystyle y\,dy=e^x\,dx$, when you integrate both sides you should get $\displaystyle \frac{1}{2}y^2=e^x+C$, and then, yes, C=7. So you have $\displaystyle \frac{1}{2}y^2=e^x+7$. Now multiply both sides by 2.

- Hollywood

5. ## Re: Derivative and differential equation problem?

Originally Posted by hollywood
From romsek's $\displaystyle y\,dy=e^x\,dx$, when you integrate both sides you should get $\displaystyle \frac{1}{2}y^2=e^x+C$, and then, yes, C=7. So you have $\displaystyle \frac{1}{2}y^2=e^x+7$. Now multiply both sides by 2.

- Hollywood
Oh! Thank you! so y^2=2e^x+14