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Vi Nguyen · May 13, 2009, 12:04 AM

How do you solve an initial value problem:

dy/dx= xe^(y-x^2) and y(0) = 0

galactus · May 13, 2009, 02:36 PM

With this one, the best way is to separate variables:

$\frac{dy}{dx}=xe^{y-x^{2}}$

Rewrite using the log laws:

$\frac{dy}{dx}=x\cdot\frac{e^{y}}{e^{x^{2}}$

Separate variables:

$\frac{dy}{e^{y}}=\frac{x}{e^{x^{2}}}dx$

Integrate:

$-e^{-y}=C-\frac{e^{-x^{2}}}{2}$

Now, can you finish by using the initial condition and solving for C? Just set y=0 and x=0 and solve for C.

You can also solve for y.

Vi Nguyen · May 14, 2009, 04:06 AM

You're the best!! Spent ages trying to work this one out. I don't know how to rate the answer, so please excuse me if I don't somehow rate it.:D

Originally Posted by galactus

With this one, the best way is to separate variables:

$\frac{dy}{dx}=xe^{y-x^{2}}$

Rewrite using the log laws:

$\frac{dy}{dx}=x\cdot\frac{e^{y}}{e^{x^{2}}$

Separate variables:

$\frac{dy}{e^{y}}=\frac{x}{e^{x^{2}}}dx$

Integrate:

$-e^{-y}=C-\frac{e^{-x^{2}}}{2}$

Now, can you finish by using the initial condition and solving for C?. Just set y=0 and x=0 and solve for C.

You can also solve for y.

galactus · May 14, 2009, 12:46 PM

To rate an answer, just click on 'Rate this Answer" directly below the post.

Glad I could help.

Cheers,
Cody