Find the unique function $\displaystyle y(x)$ satisfying the differential equation with initial condition,

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

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- May 19th 2010, 06:59 PMExotiqueDifferential equation with initial conditions
Find the unique function $\displaystyle y(x)$ satisfying the differential equation with initial condition,

$\displaystyle \frac{dy}{dx}=x^2 y$, $\displaystyle y(1)=1$ - May 19th 2010, 07:06 PM11rdc11
- May 19th 2010, 07:23 PMExotique
I'm guessing I have to solve for y in terms of x, let y = 1 and x = 1, then solve for C?

This is what I did:

$\displaystyle y = e^\frac{x^3}{3} + C$

Since $\displaystyle y(1)=1$,

$\displaystyle 1=e^\frac{1}{3} + C$

$\displaystyle C=1 - e^\frac{1}{3}$

For some reason this doesn't feel right.

Then the original equation should be:

$\displaystyle y = e^\frac{x^3}{3} + 1 - e^\frac{1}{3}$ - May 19th 2010, 07:35 PM11rdc11
- May 19th 2010, 07:46 PMExotique
How did you get from

$\displaystyle y=e^\frac{x^3}{3} e^C$

to

$\displaystyle y=Ce^\frac{x^3}{3}$ - May 19th 2010, 07:55 PMpickslides
$\displaystyle e$ to the power of an arbitrary constant remains an arbitrary constant. Therefore we can say $\displaystyle e^C = C$

- May 19th 2010, 07:57 PMExotique
Ahh, I understand now, thank you to all.

- May 19th 2010, 07:59 PMmr fantastic