Differential equation with initial conditions

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May 19th 2010, 06:59 PM
Exotique

Differential equation with initial conditions

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

$\frac{dy}{dx}=x^2 y$ , $y(1)=1$
May 19th 2010, 07:06 PM
11rdc11

Quote:

Originally Posted by Exotique

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

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

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

$\ln{|y|} = \frac{x^3}{3} + C$

Can you take it fron here?
May 19th 2010, 07:23 PM
Exotique

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:

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

Since $y(1)=1$ ,

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

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

For some reason this doesn't feel right.

Then the original equation should be:

$y = e^\frac{x^3}{3} + 1 - e^\frac{1}{3}$
May 19th 2010, 07:35 PM
11rdc11

Quote:

Originally Posted by Exotique

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:

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

Since $y(1)=1$ ,

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

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

For some reason this doesn't feel right.

Then the original equation should be:

$y = e^\frac{x^3}{3} + 1 - e^\frac{1}{3}$

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

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

$y=Ce^{\frac{x^3}{3}}$

$1=Ce^{\frac{1}{3}}$

$\frac{1}{e^{\frac{1}{3}}} = C$
May 19th 2010, 07:46 PM
Exotique

How did you get from
$y=e^\frac{x^3}{3} e^C$

to
$y=Ce^\frac{x^3}{3}$
May 19th 2010, 07:55 PM
pickslides

$e$ to the power of an arbitrary constant remains an arbitrary constant. Therefore we can say $e^C = C$
May 19th 2010, 07:57 PM
Exotique

Ahh, I understand now, thank you to all.
May 19th 2010, 07:59 PM
mr fantastic

Quote:

Originally Posted by Exotique

Ahh, I understand now, thank you to all.

This question counts towards the student's final grade. It is not MHF policy to knowingly assist with such questions. Thread closed.

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