Differential equation with initial conditions

• 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.