# Math Help - Differential equation with initial conditions

1. ## Differential equation with initial conditions

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

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

2. 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?

3. 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:

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

4. 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:

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

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

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

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

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

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

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

6. $e$ to the power of an arbitrary constant remains an arbitrary constant. Therefore we can say $e^C = C$

7. Ahh, I understand now, thank you to all.

8. 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.