Thread: Differential equation with initial conditions

1. Differential equation with initial conditions

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

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

2. Originally Posted by Exotique
Find the unique function $\displaystyle y(x)$ satisfying the differential equation with initial condition,

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

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

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:

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

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:

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

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:

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

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:

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

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

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

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

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

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

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

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

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.