1. ## IVP problem

find the exact solution of the IVP.
y'=2xy, y(0)=2

2. Originally Posted by calculuskid1
find the exact solution of the IVP.
y'=2xy, y(0)=2
separate variables ...

$\displaystyle \frac{dy}{y} = 2x \, dx$

integrate and use the initial condition to find the solution

3. Originally Posted by calculuskid1
y'=2xy, y(0)=2
This is a very simple separable DE

$\displaystyle \frac{dy}{dx} = 2xy$

$\displaystyle \frac{dy}{y} = 2x~dx$

Now tntegrate both sides.

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

4. I got that and i got ln y=x^2+c but im not sure what to do there,
ln 2=c
so then ln y=x^2+ln2
then y=e^(x^2+ln2)
can that be simplified?

5. Originally Posted by calculuskid1
I got that and i got ln y=x^2+c but im not sure what to do there,
ln 2=c
so then ln y=x^2+ln2
then y=e^(x^2+ln2)
can that be simplified?
$\displaystyle e^{x^2+\ln{2}} = e^{x^2} \cdot e^{\ln{2}} = ?$