IVP problem

**calculuskid1** · February 21st 2010, 03:25 PM

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

**skeeter** · February 21st 2010, 03:42 PM

Originally Posted by calculuskid1

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

separate variables ...

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

integrate and use the initial condition to find the solution

**pickslides** · February 21st 2010, 03:43 PM

Originally Posted by calculuskid1

y'=2xy, y(0)=2

This is a very simple separable DE

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

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

Now tntegrate both sides.

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

**calculuskid1** · February 21st 2010, 03:54 PM

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?

**skeeter** · February 21st 2010, 03:58 PM

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?

$e^{x^2+\ln{2}} = e^{x^2} \cdot e^{\ln{2}} = ?$

Thread: IVP problem