# DE Initial Value Problem

• Mar 3rd 2011, 08:54 PM
salohcinseah
DE Initial Value Problem
Question: dy/dx= e^2x+y , y(0)=3
using variable seperable
i have solved till ln y = 1/2e^2x+c then i am stuck,would any kind soul help me to further solve it?
• Mar 3rd 2011, 09:14 PM
pickslides
Is your equation $\displaystyle \displaystyle \frac{dy}{dx}= e^{2x}+y$

or $\displaystyle \displaystyle \frac{dy}{dx}= e^{2x+y}$ ??
• Mar 3rd 2011, 09:27 PM
salohcinseah
oh it the first 1
• Mar 3rd 2011, 09:38 PM
Prove It
$\displaystyle \displaystyle \frac{dy}{dx} = e^{2x} + y$

$\displaystyle \displaystyle \frac{dy}{dx} - y = e^{2x}$.

This is first order linear, so multiplying both sides by the integrating factor $\displaystyle \displaystyle e^{\int{-1\,dx}} = e^{-x}$ gives

$\displaystyle \displaystyle e^{-x}\,\frac{dy}{dx} - e^{-x}\,y = e^{x}$

$\displaystyle \displaystyle \frac{d}{dx}\left(e^{-x}\,y\right) = e^x$

$\displaystyle \displaystyle e^{-x}\,y = \int{e^x\,dx}$

$\displaystyle \displaystyle e^{-x}\,y = e^x + C$

$\displaystyle \displaystyle y = e^{2x} + Ce^x$.

Now, you know from your initial condition that when $\displaystyle \displaystyle x = 0, y = 3$, so

$\displaystyle \displaystyle 3 = e^{2\cdot 0} + Ce^0$

$\displaystyle \displaystyle 3 = 1 + C$

$\displaystyle \displaystyle C = 2$.

Therefore the solution is

$\displaystyle \displaystyle y = e^{2x} + 2e^{x}$.
• Mar 3rd 2011, 10:21 PM
salohcinseah
i see thanks