# differential equations Q2

• March 1st 2009, 01:24 AM
twilightstr
differential equations Q2
losve the differnetial equation
y' + 2y =2e^x
• March 1st 2009, 05:00 PM
Prove It
Quote:

Originally Posted by twilightstr
is the answer y= 4/3e^4x + 2Ce^x

I get $\frac{2}{3}e^x + Ce^{-2x}$

Looking at the equation we have

$\frac{dy}{dx} + 2y = 2e^x$

So the integrating factor is $e^{\int{2\,dx}} = e^{2x}$.

Multiplying both sides of the equation by the integrating factor gives

$e^{2x}\frac{dy}{dx} + 2e^{2x}y= 2e^{3x}$

The left hand side is a product rule expansion of $e^{2x}y$ so we rewrite the equation as

$\frac{d}{dx}(e^{2x}y) = 2e^{3x}$

$e^{2x}y = \int{2e^{3x}\,dx}$

$e^{2x}y = \frac{2}{3}e^{3x} + C$

$y = \frac{2}{3}e^x + Ce^{-2x}$.
• March 1st 2009, 05:23 PM
twilightstr
i messed up on the algebra
thanks