# Math Help - differential equations Q2

1. ## differential equations Q2

losve the differnetial equation
y' + 2y =2e^x

2. 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

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

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

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

$y = \frac{2}{3}e^x + Ce^{-2x}$.

3. i messed up on the algebra
thanks