losve the differnetial equation

y' + 2y =2e^x

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- Mar 1st 2009, 12:24 AMtwilightstrdifferential equations Q2
losve the differnetial equation

y' + 2y =2e^x - Mar 1st 2009, 04:00 PMProve It
I get $\displaystyle \frac{2}{3}e^x + Ce^{-2x}$

Looking at the equation we have

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

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

Multiplying both sides of the equation by the integrating factor gives

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

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

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

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

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

$\displaystyle y = \frac{2}{3}e^x + Ce^{-2x}$. - Mar 1st 2009, 04:23 PMtwilightstr
i messed up on the algebra

thanks