# How do you solve the differential equation y'=e^(-y)(2x-4), where y5)=0 ?

Sep 8, 2014

This is an example of a separable equation with an initial value.
$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{- y} \left(2 x - 4\right)$
$y \left(5\right) = 0$

by multiplying by ${e}^{y}$ and by $\mathrm{dx}$,
$R i g h t a r r o w {e}^{y} \mathrm{dy} = \left(2 x - 4\right) \mathrm{dx}$
by integrating,
$R i g h t a r r o w \int {e}^{y} \mathrm{dy} = \int \left(2 x - 4\right) \mathrm{dx}$
$R i g h t a r r o w {e}^{y} = {x}^{2} - 4 x + C$
by taking the natural log,
$R i g h t a r r o w y = \ln \left({x}^{2} - 4 x + C\right)$

Now, we need to find $C$ using $y \left(5\right) = 0$.
$y \left(5\right) = \ln \left({\left(5\right)}^{2} - 4 \left(5\right) + C\right) = \ln \left(5 + C\right) = 0$
$R i g h t a r r o w 5 + C = 1 R i g h t a r r o w C = - 4$

Hence, the solution is $y = \ln \left({x}^{2} - 4 x - 4\right)$.
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