# What is a solution to the differential equation y'=3^-y(2x-4) and y(5)=0?

Jul 25, 2016

$y = \frac{\ln \left(\ln 3 \left({x}^{2} - 4 x - 5\right) + 1\right)}{\ln} 3$

#### Explanation:

$y ' = {3}^{-} y \left(2 x - 4\right)$

this is separable

${3}^{y} \setminus y ' = 2 x - 4$

$\int \setminus {3}^{y} \setminus y ' \setminus \mathrm{dx} = \int \setminus 2 x - 4 \setminus \mathrm{dx}$

we can explore the integral on the LHS a little

consider $u = {C}^{v}$
$\ln u = v \ln C$
$\frac{1}{u} \frac{\mathrm{du}}{\mathrm{dv}} = \ln C$
$\implies \frac{d}{\mathrm{dv}} \left({C}^{v}\right) = \ln C \setminus {C}^{v}$
so $\frac{d}{\mathrm{dx}} \left({3}^{y}\right) = \ln 3 \setminus {3}^{y} \setminus y '$

and so the integral is
$\int \setminus \frac{d}{\mathrm{dx}} \left(\frac{1}{\ln 3} {3}^{y}\right) \setminus \mathrm{dx} = \int \setminus 2 x - 4 \setminus \mathrm{dx}$

$\frac{1}{\ln 3} {3}^{y} = {x}^{2} - 4 x + C$

${3}^{y} = \ln 3 \left({x}^{2} - 4 x\right) + C$

$y \ln 3 = \ln \left(\ln 3 \left({x}^{2} - 4 x\right) + C\right)$

$y = \frac{\ln \left(\ln 3 \left({x}^{2} - 4 x\right) + C\right)}{\ln} 3$

applying the IV: $y \left(5\right) = 0$

$0 = \frac{\ln \left(\ln 3 \left(25 - 20\right) + C\right)}{\ln} 3$

$\implies \ln 3 \left(5\right) + C = 1$

$y = \frac{\ln \left(\ln 3 \left({x}^{2} - 4 x - 5\right) + 1\right)}{\ln} 3$