# What is a solution to the differential equation dy/dx=e^(y+2x)?

Jul 16, 2016

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

#### Explanation:

it's separable!!

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

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

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

$\int \setminus {e}^{- y} \frac{\mathrm{dy}}{\mathrm{dx}} \setminus \mathrm{dx} = \int \setminus {e}^{2 x} \setminus \mathrm{dx}$

$\int \setminus {e}^{- y} \setminus \mathrm{dy} = \int \setminus {e}^{2 x} \setminus \mathrm{dx}$

$- {e}^{- y} = \frac{1}{2} {e}^{2 x} + C$

${e}^{- y} = C - \frac{1}{2} {e}^{2 x}$

$\ln \left({e}^{- y}\right) = \ln \left(C - \frac{1}{2} {e}^{2 x}\right)$

$- y = \ln \left(C - \frac{1}{2} {e}^{2 x}\right)$

$y = - \ln \left(C - \frac{1}{2} {e}^{2 x}\right)$

$y = \ln \left(\frac{1}{C - \frac{1}{2} {e}^{2 x}}\right)$

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

${e}^{2 x} + 2 {e}^{- y} + C = 0$

#### Explanation:

The given differential equation is

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

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

Multiplying both sides by $\mathrm{dx}$

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

$\frac{\mathrm{dy}}{\cancel{\mathrm{dx}}} \cdot \cancel{\mathrm{dx}} = {e}^{y} \cdot {e}^{2 x} \cdot \mathrm{dx}$

$\mathrm{dy} = {e}^{y} \cdot {e}^{2 x} \cdot \mathrm{dx}$

Dividing both sides by ${e}^{y}$

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

$\frac{\mathrm{dy}}{e} ^ y = \frac{\cancel{{e}^{y}} \cdot {e}^{2 x} \cdot \mathrm{dx}}{\cancel{{e}^{y}}}$

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

${e}^{- y} \cdot \mathrm{dy} - {e}^{2 x} \cdot \mathrm{dx} = 0$

Integrating both sides of the equation

$\int {e}^{- y} \cdot \mathrm{dy} - \int {e}^{2 x} \cdot \mathrm{dx} = \int 0$

$- 1 \cdot \int {e}^{- y} \cdot \left(- 1\right) \mathrm{dy} - \frac{1}{2} \cdot \int {e}^{2 x} \cdot 2 \cdot \mathrm{dx} = \int 0$

$- 1 \cdot {e}^{- y} - \frac{1}{2} \cdot {e}^{2 x} = {C}_{1}$

${e}^{- y} + \frac{1}{2} \cdot {e}^{2 x} + {C}_{1} = 0$

${e}^{2 x} + 2 {e}^{- y} + C = 0$

God bless...I hope the explanation is useful.