Question

Find the general solution of the ODE:

y' − e^y − 2xe^y = 0

Answer 1

Answer:

The general solution of the ODE is $y=-\ln(-x-x^2-C)$.

Step-by-step explanation:

The given differential equation is

$y'-e^y-2xe^y=0$

We need to find the general solution of the given ODE.

Use variable separable method to solve the above equation.

$y'=e^y+2xe^y$

It can be written as

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

$\frac{dy}{dx}=e^y(1+2x)$

Separate the variables.

$\frac{dy}{e^y}=(1+2x)dx$

Integrate both sides.

$\int \frac{dy}{e^y}=\int (1+2x)dx$

$\int e^{-y}dy=\int 1dx+\int 2xdx$

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

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

Taking ln both sides.

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

$-y=\ln (-x-x^2-C)$                 $[\because \ln e^x=x]$

$y=-\ln (-x-x^2-C)$

Therefore the general solution of the ODE is $y=-\ln(-x-x^2-C)$.