Question

Find the general solution of the ODE:

y' − e^y − 2xe^y = 0

Answer 1

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

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

`(dy)/(dx)=e^y+2xe^y`

`(dy)/(dx)=e^y(1+2x)`

Separate the variables.

`(dy)/(e^y)=(1+2x)dx`

Integrate both sides.

`\int (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)`.