Question

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

Answer 1

`y=ln (2/(C - e^(2x)))`

Explanation:

it's separable!!

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

`dy/dx=e^ye^(2x)`

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

`int\ e^(-y) dy/dx \ dx=int \ e^(2x) \ dx`

`int\ e^(-y) \dy=int \ e^(2x) \ dx`

`-e^(-y) =1/2 e^(2x) +C`

`e^(-y) =C - 1/2 e^(2x)`

`ln(e^(-y)) =ln (C - 1/2 e^(2x))`

`-y =ln (C - 1/2 e^(2x))`

`y =-ln (C - 1/2 e^(2x))`

`y =ln (1/(C - 1/2 e^(2x)))`

`y=ln (2/(C - e^(2x)))`

Answer 2

`e^(2x)+2e^(-y)+C=0`

Explanation:

The given differential equation is

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

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

Multiplying both sides by `dx`

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

`dy/cancel(dx)*cancel(dx)=e^(y)*e^(2x)*dx`

`dy=e^(y)*e^(2x)*dx`

Dividing both sides by `e^y`

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

`dy/e^y=(cancel(e^(y))*e^(2x)*dx)/cancel(e^y)`

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

`e^(-y)*dy-e^(2x)*dx=0`

Integrating both sides of the equation

`int e^(-y)*dy-int e^(2x)*dx=int 0`

`-1*int e^(-y)*(-1)dy-1/2*int e^(2x)*2*dx=int 0`

`-1* e^(-y)-1/2* e^(2x)=C_1`

`e^(-y)+1/2* e^(2x)+C_1=0`

`e^(2x)+2e^(-y)+C=0`

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