Question

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

Answer 1

`y = -ln(-e^x + C)`, or `ln(1/(C-e^x))`

Explanation:

`dy/dx = e^(x+y)`
`:. dy/dx = e^xe^y`

So we can identify this as a First Order Separable Differential Equation. We can therefore "separate the variables" to give:

`int 1/e^y dy = int e^x dx`
`:. int e^-y dy = int e^x dx`

Integrating gives us:

`-e^-y = e^x + C'`
`:. e^-y = -e^x + C`
`:. -y = ln(-e^x + C)`
`:. y = -ln(C-e^x)`, or `ln(1/(C-e^x))`