# How tosolve this problem? (x+e^y)dy - dx = 0

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### 1 Answer

We are going to find `(dx)/(dy) = x + e^y` rewritiing as

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

This is a linear ode class 1. We find `f(y)=-1` , `int f(y) dy = -y`

So we multiply both sides by `e^(-y)` and we get

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

Now `(d)/(dy)xe^(-y) = e^(-y)(dx)/(dy)-e^(-y)x` which is the left side of our equation. so

`(d(xe^(-y)))/(dy) = 1`

Integrating both sides with respect to y gives us

`xe^(-y) = y + C`

Solving for `x = e^y(y+C)` we get our solution.

We can see this is a solution because

`(dx)/(dy)=e^(y)(y+C)+e^y=x+e^y` which, multiplying both sides by dy and subtracting dx from both sides is what we started with.