# solution dy/dx=e^y+x+e^y-xdy/dx equls to e two the power y+x plus e two the power y-x

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### 2 Answers

You need to factor out `e^y` to the right side such that:

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

You need to separate the variables x and y, hence, you need to divide by `e^y` and you need to multiply by `dx` such that:

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

You need to integrate both sides such that:

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

You need to use the property of linearity of integral to the right such that:

`-e^(-y) = int e^x dx + int e^(-x) dx`

`-e^(-y) = e^x - e^(-x) + c`

`e^(-y) =e^(-x) - e^x - c`

You need to solve for y, hence, you need to take logarithms both sides such that:

`ln (e^(-y)) = ln (e^(-x) - e^x - c)`

`-y*ln e = ln (e^(-x) - e^x - c)`

Since `ln e = 1` yields:

`-y = ln (e^(-x) - e^x - c) => y = -ln (e^(-x) - e^x - c)`

`y = ln (e^(-x) - e^x - c)^(-1) => y = ln (1/(e^(-x) - e^x - c))`

**Hence, evaluating the solution to the first order non-linear differential equation yields `y = ln (1/(e^(-x) - e^x - c)).` **

### User Comments

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

= e^y * (e^x + e^(-x))

= e^y * 2 Cos(x)

[Where I have used, Cos(x) = [e^x + e^(-x)]/2]

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

=> - (d/dx) e^(-y) = 2 Cos(x)

=> - Integral[ (d/dx) e^(-y) dx] = 2 Integral[ Cos(x) dx] + Constant

=> - e^(-y) = 2 Sin(x) + Constant

=> e^y = - 2 Sin(x) - Constant

=> y = Log[ - 2 Sin(x) - Constant]

So the answer is

**y = Log[ - 2 Sin(x) - Constant] **