# How to solve this ((2x(1-e^y))/(1+x^2)^2)dx +(e^y)/(1+x^2)dy=0

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### 1 Answer

You need to divide by dx such that:

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

You need to solve for `(dy)/(dx), ` hence, you need to isolate the term that contains `(dy)/(dx) ` to the left side such that:

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

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

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

Reducing by `1-e^(-y)` yields:

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

Multiply by dx both sides such that:

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

Integrating both sides yields:

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

You should use substitution to solve the integral to the right such that:

`x^2+1= t => 2xdx= dt`

`int (2x)/(x^2+1) dx = int (dt)/t = ln|t| + c`

`int (2x)/(x^2+1) dx =ln(x^2+1) + c`

`ln(1 - e^y) = ln(x^2+1) + c =>1 -e^y = e^(ln(x^2+1) + c)`

`e^y = 1 -e^(ln(x^2+1) + c) => y = ln(1 - e^(ln(x^2+1) + c))`

**Hence, evaluating the solution to the first order non-liear differential equation yields `y = ln(1 - e^(ln(x^2+1) + c)).` **