`xdx + (y+e^y)(x^2+1)dy = 0` Solve the first-order differential equation by any appropriate method

Given` xdx + (y+e^y)(x^2+1)dy = 0`

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

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

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

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

by integrating on both sides we get ,

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

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

=> `y^2+2e^y = -ln(x^2+1)+C` where C is an arbitrary constant

Comments

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