How do you find the general solution to $\frac{d y}{d x} + e^{x + y} = 0$ $dy/dx+e^(x+y)=0$ ?

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Answer 1 · 2016-07-24T17:34:53

$y = ln (\frac{1}{e^{x} + C})$ $y = ln ( 1/(e^x + C) )$

$\frac{d y}{d x} + e^{x + y} = 0$ $dy/dx+e^(x+y)=0$

this is separable

$\frac{d y}{d x} = - e^{x + y}$ $dy/dx= - e^(x+y)$

$\frac{d y}{d x} = - e^{x} e^{y}$ $dy/dx= - e^x e^y$

$e^{- y} \frac{d y}{d x} = - e^{x}$ $e^(-y) dy/dx= - e^x$

$int\ e^(-y) dy/dx \ dx=int - e^x \ dx$

$int\ e^(-y) \ dy=- int e^x \ dx$

$-e^(-y) =- e^x + C$

$e^(-y) = e^x + C$

$e^(y) = 1/(e^x + C)$

$y = ln ( 1/(e^x + C) )$

How do you find the general solution to dy/dx+e^(x+y)=0dydx+ex+y=0dy/dx+e^(x+y)=0?