What is a solution to the differential equation $\frac{d y}{d x} = e^{x + y}$ $dy/dx=e^(x+y)$ ?

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Answer 1 · 2016-11-15T01:24:23

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

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

So we can identify this as a First Order Separable Differential Equation. We can therefore "separate the variables" to give:

$int 1/e^y dy = int e^x dx$
$:. int e^-y dy = int e^x dx$

Integrating gives us:

$-e^-y = e^x + C'$
$:. e^-y = -e^x + C$
$:. -y = ln(-e^x + C)$
$:. y = -ln(C-e^x)$ , or $ln(1/(C-e^x))$

What is a solution to the differential equation dy/dx=e^(x+y)dydx=ex+ydy/dx=e^(x+y)?