What is the solution for 1/x (dy)/(dx)=e^{x+y} ?

The solution for 1/x (dy)/(dx)=e^{x+y} is y=-ln(-e^xx+e^x-c_{1})

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1/x (dy)/(dx)=e^{x+y}

\frac{1}{x} \frac{d y}{d x} = e^{x + y}

y = - ln (- e^{x} x + e^{x} - c_{1})

Solution steps

\frac{1}{x} \frac{d y}{d x} = e^{x + y}

Solve separable ODE:

y = - ln (- e^{x} x + e^{x} - c_{1})

y = - ln (- e^{x} x + e^{x} - c_{1})

\frac{d}{d x} (\frac{1}{2 x ln ( x ) + 1})

\int cot^{3} (x) \cdot csc^{3} (x) d x

n = 1 \sum \infty \frac{2 n !}{n ^{n}}

r (2 r + 1)

\frac{d y}{d t} = 5 (y - t^{2})

What is the solution for 1/x (dy)/(dx)=e^{x+y} ?
The solution for 1/x (dy)/(dx)=e^{x+y} is y=-ln(-e^xx+e^x-c_{1})