What is the solution for (dy)/(dx)=2e^{x-y},y(0)=ln(5) ?

The solution for (dy)/(dx)=2e^{x-y},y(0)=ln(5) is y=ln(2e^x+3)

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(dy)/(dx)=2e^{x-y},y(0)=ln(5)

\frac{d y}{d x} = 2 e^{x - y}, y (0) = ln (5)

y = ln (2 e^{x} + 3)

Solution steps

\frac{d y}{d x} = 2 e^{x - y}

Solve separable ODE:

y = ln (2 e^{x} + 3)

y = ln (2 e^{x} + 3)

y^{^{'}} = e^{2 y} sin (x), y (0) = 0

\frac{d y}{y} = - \frac{2 x}{x ^{2} - 1} d x

2 x d x + sin (y) d y = 0

\frac{d y}{d x} = \frac{x + 1}{y ^{2}}

y^{^{'}} = 2 y + 1 cos (x)

What is the solution for (dy)/(dx)=2e^{x-y},y(0)=ln(5) ?
The solution for (dy)/(dx)=2e^{x-y},y(0)=ln(5) is y=ln(2e^x+3)