What is the solution for (dy)/(dx)=2e^{x-y},y(0)=ln(4) ?

The solution for (dy)/(dx)=2e^{x-y},y(0)=ln(4) is y=ln(2e^x+2)

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(dy)/(dx)=2e^{x-y},y(0)=ln(4)

\frac{d y}{d x} = 2 e^{x - y}, y (0) = ln (4)

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

Solution steps

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

Solve separable ODE:

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

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

\frac{d y}{d x} = \frac{4 y}{1 + x}

y^{^{'}} = \frac{( 2 y + 3 )}{x ^{2}}

\frac{d y}{d x} = 2 \cdot (x^{2}) \cdot (y^{2})

d x - \frac{d y}{1 + y ^{2}} = 0

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

What is the solution for (dy)/(dx)=2e^{x-y},y(0)=ln(4) ?
The solution for (dy)/(dx)=2e^{x-y},y(0)=ln(4) is y=ln(2e^x+2)