What is the solution for (dy)/(dx)=3e^{x-y},y(0)=ln(6) ?

The solution for (dy)/(dx)=3e^{x-y},y(0)=ln(6) is y=ln(3e^x+3)

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(dy)/(dx)=3e^{x-y},y(0)=ln(6)

\frac{d y}{d x} = 3 e^{x - y}, y (0) = ln (6)

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

Solution steps

\frac{d y}{d x} = 3 e^{x - y}

Solve separable ODE:

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

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

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

y = x y^{^{'}} - \frac{1}{y}

y^{^{'}} = \frac{2 e ^{x}}{e ^{y}}

(y + 1) cos (x) d x - d y = 0

\frac{d}{d x} (x^{4} + y^{4}) = 4, x^{3} + 4 y^{3} y^{^{'}} = 0

What is the solution for (dy)/(dx)=3e^{x-y},y(0)=ln(6) ?
The solution for (dy)/(dx)=3e^{x-y},y(0)=ln(6) is y=ln(3e^x+3)