What is the solution for y^'=6e^{(2x-y)},y(0)=0 ?

The solution for y^'=6e^{(2x-y)},y(0)=0 is y=ln(3e^{2x}-2)

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y^'=6e^{(2x-y)},y(0)=0

y^{^{'}} = 6 e^{(2 x - y)}, y (0) = 0

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

Solution steps

y^{'} (x) = 6 e^{(2 x - y)}

Solve separable ODE:

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

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

\int \frac{( 1 + e ^{x} ) ^{2}}{e ^{x}} d x

y = 4 - x^{2}, y = 2 - x

\frac{d}{d x} (x csc (2 x))

\frac{1}{x} - \frac{2}{x ^{2}}

y = (5 x - 3)^{4} (2 x - 7)^{- 3}

What is the solution for y^'=6e^{(2x-y)},y(0)=0 ?
The solution for y^'=6e^{(2x-y)},y(0)=0 is y=ln(3e^{2x}-2)