What is the solution for 2y^'-4e^{2x}y=0,y(0)=6 ?

The solution for 2y^'-4e^{2x}y=0,y(0)=6 is y=6e^{e^{2x}-1}

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2y^'-4e^{2x}y=0,y(0)=6

2 y^{^{'}} - 4 e^{2 x} y = 0, y (0) = 6

y = 6 e^{e^{2 x} - 1}

Solution steps

2 y^{'} (x) - 4 e^{2 x} y = 0

Solve separable ODE:

y = 6 e^{e^{2 x} - 1}

y = 6 e^{e^{2 x} - 1}

e^{x} y^{^{'}} - e^{x} \cdot y = e^{x}

\frac{d}{d x} (x - y^{10}) = \frac{d}{d x} (8)

\frac{d y}{d x} y = \frac{2 x + 3}{x + 1}

(x + 3 x y^{2}) d x + e^{x^{2}} y d y = 0

\frac{d}{d θ} (r (θ)) r (θ) = 5 - θ^{3} sin (θ)

What is the solution for 2y^'-4e^{2x}y=0,y(0)=6 ?
The solution for 2y^'-4e^{2x}y=0,y(0)=6 is y=6e^{e^{2x}-1}