What is the solution for 3y^'-3e^{2x}y=0,y(0)=1 ?

The solution for 3y^'-3e^{2x}y=0,y(0)=1 is y=e^{1/2 e^{2x}-1/2}

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3y^'-3e^{2x}y=0,y(0)=1

3 y^{^{'}} - 3 e^{2 x} y = 0, y (0) = 1

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

Solution steps

3 y^{'} (x) - 3 e^{2 x} y = 0

Solve separable ODE:

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

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

\frac{d x}{d t} = t (x - 2)

\frac{d y}{d x} = \frac{8 sin ( x )}{sin ( y )}, y (0) = \frac{π}{2}

(x + 3)^{2} y + (x + 3) y^{^{'}} - y = 0

\frac{d x}{d t} = 2 t x

\frac{d}{d t} ((y) (8 y - y^{2})) = - \frac{8}{144} y, y (0) = 8

What is the solution for 3y^'-3e^{2x}y=0,y(0)=1 ?
The solution for 3y^'-3e^{2x}y=0,y(0)=1 is y=e^{1/2 e^{2x}-1/2}