What is the solution for 4y^'-4e^{2x}y=0,y(0)=3 ?

The solution for 4y^'-4e^{2x}y=0,y(0)=3 is y=3e^{1/2 e^{2x}-1/2}

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4y^'-4e^{2x}y=0,y(0)=3

4 y^{^{'}} - 4 e^{2 x} y = 0, y (0) = 3

y = 3 e^{\frac{1}{2} e^{2 x} - \frac{1}{2}}

Solution steps

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

Solve separable ODE:

y = 3 e^{\frac{1}{2} e^{2 x} - \frac{1}{2}}

y = 3 e^{\frac{1}{2} e^{2 x} - \frac{1}{2}}

4 (1 + y^{2}) d x - x y d y = 0

\frac{d y}{d t} + y t sin (t) = 0

\frac{d y}{d x} = \frac{3 x ^{2} + 4 x + 2}{2 y - 2}

\frac{d y}{d x} = - \frac{7}{4} x^{8} y^{6}

y^{^{'}} = y tan (x) - x^{2} y

What is the solution for 4y^'-4e^{2x}y=0,y(0)=3 ?
The solution for 4y^'-4e^{2x}y=0,y(0)=3 is y=3e^{1/2 e^{2x}-1/2}