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

The solution for y^'-2y=2y^{1/2},y(0)=1 is y=4e^{2t}-4e^t+1

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y^'-2y=2y^{1/2},y(0)=1

y^{^{'}} - 2 y = 2 y^{\frac{1}{2}}, y (0) = 1

y = 4 e^{2 t} - 4 e^{t} + 1

Solution steps

y^{'} (t) - 2 y = 2 y^{\frac{1}{2}}

Solve separable ODE:

y = 4 e^{2 t} - 4 e^{t} + 1

y = 4 e^{2 t} - 4 e^{t} + 1

\frac{d P}{d t} = 8 Pt, P (1) = 2

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

\frac{d y}{d x} + 3 e^{x + y} = 0

y^{^{'}} = sin (x) cos^{2} (x) tan (y)

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

What is the solution for y^'-2y=2y^{1/2},y(0)=1 ?
The solution for y^'-2y=2y^{1/2},y(0)=1 is y=4e^{2t}-4e^t+1