What is the solution for y^'+y=2y^{1/2} ?

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

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y^'+y=2y^{1/2}

y^{^{'}} + y = 2 y^{\frac{1}{2}}

y = e^{- t - c_{1}} - 4 e^{\frac{- t - c _{1}}{2}} + 4

Solution steps

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

Solve separable ODE:

y = e^{- t - c_{1}} - 4 e^{\frac{- t - c _{1}}{2}} + 4

y = e^{- t - c_{1}} - 4 e^{\frac{- t - c _{1}}{2}} + 4

\frac{d}{d θ} (r (θ)) + r (θ) tan (θ) = 3 tan (θ)

(x + 1) \frac{d y}{d x} = 2 y

\frac{d v}{d t} = - 0.4 \cdot v

\frac{d y}{d t} = \frac{2 t}{y + y t ^{2}}

y^{^{'}} = x^{3} e^{- y}

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