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

The solution for y^'+y=y^{-1/2} is y=(-e^{-(3t+c_{1})/2}+1)^{2/3}

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

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

y = (- e^{- \frac{3 t + c _{1}}{2}} + 1)^{\frac{2}{3}}

Solution steps

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

Solve separable ODE:

y = (- e^{- \frac{3 t + c _{1}}{2}} + 1)^{\frac{2}{3}}

y = (- e^{- \frac{3 t + c _{1}}{2}} + 1)^{\frac{2}{3}}

\int_{- 2}^{1} \frac{1}{( x - 1 ) ^{3}} d x

f (x) = 1 - 2 x

\frac{d}{d x} (e^{- 3 x} sin (3 x))

y = x^{2} + x + 1

\frac{\partial}{\partial y} (x + \frac{1}{2} ln (x^{2} + y^{2}))

What is the solution for y^'+y=y^{-1/2} ?
The solution for y^'+y=y^{-1/2} is y=(-e^{-(3t+c_{1})/2}+1)^{2/3}