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

The solution for y^'+xy=xy^{-1/2} is y=\sqrt[3]{e^{(-3x^2-6c_{1})/2}-2e^{-(3x^2)/4-(3c_{1})/2}+1}

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

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

y = 3 e^{\frac{- 3 x ^{2} - 6 c _{1}}{2}} - 2 e^{- \frac{3 x ^{2}}{4} - \frac{3 c _{1}}{2}} + 1

Solution steps

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

Solve separable ODE:

y = 3 e^{\frac{- 3 x ^{2} - 6 c _{1}}{2}} - 2 e^{- \frac{3 x ^{2}}{4} - \frac{3 c _{1}}{2}} + 1

y = 3 e^{\frac{- 3 x ^{2} - 6 c _{1}}{2}} - 2 e^{- \frac{3 x ^{2}}{4} - \frac{3 c _{1}}{2}} + 1

\frac{d z}{d t} + 2 e^{t + z} = 0

\frac{d y}{d x} = 3 x^{2} y, y (1) = 1

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

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

\frac{d v}{d t} = g - k v (t)

What is the solution for y^'+xy=xy^{-1/2} ?
The solution for y^'+xy=xy^{-1/2} is y=\sqrt[3]{e^{(-3x^2-6c_{1})/2}-2e^{-(3x^2)/4-(3c_{1})/2}+1}