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

The solution for y^'=xy^{1/2} is y=(x^4}{16}+\frac{c_{1}x^2)/4+(c_{1}^2)/4

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y^'=xy^{1/2}

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

y = \frac{x ^{4}}{16} + \frac{c _{1} x ^{2}}{4} + \frac{c _{1}^{2}}{4}

Solution steps

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

Solve separable ODE:

y = \frac{x ^{4}}{16} + \frac{c _{1} x ^{2}}{4} + \frac{c _{1}^{2}}{4}

y = \frac{x ^{4}}{16} + \frac{c _{1} x ^{2}}{4} + \frac{c _{1}^{2}}{4}

x y d y + y d x + 4 y d y + 3 d x = 8 d y + 3 x d x + 2 x d y + x y d x

f (x) = \frac{df}{d x}

\frac{d y}{d x} = \frac{x ^{3}}{2 y}

x y^{^{'}} = (x + 1) y

\frac{d y}{d x} + y = 1, y (0) = 0

What is the solution for y^'=xy^{1/2} ?
The solution for y^'=xy^{1/2} is y=(x^4}{16}+\frac{c_{1}x^2)/4+(c_{1}^2)/4