What is the solution for y^'=xsqrt(y) ?

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

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y^'=xsqrt(y)

y^{^{'}} = x y

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

Solution steps

y^{'} (x) = x y

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}

y = ln (x + x^{2} - 1)

x \to 0 + lim (ln (\frac{1}{x}))

x \to 1 lim (7 ∣ x + 1 ∣ - 4)

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

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

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