What is the solution for y^'=x*sqrt(y) ?

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

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y^'=x*sqrt(y)

y^{^{'}} = x \cdot y

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

Solution steps

y^{'} (x) = x \cdot 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}

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

y^{^{''}} + 4 y = tan (2 x)

x \to \frac{π}{4} lim (cos (2 x))

\int \frac{1}{81 + x ^{2}} d x

\frac{\partial}{\partial y} (\frac{2 x}{y ^{3}})

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