What is the solution for (dy)/(y^{1/2)}=xdx ?

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

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(dy)/(y^{1/2)}=xdx

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

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

Solution steps

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

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

x^{^{'}} = 1 - 3 x + 2 x^{2}, x (0) = 2

3 y + 5 + (3 x - 7) \frac{d y}{d x} = 0

y^{^{'}} = \frac{7 x ^{2} - 1}{7 + 4 y}

\frac{d y}{d x} y = ln (arccos (x^{2} + 4))

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