What is the solution for y^'-xsqrt(y)=0 ?

The solution for y^'-xsqrt(y)=0 is y=(x^4}{16}+\frac{c_{1}x^2)/4+(c_{1}^2)/4

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y^'-xsqrt(y)=0

y^{^{'}} - x y = 0

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

Solution steps

y^{'} (x) - x y = 0

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{2 y}{( x + 1 ) + ( x + 1 ) ^{\frac{5}{2}}}

y^{^{'}} + 4 y = 0, y (0) = cos (2 t), y (0) = sin (2 t)

\frac{d y}{d x} = \frac{cos ( x )}{2 y}

- y = k \frac{d y}{d x}

e^{2 t} y^{^{'}} = y^{2}

What is the solution for y^'-xsqrt(y)=0 ?
The solution for y^'-xsqrt(y)=0 is y=(x^4}{16}+\frac{c_{1}x^2)/4+(c_{1}^2)/4