What is the solution for (dy}{dx}=\frac{sqrt(y)ln(x))/x ?

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

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(dy}{dx}=\frac{sqrt(y)ln(x))/x

\frac{d y}{d x} = \frac{y ln ( x )}{x}

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

Solution steps

\frac{d y}{d x} = \frac{y ln ( x )}{x}

Solve separable ODE:

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

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

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

y^{^{'}} + ln (x) y = 3 ln (x)

x^{2} y^{^{'}} + y = - 2

3 y^{2} \frac{d y}{d x} = 4 x

x y^{3} d x + e^{x} d y = 0

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