What is the solution for (dy}{dx}=\frac{y^{1/2}ln(x))/x ?

The solution for (dy}{dx}=\frac{y^{1/2}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{y^{1/2}ln(x))/x

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

\frac{d y}{d t} = k (y - 2)

y^{^{'}} = \frac{( x y + 3 y )}{x - 3}

\frac{d}{d x} (ln (y)) = \frac{1}{3} ln (x^{2} - 9)

y^{^{'}} = sin (x) sin (y)

\frac{d Q}{d x} = 5 x^{4} Q

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