What is the solution for (dy}{dx}=\frac{y^{1/2}ln(x))/x ,y(1)=0 ?

The solution for (dy}{dx}=\frac{y^{1/2}ln(x))/x ,y(1)=0 is y=(ln^4(x))/(16)

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(dy}{dx}=\frac{y^{1/2}ln(x))/x ,y(1)=0

\frac{d y}{d x} = \frac{y ^{\frac{1}{2}} ln ( x )}{x}, y (1) = 0

y = \frac{ln ^{4} ( x )}{16}

Solution steps

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

Solve separable ODE:

y = \frac{ln ^{4} ( x )}{16}

y = \frac{ln ^{4} ( x )}{16}

y d x - x d y = x (d y - y d x)

x^{2} y^{^{'}} - 2 - 3 y = 0

y^{^{'}} + x^{2} y = x^{2} y^{- 3}

(1 - 3 y^{2}) d y - x sin (x) d x = 0, y (\frac{π}{2}) = 2

\frac{d y}{d x} = 6 x 25 - y^{2}

What is the solution for (dy}{dx}=\frac{y^{1/2}ln(x))/x ,y(1)=0 ?
The solution for (dy}{dx}=\frac{y^{1/2}ln(x))/x ,y(1)=0 is y=(ln^4(x))/(16)