What is the solution for 1/y dx-x/((ln(x))^3)dy=0 ?

The solution for 1/y dx-x/((ln(x))^3)dy=0 is y=sqrt((ln^4(x)+c_{1))/2},y=-sqrt((ln^4(x)+c_{1))/2}

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1/y dx-x/((ln(x))^3)dy=0

\frac{1}{y} d x - \frac{x}{( ln ( x ) ) ^{3}} d y = 0

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

Solution steps

\frac{1}{y} d x - \frac{x}{( ln ( x ) ) ^{3}} d y = 0

Solve separable ODE:

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

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

\frac{1}{y ^{2}} - \frac{3}{y ^{4}} (y + 5 y^{3})

\int 3 x^{- 3} + sin (x) d x

\frac{d y}{d x} = \frac{y}{x} + \frac{x}{y}

\frac{\partial}{\partial θ} (r (θ) \cdot cos (θ))

e^{r t}

What is the solution for 1/y dx-x/((ln(x))^3)dy=0 ?
The solution for 1/y dx-x/((ln(x))^3)dy=0 is y=sqrt((ln^4(x)+c_{1))/2},y=-sqrt((ln^4(x)+c_{1))/2}