What is the solution for x^4y^'+y^3x^3=0 ?

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

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x^4y^'+y^3x^3=0

x^{4} y^{^{'}} + y^{3} x^{3} = 0

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

Solution steps

x^{4} y^{'} (x) + y^{3} x^{3} = 0

Solve separable ODE:

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

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

1 + tan (y) y^{^{'}} = x^{2} + 1

\frac{d v}{d t} = 30 - \frac{9}{100} v, v = 0, t = 2

\frac{d y}{d x} y = \frac{tan ( x )}{1 + tan ( x )}

\frac{d y}{d x} = \frac{y}{x}, y = 1, x = 1

3 x \frac{d y}{d x} + y = \frac{8}{y ^{2}}

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