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

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

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(4+x)y^'=y^3

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

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

Solution steps

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

Solve separable ODE:

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

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

x \to 2 lim (\frac{1}{x - 1})

i = 1 \sum \infty (i) \cdot \frac{1}{6 ^{i}}

x \to 2 lim (\frac{x + 2}{x})

\frac{\partial}{\partial y} (x y d x)

\int 3 + x d x

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