What is the solution for xy^'+y=6xy^2,y(1)=-4 ?

The solution for xy^'+y=6xy^2,y(1)=-4 is y=-4/(x(24ln(x)+1))

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xy^'+y=6xy^2,y(1)=-4

x y^{^{'}} + y = 6 x y^{2}, y (1) = - 4

y = - \frac{4}{x ( 24 ln ( x ) + 1 )}

Solution steps

x y^{'} (x) + y = 6 x y^{2}

Solve Bernoulli ODE:

y = - \frac{4}{x ( 24 ln ( x ) + 1 )}

y = - \frac{4}{x ( 24 ln ( x ) + 1 )}

1 - 2 sin^{3} (x)

\int \frac{x ^{3}}{e ^{x}} d x

\frac{d}{d x} (\frac{4 x ^{5} + 6}{x ^{3}})

- \frac{5}{x}

x, x^{2}, [0, 1]

What is the solution for xy^'+y=6xy^2,y(1)=-4 ?
The solution for xy^'+y=6xy^2,y(1)=-4 is y=-4/(x(24ln(x)+1))