What is the solution for xy^'+y=-9xy^2,y(1)=-4 ?

The solution for xy^'+y=-9xy^2,y(1)=-4 is y= 4/(x(36ln(x)-1))

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xy^'+y=-9xy^2,y(1)=-4

x y^{^{'}} + y = - 9 x y^{2}, y (1) = - 4

y = \frac{4}{x ( 36 ln ( x ) - 1 )}

Solution steps

x y^{'} (x) + y = - 9 x y^{2}

Solve Bernoulli ODE:

y = \frac{4}{x ( 36 ln ( x ) - 1 )}

y = \frac{4}{x ( 36 ln ( x ) - 1 )}

\int_{0}^{\frac{π}{5}} sin^{4} (5 x) d x

y^{^{'}} + y = \frac{1}{1 + t ^{2}}

x \to 0 lim (x^{2} ln (4 x^{2}))

\frac{1}{4 s ^{3}}

\frac{\partial}{\partial x} (1 - x^{2} - x - 3 y^{2} + y)

What is the solution for xy^'+y=-9xy^2,y(1)=-4 ?
The solution for xy^'+y=-9xy^2,y(1)=-4 is y= 4/(x(36ln(x)-1))