What is the solution for xy^'+y=-5xy^2,y(1)=3 ?

The solution for xy^'+y=-5xy^2,y(1)=3 is y= 3/(x(15ln(x)+1))

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xy^'+y=-5xy^2,y(1)=3

x y^{^{'}} + y = - 5 x y^{2}, y (1) = 3

y = \frac{3}{x ( 15 ln ( x ) + 1 )}

Solution steps

x y^{'} (x) + y = - 5 x y^{2}

Solve Bernoulli ODE:

y = \frac{3}{x ( 15 ln ( x ) + 1 )}

y = \frac{3}{x ( 15 ln ( x ) + 1 )}

y^{^{''}} + 2 y^{^{'}} = 2 t + 3

\int_{1}^{4} \frac{7}{x} d x

\frac{60 t ^{2}}{t ^{2} + 150}

\frac{\partial}{\partial x} (8 x y - \frac{1}{4} (x + y)^{4})

\frac{d}{d x} (cos (\frac{π}{x}))

What is the solution for xy^'+y=-5xy^2,y(1)=3 ?
The solution for xy^'+y=-5xy^2,y(1)=3 is y= 3/(x(15ln(x)+1))