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Popular >

xy^'+y=4xy^2

Solution

x y^{^{'}} + y = 4 x y^{2}

Solution

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

Solution steps

x y^{'} (x) + y = 4 x y^{2}

Solve Bernoulli ODE:

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

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

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\int 3 x^{2} (x^{3} + 1)^{10} d x

limit as x approaches 4 of 9-x-5

x \to 4 lim (9 - x - 5)

Frequently Asked Questions (FAQ)

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