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Popular >

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

Solution

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

Solution

y = \frac{2}{x ( 4 ln ( x ) + 1 )}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{2}{x ( 4 ln ( x ) + 1 )}

y = \frac{2}{x ( 4 ln ( x ) + 1 )}

Popular Examples

(\partial)/(\partial y)(y^2sin(xy))

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limit as x approaches 5 of 2/((x-5)^6)

x \to 5 lim (\frac{2}{( x - 5 ) ^{6}})

limit as x approaches 1 of 2x^{-1}

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

laplacetransform te^{-5t}sin(5t)laplace transform

t e^{- 5 t} sin (5 t)

integral of 6e^7

\int 6 e^{7} d x

Frequently Asked Questions (FAQ)

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