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Popular >

xy^'+y=2xy^2

Solution

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

Solution

y = \frac{1}{x ( - 2 ln ( x ) + c _{1} )}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{1}{x ( - 2 ln ( x ) + c _{1} )}

y = \frac{1}{x ( - 2 ln ( x ) + c _{1} )}

Popular Examples

derivative of 2^{2x+2}

\frac{d}{d x} (2^{2 x + 2})

integral of (x(3-5x))/((3x-1)(x-1)^2)

\int \frac{x ( 3 - 5 x )}{( 3 x - 1 ) ( x - 1 ) ^{2}} d x

limit as x approaches 3-of 1/((x-3)^3)

x \to 3 - lim (\frac{1}{( x - 3 ) ^{3}})

area 7x+y^2=15,x=2y area

7 x + y^{2} = 15, x = 2 y

area y=cos(3x),(0.4,0.6)area

y = cos (3 x), (0.4, 0.6)

Frequently Asked Questions (FAQ)

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