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Popular >

y^'=xy^2+2y^2,y(0)=1

Solution

y^{^{'}} = x y^{2} + 2 y^{2}, y (0) = 1

Solution

y = - \frac{2}{x ^{2} + 4 x - 2}

Solution steps

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

Solve separable ODE:

y = - \frac{2}{x ^{2} + 4 x - 2}

y = - \frac{2}{x ^{2} + 4 x - 2}

Popular Examples

derivative of f(x)=(2x^2-6x+5)/(2(2x-3))derivative of

f (x) = \frac{2 x ^{2} - 6 x + 5}{2 ( 2 x - 3 )}

(dy)/(dx)=4xe^{3y}

\frac{d y}{d x} = 4 x e^{3 y}

xy^'+2y=4cos(x)

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

(dy)/(dx)=sec^2(x)(2+y)^2

\frac{d y}{d x} = sec^{2} (x) (2 + y)^{2}

inverse oflaplace (4e^{-2s})/(3s^2+48)inverse laplace

\frac{4 e ^{- 2 s}}{3 s ^{2} + 48}

Frequently Asked Questions (FAQ)

What is the solution for y^'=xy^2+2y^2,y(0)=1 ?
The solution for y^'=xy^2+2y^2,y(0)=1 is y=-2/(x^2+4x-2)