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Popular >

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

Solution

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

Solution

y = - \frac{2}{x ^{2} + 2 x - 1}

Solution steps

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

Solve separable ODE:

y = - \frac{2}{x ^{2} + 2 x - 1}

y = - \frac{2}{x ^{2} + 2 x - 1}

Popular Examples

(1+x^2)y^'=sin(y)

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

(3dy)/(2dx)+(4x+2xy^2)/(2y+x^2y)=0

\frac{3 d y}{2 d x} + \frac{4 x + 2 x y ^{2}}{2 y + x ^{2} y} = 0

y^{^{'}} - y = 3, y (0) = 0

y^'= x/(3-y),y(2)=0

y^{^{'}} = \frac{x}{3 - y}, y (2) = 0

(dy)/(dx)y=x^{ln(x)}

\frac{d y}{d x} y = x^{l n (x)}

Frequently Asked Questions (FAQ)

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