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Popular >

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

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Popular Examples

integral of (x^3)/(x^2+5)

\int \frac{x ^{3}}{x ^{2} + 5} d x

derivative of (dy/(dx))=4x

\frac{d}{d x} (\frac{d y}{d x}) = 4 x

tangent of f(x)=-3e^{-2x}+3 tangent of

f (x) = - 3 e^{- 2 x} + 3

integral of 1/((x^2+2)^{5/2)}

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

integral of (u-1)/(u^2)

\int \frac{u - 1}{u ^{2}} d u

Frequently Asked Questions (FAQ)

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