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Popular >

x^'=x^2+x

Solution

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

Solution

x = \frac{e ^{t + c_{1}}}{1 - e ^{t + c_{1}}}

Solution steps

x^{'} (t) = x^{2} + x

Solve separable ODE:

x = \frac{e ^{t + c_{1}}}{1 - e ^{t + c_{1}}}

x = \frac{e ^{t + c_{1}}}{1 - e ^{t + c_{1}}}

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

laplacetransform (1+t)^3 laplace transform

(1 + t)^{3}

integral of sqrt(2e^{2t)}

\int 2 e^{2 t} d t

derivative of (x^2+2x^4(x^2+3x^2+7)^3)

\frac{d}{d x} ((x^{2} + 2 x)^{4} (x^{2} + 3 x^{2} + 7)^{3})

Frequently Asked Questions (FAQ)

What is the solution for x^'=x^2+x ?
The solution for x^'=x^2+x is x=(e^{t+c_{1}})/(1-e^{t+c_{1)}}