What is the solution for x^'=x+xa^2 ?

The solution for x^'=x+xa^2 is x=(e^{t+a^2t+c_{1}+c_{1}a^2})/(1+a^2)

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x^'=x+xa^2

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

x = \frac{e ^{t + a^{2} t + c_{1} + c_{1} a^{2}}}{1 + a ^{2}}

Solution steps

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

Solve separable ODE:

x = \frac{e ^{t + a^{2} t + c_{1} + c_{1} a^{2}}}{1 + a ^{2}}

x = \frac{e ^{t + a^{2} t + c_{1} + c_{1} a^{2}}}{1 + a ^{2}}

x 1 + y^{2} + x y^{^{'}} 1 + x^{2} = 0

\frac{d y}{d x} = 5 x^{2} y^{3}

\frac{d y}{d x} (x^{2} + 2) + 4 x (y^{2} + 2 y + 1) = 0

x \cdot \frac{d y}{d x} = 3 y

\frac{d p}{d x} = p

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