What is the solution for x^'+x^2=1 ?

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

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x^'+x^2=1

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

x = - \frac{e ^{2 t + 2 c_{1}}}{1 - e ^{2 (t + c_{1})}} - \frac{1}{1 - e ^{2 (t + c_{1})}}

Solution steps

x^{'} (t) + x^{2} = 1

Solve separable ODE:

x = - \frac{e ^{2 t + 2 c_{1}}}{1 - e ^{2 (t + c_{1})}} - \frac{1}{1 - e ^{2 (t + c_{1})}}

x = - \frac{e ^{2 t + 2 c_{1}}}{1 - e ^{2 (t + c_{1})}} - \frac{1}{1 - e ^{2 (t + c_{1})}}

\frac{d y}{d t} = (t + 1) y, y (0) = \frac{1}{2}

\frac{d y}{d x} = - 1 y + 3 y

y^{^{'}} = 3 x^{2} \cdot y^{3}, y (0) = - 5

\frac{y}{ln ( x )} d x + x d y = 0

\frac{d y}{d x} = \frac{9}{2} y

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