What is the solution for x^'=(1-x)(1-cos(pi)x) ?

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

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x^'=(1-x)(1-cos(pi)x)

x^{^{'}} = (1 - x) (1 - cos (π) x)

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) = (1 - x) (1 - cos (π) x)

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})}}

y^{^{'}} = - \frac{0.0491}{c} \cdot 19.62 (y + 1)

(y)^{6} (\frac{\frac{d y}{d x} - 5}{3}) = tan (3 x) - \frac{5}{3} (y)^{6}

\frac{d p}{d t} = 3 p - p^{2}

\frac{d y}{d x} = \frac{ln ( x ^{6} )}{x y ^{2}}, y (1) = 5

y - 2 y^{^{'}} = 5

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