t^'=(2t-1)^3

Lösung

t^{^{'}} = (2 t - 1)^{3}

t (x) = \frac{- \frac{1}{4 ( x + c _{1} )}}{2} + \frac{1}{2}, t (x) = - \frac{- \frac{1}{4 ( x + c _{1} )}}{2} + \frac{1}{2}

Schritte zur Lösung

t^{'} (x) = (2 t (x) - 1)^{3}

Solve separable ODE:

t (x) = \frac{- \frac{1}{4 ( x + c _{1} )}}{2} + \frac{1}{2}, t (x) = - \frac{- \frac{1}{4 ( x + c _{1} )}}{2} + \frac{1}{2}

t (x) = \frac{- \frac{1}{4 ( x + c _{1} )}}{2} + \frac{1}{2}, t (x) = - \frac{- \frac{1}{4 ( x + c _{1} )}}{2} + \frac{1}{2}

\frac{d y}{d x} = 10 - \frac{y}{50}

\frac{d y}{d x} x^{2} + x^{2} y^{2} + y^{2} = 0

\frac{d y}{d x} = \frac{( x y - 4 x + y - 4 )}{( x y - 6 x - 3 y + 18 )}

\frac{d y}{d t} = 16 y

2 y y^{^{'}} = 12 t^{2}