x^'=x^2

Lösung

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

x = - \frac{1}{t + c _{1}}

Schritte zur Lösung

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

Solve separable ODE:

x = - \frac{1}{t + c _{1}}

x = - \frac{1}{t + c _{1}}

ma c l a u r in \frac{1}{4 + x}

z \to 0 lim (\frac{8}{z ^{3}} + \frac{5}{z} + 2)

im pl i c i t \frac{d y}{d x}, y = \frac{( x ^{2} + 2 x )}{4 - 5 x}

\frac{d}{d x} (\frac{1}{4 x ^{\frac{3}{4}}})

\frac{\partial}{\partial x} (1.85 + 2.05 x + 0.656 y)