y^'=1

Lösung

y^{^{'}} = 1

y = t + c_{1}

Schritte zur Lösung

y^{'} (t) = 1

Wenn

f^{^{'}} (x) = g (x)

dann

f (x) = \int g (x) d x

y = \int 1 d t

\int 1 d t = t + c_{1}

y = t + c_{1}

\int 5 cos (v) - \frac{7}{1 - v ^{2}} d v

\frac{\partial}{\partial y} (4 - x^{2} - 5 y^{2})

\frac{\partial}{\partial x} (5 e^{x})

in v erse l a pl a ce \frac{s - 3}{s ^{2} - 4 s + 3}

x \to 0 lim (\frac{ln ( x ^{2} + 1 )}{x})