y^'+a*y=b

Lösung

y^{^{'}} + a \cdot y = b

y = - \frac{e ^{- a t - c_{1} a}}{a} + \frac{b}{a}

Schritte zur Lösung

y^{'} (t) + a \cdot y = b

Solve separable ODE:

y = - \frac{e ^{- a t - c_{1} a}}{a} + \frac{b}{a}

y = - \frac{e ^{- a t - c_{1} a}}{a} + \frac{b}{a}

\frac{d}{d x} (2 x + 2 (cos (x) + 1) sin (x))

\int sec^{6} (3 x) tan^{2} (3 x) d x

d er i v a t i v e f (x) = \frac{1 + 4 x}{9 - 6 x}

\int - x^{2} + 5 x d x

\frac{\partial}{\partial x} (\frac{3 y - 6 x}{y ^{2} + 7 x ^{2}})