x^'=ax+b

Lösung

x^{^{'}} = a x + b

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

Schritte zur Lösung

x^{'} (t) = a x + b

Solve separable ODE:

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

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

\int x e^{\frac{- x}{4}} d x

\frac{d}{d x} (\frac{x ^{5} - 4 x ^{2}}{x ^{3}})

f (x) = cos ((ln (x + 1))^{2} - e^{x^{2}})

\int \frac{10}{x ^{2} + 16} d x

\frac{d}{d x} (\frac{1}{15} cos^{3} (x) (3 cos^{2} (x) - 5))