0.125y^{''}+4y^'+28y=0

Lösung

0.125 y^{^{''}} + 4 y^{^{'}} + 28 y = 0

y = c_{1} e^{4 (2 - 4) t} + c_{2} e^{- 4 (4 + 2) t}

Schritte zur Lösung

0.125 y^{''} (t) + 4 y^{'} (t) + 28 y = 0

Solve linear ODE:

y = c_{1} e^{4 (2 - 4) t} + c_{2} e^{- 4 (4 + 2) t}

y = c_{1} e^{4 (2 - 4) t} + c_{2} e^{- 4 (4 + 2) t}

\frac{d}{d x} (6 x^{3} cos (x))

4 x y^{^{'}} + 2 y = 10 x^{6}

\int \frac{x ^{3} + x ^{2}}{x ^{5}} d x

d er i v a t i v e (x^{4} - 3 x)^{2}

d er i v a t i v e 13 x