y^{''}-2y^'-3y=24te^{2t},y(0)=5,y^'(0)=0

Lösung

y^{^{''}} - 2 y^{^{'}} - 3 y = 24 t e^{2 t}, y (0) = 5, y^{^{'}} (0) = 0

y = (- \frac{37}{12} + \frac{31}{3}) e^{3 t} + \frac{37}{12} e^{- t} - 8 t e^{2 t} - \frac{16}{3} e^{2 t}

Schritte zur Lösung

y^{''} (t) - 2 y^{'} (t) - 3 y = 24 t e^{2 t}

Solve linear ODE:

y = (- \frac{37}{12} + \frac{31}{3}) e^{3 t} + \frac{37}{12} e^{- t} - 8 t e^{2 t} - \frac{16}{3} e^{2 t}

y = (- \frac{37}{12} + \frac{31}{3}) e^{3 t} + \frac{37}{12} e^{- t} - 8 t e^{2 t} - \frac{16}{3} e^{2 t}

d er i v a t i v e f (x) = \frac{1 + 3 x}{1 - 3 x}

\int \frac{x ^{3}}{x ^{2} - 9} d x

t a y l or sin (2 x), 0

\frac{d}{d t} (\frac{e ^{- t}}{1 + t ^{2}})

d er i v a t i v e arcsec (x) \cdot arccos (x)