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Beliebt Rechnen >

linear y^{''}+4y=e^{3t},y(0)=0,y^'(0)=0

Lösung

linear

y^{^{''}} + 4 y = e^{3 t}, y (0) = 0, y^{^{'}} (0) = 0

Lösung

y = - \frac{1}{13} cos (2 t) - \frac{3}{26} sin (2 t) + \frac{1}{13} e^{3 t}

Schritte zur Lösung

y^{''} (t) + 4 y = e^{3 t}

Solve linear ODE:

y = - \frac{1}{13} cos (2 t) - \frac{3}{26} sin (2 t) + \frac{1}{13} e^{3 t}

y = - \frac{1}{13} cos (2 t) - \frac{3}{26} sin (2 t) + \frac{1}{13} e^{3 t}

Graph

Beliebte Beispiele

(D^3+D^2+D+1)y=xe^x

(D^{3} + D^{2} + D + 1) y = x e^{x}

30y^{''}+25y^'+4y=40

30 y^{^{''}} + 25 y^{^{'}} + 4 y = 40

y^{''}-y=sec^3(x)-sec(x)

y^{^{''}} - y = sec^{3} (x) - sec (x)

y^{''''}-5y^{''}+4y=8cos(2x)-12e^{-x}

y^{^{''''}} - 5 y^{^{''}} + 4 y = 8 cos (2 x) - 12 e^{- x}

x^{''}+x=9cos(8t)

x^{^{''}} + x = 9 cos (8 t)