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Beliebt Rechnen >

y^{''}+2y^'=cos(t),y(0)=0

Lösung

y^{^{''}} + 2 y^{^{'}} = cos (t), y (0) = 0

Lösung

y = - c_{2} + \frac{1}{5} + c_{2} e^{- 2 t} + \frac{2}{5} sin (t) - \frac{1}{5} cos (t)

Schritte zur Lösung

y^{''} (t) + 2 y^{'} (t) = cos (t)

Solve linear ODE:

y = - c_{2} + \frac{1}{5} + c_{2} e^{- 2 t} + \frac{2}{5} sin (t) - \frac{1}{5} cos (t)

y = - c_{2} + \frac{1}{5} + c_{2} e^{- 2 t} + \frac{2}{5} sin (t) - \frac{1}{5} cos (t)

Graph

Beliebte Beispiele

y^{''}+7y^'=cos(2x)+8e^{-5x}

y^{^{''}} + 7 y^{^{'}} = cos (2 x) + 8 e^{- 5 x}

y^{''}+3y^'-4y=2,y^'(0)=0,y(0)=0

y^{^{''}} + 3 y^{^{'}} - 4 y = 2, y^{^{'}} (0) = 0, y (0) = 0

y^{''}+3y^'-4y=x^2e^x

y^{^{''}} + 3 y^{^{'}} - 4 y = x^{2} e^{x}

y^{''}-y^'-6y=e^{-x}+12x

y^{^{''}} - y^{^{'}} - 6 y = e^{- x} + 12 x

5y^{''}-15y=40e^x

5 y^{^{''}} - 15 y = 40 e^{x}