y^{''}+25y=cos(5x)csc^2(5x)

Lösung

y^{^{''}} + 25 y = cos (5 x) csc^{2} (5 x)

y = c_{1} cos (5 x) + c_{2} sin (5 x) + \frac{- ln ( sin ( 5 x ) ) cos ( 5 x ) + sin ( 5 x ) ( - 5 x - cot ( 5 x ) )}{25}

Schritte zur Lösung

y^{''} (x) + 25 y = cos (5 x) csc^{2} (5 x)

Solve linear ODE:

y = c_{1} cos (5 x) + c_{2} sin (5 x) + \frac{- ln ( sin ( 5 x ) ) cos ( 5 x ) + sin ( 5 x ) ( - 5 x - cot ( 5 x ) )}{25}

y = c_{1} cos (5 x) + c_{2} sin (5 x) + \frac{- ln ( sin ( 5 x ) ) cos ( 5 x ) + sin ( 5 x ) ( - 5 x - cot ( 5 x ) )}{25}

y^{^{''}} + 6 y^{^{'}} + 9 y = 2 e^{- x}

y^{^{''}} - 2 y^{^{'}} - 3 y = 8 e^{x} - 6

y^{^{''}} + 7 y^{^{'}} + 12 y = e^{t}, y (0) = 0, y^{^{'}} (0) = 1

y^{^{'''}} - 2 y^{^{''}} - y^{^{'}} + 2 y = e^{7 t}

y^{^{''}} + 3 y^{^{'}} - 18 y = - 66 t + 36 t^{2}