y^{''}+25y=3csc^2(5t)

Lösung

y^{^{''}} + 25 y = 3 csc^{2} (5 t)

y = c_{1} cos (5 t) + c_{2} sin (5 t) + \frac{- 3 ln ( tan ( \frac{5 t}{2} ) ) cos ( 5 t ) - 3}{25}

Schritte zur Lösung

y^{''} (t) + 25 y = 3 csc^{2} (5 t)

Solve linear ODE:

y = c_{1} cos (5 t) + c_{2} sin (5 t) + \frac{- 3 ln ( tan ( \frac{5 t}{2} ) ) cos ( 5 t ) - 3}{25}

y = c_{1} cos (5 t) + c_{2} sin (5 t) + \frac{- 3 ln ( tan ( \frac{5 t}{2} ) ) cos ( 5 t ) - 3}{25}

\frac{d ^{2} y}{d x ^{2}} - y = x

y^{^{''}} - 3 y^{^{'}} + 2 y = 12

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

(D^{2} - 4 D + 4) y = e^{x}

y^{^{''}} - 3 y^{^{'}} + 2 y = 12 e^{- 2 t}