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

Lösung

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

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

Schritte zur Lösung

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

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} + y = \frac{e ^{x}}{x ^{5}}

y^{^{''}} + 6 y^{^{'}} + 9 y = t

y^{^{''}} - y = 5 e^{4 t} + 6 t e^{4 t} + 3 t^{2} e^{4 t}

y^{^{''}} + 42 y = 42 cos (42 t)

y^{^{''}} - 4 y^{^{'}} + 3 y = 2 e^{5 x}, y (0) = 1, y (0) = - 2