y^{''}+4y=13sec^2(2t)

Lösung

y^{^{''}} + 4 y = 13 sec^{2} (2 t)

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{- 13 cos ( 2 t ) ( cos ( 2 t ) + sin ( 2 t ) tan ( 2 t ) ) + 13 ln ( tan ( 2 t ) + sec ( 2 t ) ) sin ( 2 t )}{4}

Schritte zur Lösung

y^{''} (t) + 4 y = 13 sec^{2} (2 t)

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{- 13 cos ( 2 t ) ( cos ( 2 t ) + sin ( 2 t ) tan ( 2 t ) ) + 13 ln ( tan ( 2 t ) + sec ( 2 t ) ) sin ( 2 t )}{4}

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{- 13 cos ( 2 t ) ( cos ( 2 t ) + sin ( 2 t ) tan ( 2 t ) ) + 13 ln ( tan ( 2 t ) + sec ( 2 t ) ) sin ( 2 t )}{4}

y^{^{''}} + y^{^{'}} = 3 + e^{x}

y^{^{''}} + 3 y^{^{'}} + 2 y = t, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 9 y = x^{2} e^{3 x} + 6

y^{^{''}} - y^{^{'}} - 2 y = 4 x^{2}, y (0) = 1, y^{^{'}} (0) = 4

y^{^{''}} - 2 y^{^{'}} - 3 y = 4 e x - 12