y^{''}+4y=17sec^2(2t)

Lösung

y^{^{''}} + 4 y = 17 sec^{2} (2 t)

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{- 17 cos ( 2 t ) ( cos ( 2 t ) + sin ( 2 t ) tan ( 2 t ) ) + 17 ln ( tan ( 2 t ) + sec ( 2 t ) ) sin ( 2 t )}{4}

Schritte zur Lösung

y^{''} (t) + 4 y = 17 sec^{2} (2 t)

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{- 17 cos ( 2 t ) ( cos ( 2 t ) + sin ( 2 t ) tan ( 2 t ) ) + 17 ln ( tan ( 2 t ) + sec ( 2 t ) ) sin ( 2 t )}{4}

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{- 17 cos ( 2 t ) ( cos ( 2 t ) + sin ( 2 t ) tan ( 2 t ) ) + 17 ln ( tan ( 2 t ) + sec ( 2 t ) ) sin ( 2 t )}{4}

y^{^{''}} + b y^{^{'}} = a

4 y^{^{''''}} - y^{^{''}} = 6 x - 1

y^{^{''}} + y^{^{'}} - 2 y = x^{2} + cos (3 x)

(d^{2} + 1) y = csc (x)

y^{^{''}} + 1.44 y = 24 cos (1.2 x)