3y^{''}+2g^2y=4.9gcos(8t)

Lösung

3 y^{^{''}} + 2 g^{2} y = 4.9 g cos (8 t)

y = c_{1} cos (\frac{2 g t}{3}) + c_{2} sin (\frac{2 g t}{3}) + \frac{49 g cos ( 8 t )}{20 ( g ^{2} - 96 )}

Schritte zur Lösung

3 y^{''} (t) + 2 g^{2} y = 4.9 g cos (8 t)

Solve linear ODE:

y = c_{1} cos (\frac{2 g t}{3}) + c_{2} sin (\frac{2 g t}{3}) + \frac{49 g cos ( 8 t )}{20 ( g ^{2} - 96 )}

y = c_{1} cos (\frac{2 g t}{3}) + c_{2} sin (\frac{2 g t}{3}) + \frac{49 g cos ( 8 t )}{20 ( g ^{2} - 96 )}