y^{''}-2y^'+10y=excos(3x)

Lösung

y^{^{''}} - 2 y^{^{'}} + 10 y = e x cos (3 x)

y = e^{x} (c_{1} cos (3 x) + c_{2} sin (3 x)) - \frac{6 e x sin ( 3 x )}{37} + \frac{e x cos ( 3 x )}{37} - \frac{234 e}{1369} sin (3 x) + \frac{2 e}{1369} cos (3 x)

Schritte zur Lösung

y^{''} (x) - 2 y^{'} (x) + 10 y = e x cos (3 x)

Solve linear ODE:

y = e^{x} (c_{1} cos (3 x) + c_{2} sin (3 x)) - \frac{6 e x sin ( 3 x )}{37} + \frac{e x cos ( 3 x )}{37} - \frac{234 e}{1369} sin (3 x) + \frac{2 e}{1369} cos (3 x)

y = e^{x} (c_{1} cos (3 x) + c_{2} sin (3 x)) - \frac{6 e x sin ( 3 x )}{37} + \frac{e x cos ( 3 x )}{37} - \frac{234 e}{1369} sin (3 x) + \frac{2 e}{1369} cos (3 x)

5 y^{^{''}} - 2 y^{^{'}} + 80 y = 24 sin (4 t)

y^{^{''}} - 4 y = - 4

x^{^{''}} + x^{^{'}} = e^{- t}

y^{^{''}} + 49 y = 8 cos (7 x) (3 x + 2)

y^{^{''}} + y^{^{'}} - 6 y = t - 2 π, y (0) = 1, y^{^{'}} (0) = 2