y^{''}+2y^'+2y=acos(bt)

解

y^{^{''}} + 2 y^{^{'}} + 2 y = a cos (b t)

y = e^{- t} (c_{1} cos (t) + c_{2} sin (t)) + \frac{2 ab sin ( b t )}{b ^{4} + 4} - \frac{a cos ( b t ) ( b + 2 ) ( b - 2 )}{( b ^{2} + 2 b + 2 ) ( b ^{2} - 2 b + 2 )}

解答ステップ

y^{''} (t) + 2 y^{'} (t) + 2 y = a cos (b t)

線形 ODE を解く:

y = e^{- t} (c_{1} cos (t) + c_{2} sin (t)) + \frac{2 ab sin ( b t )}{b ^{4} + 4} - \frac{a cos ( b t ) ( b + 2 ) ( b - 2 )}{( b ^{2} + 2 b + 2 ) ( b ^{2} - 2 b + 2 )}

y = e^{- t} (c_{1} cos (t) + c_{2} sin (t)) + \frac{2 ab sin ( b t )}{b ^{4} + 4} - \frac{a cos ( b t ) ( b + 2 ) ( b - 2 )}{( b ^{2} + 2 b + 2 ) ( b ^{2} - 2 b + 2 )}