y^{''}+2ay^'+b^2y=sin(ct)

解

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

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

解答ステップ

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

線形 ODE を解く:

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

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