a^2y+y^{''}=cos(bt),y(0)=0,y^'(0)=0

解

a^{2} \cdot y + y^{^{''}} = cos (b \cdot t), y (0) = 0, y^{^{'}} (0) = 0

y = \frac{cos ( b t )}{a ^{2} - b ^{2}} - \frac{cos ( a t )}{a ^{2} - b ^{2}}

解答ステップ

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

線形 ODE を解く:

y = \frac{cos ( b t )}{a ^{2} - b ^{2}} - \frac{cos ( a t )}{a ^{2} - b ^{2}}

y = \frac{cos ( b t )}{a ^{2} - b ^{2}} - \frac{cos ( a t )}{a ^{2} - b ^{2}}