y^{''}+49y=-5cos(7t+pi^3),y(0)=0,y^'(0)=0

解

y^{^{''}} + 49 y = - 5 cos (7 t + π^{3}), y (0) = 0, y^{^{'}} (0) = 0

y = - \frac{5 t sin ( 7 t + π ^{3} )}{14} + \frac{5 sin ( π ^{3} ) sin ( 7 t )}{98}

解答ステップ

y^{''} (t) + 49 y = - 5 cos (7 t + π^{3})

線形 ODE を解く:

y = - \frac{5 t sin ( 7 t + π ^{3} )}{14} + \frac{5 sin ( π ^{3} ) sin ( 7 t )}{98}

y = - \frac{5 t sin ( 7 t + π ^{3} )}{14} + \frac{5 sin ( π ^{3} ) sin ( 7 t )}{98}