y^{''}+4y=10x(4-x),y(0)=-5,y(4)=10

解

y^{^{''}} + 4 y = 10 x (4 - x), y (0) = - 5, y (4) = 10

y = - \frac{25}{4} cos (2 x) + \frac{( 35 + 25 cos ( 8 ) ) sin ( 2 x )}{4 sin ( 8 )} - \frac{5 x ^{2}}{2} + 10 x + \frac{5}{4}

解答ステップ

y^{''} (x) + 4 y = 10 x (4 - x)

線形 ODE を解く:

y = - \frac{25}{4} cos (2 x) + \frac{( 35 + 25 cos ( 8 ) ) sin ( 2 x )}{4 sin ( 8 )} - \frac{5 x ^{2}}{2} + 10 x + \frac{5}{4}

y = - \frac{25}{4} cos (2 x) + \frac{( 35 + 25 cos ( 8 ) ) sin ( 2 x )}{4 sin ( 8 )} - \frac{5 x ^{2}}{2} + 10 x + \frac{5}{4}