y^{''}+2y^'+2y=0,y(pi/2)=0,y^'(pi/2)=4

解答

y^{^{''}} + 2 y^{^{'}} + 2 y = 0, y (\frac{π}{2}) = 0, y^{^{'}} (\frac{π}{2}) = 4

y = - 4 e^{\frac{π}{2} - t} cos (t)

求解步骤

y^{''} (t) + 2 y^{'} (t) + 2 y = 0

求解线性方程:

y = - 4 e^{\frac{π}{2} - t} cos (t)

y = - 4 e^{\frac{π}{2} - t} cos (t)

33 y^{^{''''}} + 188 y^{^{''}} + 64 y = 0

\frac{d ^{2} y}{d x ^{2}} + 3 \frac{d y}{d x} + y = 0

y^{^{''}} - 6 y^{^{'}} + 34 y = 0, y (0) = 5, y^{^{'}} (0) = - 5

y^{^{''}} - 10 y^{^{'}} + 50 y = 0, y (0) = - 4, y^{^{'}} (0) = - 5

9 y^{^{''}} - 12 y^{^{'}} + 4 y = 0, y (0) = e^{(\frac{2 x}{3})}