u^{''}+\H(t)=0,\H(0)=0,u^'(0)=2

解答

u^{^{''}} + H (t) = 0, u^{^{'}} (0) = 2

u (t) = \int \int - H (t) d t d t

求解步骤

u^{''} (t) + H (t) = 0

整理

u^{''} (t) : u^{''} (t) = - H (t)

解

u^{''} (t) = - H (t) : u^{'} (t) = \int - H (t) d t

解

u^{'} (t) = \int - H (t) d t : u (t) = \int \int - H (t) d t d t

u (t) = \int \int - H (t) d t d t

2 y^{^{''}} + 2 y^{^{'}} + 8 y = 0

\frac{d ^{2} y}{d x ^{2}} + 7 \frac{d y}{d x} - 8 y = 0

y^{^{''}} - 16 y^{^{'}} + 113 y = 0

\frac{d ^{4} x}{d t ^{4}} - x = 0

y^{^{''''}} - 44 y^{^{'''}} + 484 y^{^{''}} = 0