4y^{''}-21y^'+20y=0,y(0)=0

解答

4 y^{^{''}} - 21 y^{^{'}} + 20 y = 0, y (0) = 0

y = - c_{2} e^{4 t} + c_{2} e^{\frac{5 t}{4}}

求解步骤

4 y^{''} (t) - 21 y^{'} (t) + 20 y = 0

求解线性方程:

y = - c_{2} e^{4 t} + c_{2} e^{\frac{5 t}{4}}

y = - c_{2} e^{4 t} + c_{2} e^{\frac{5 t}{4}}

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} - 5 y = 0

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

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