9x^{''}+9x^'-4x=0

解答

9 x^{^{''}} + 9 x^{^{'}} - 4 x = 0

x = c_{1} e^{\frac{t}{3}} + c_{2} e^{- \frac{4 t}{3}}

求解步骤

9 x^{''} (t) + 9 x^{'} (t) - 4 x = 0

求解线性方程:

x = c_{1} e^{\frac{t}{3}} + c_{2} e^{- \frac{4 t}{3}}

x = c_{1} e^{\frac{t}{3}} + c_{2} e^{- \frac{4 t}{3}}

y^{^{''}} + y = 0, y (π) = 1, y (0) = 1

a \cdot y^{^{''}} + b \cdot y = 0

y^{^{''}} - 8 y^{^{'}} + 16 y = 0, y (0) = 4, y^{^{'}} (0) = 4

y^{^{''}} + 8 y^{^{'}} + 15 y = 0, y (0) = 12, y^{^{'}} (0) = - 52

3 (\frac{d ^{2} y}{d x ^{2}}) + 19 (\frac{d y}{d x}) - 14 y = 0