4y^{''}-28y^'+49y=0

解答

4 y^{^{''}} - 28 y^{^{'}} + 49 y = 0

y = c_{1} e^{\frac{7 t}{2}} + c_{2} e^{\frac{7 t}{2}} t

求解步骤

4 y^{''} (t) - 28 y^{'} (t) + 49 y = 0

求解线性方程:

y = c_{1} e^{\frac{7 t}{2}} + c_{2} e^{\frac{7 t}{2}} t

y = c_{1} e^{\frac{7 t}{2}} + c_{2} e^{\frac{7 t}{2}} t

y^{^{''}} + 6 y = 0, y (0) = 1, y^{^{'}} (0) = 6

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

\frac{d ^{2} f}{d x ^{2}} = - a^{2} f (x)

(D^{6} + 6 D^{4} + 9 D^{2}) y = 0

y^{^{''}} - 2 y^{^{'}} - 3 y = 0, y (0) = 5