4y^{''}+20y^'+29y=0,y(1)=8

解答

4 y^{^{''}} + 20 y^{^{'}} + 29 y = 0, y (1) = 8

y = e^{- \frac{5 t}{2}} \frac{cos ( t ) ( 8 e ^{\frac{5}{2}} - sin ( 1 ) c _{2} )}{cos ( 1 )} + c_{2} sin (t)

求解步骤

4 y^{''} (t) + 20 y^{'} (t) + 29 y = 0

求解线性方程:

y = e^{- \frac{5 t}{2}} \frac{cos ( t ) ( 8 e ^{\frac{5}{2}} - sin ( 1 ) c _{2} )}{cos ( 1 )} + c_{2} sin (t)

y = e^{- \frac{5 t}{2}} \frac{cos ( t ) ( 8 e ^{\frac{5}{2}} - sin ( 1 ) c _{2} )}{cos ( 1 )} + c_{2} sin (t)

y^{^{''}} - 3 y^{^{'}} + α y = 0, y (0) = 1, y^{^{'}} (0) = 5

6 y^{^{''}} + y^{^{'}} - 2 y = 0, y (1) = 2, y^{^{'}} (1) = \frac{1}{6}

x^{^{''}} - 0.1 x^{^{'}} + 1.0025 x = 0

y^{^{'''}} + 6 y = 0

y^{^{''}} + 120 y^{^{'}} + 10000 y = 0