y^{''}+y=(t-u),y(0)=10,y^'(0)=0

解答

y^{^{''}} + y = (t - u), y (0) = 10, y^{^{'}} (0) = 0

y = cos (t) (10 + u) - sin (t) + t - u

求解步骤

y^{''} (t) + y = (t - u)

求解线性方程:

y = cos (t) (10 + u) - sin (t) + t - u

y = cos (t) (10 + u) - sin (t) + t - u

y^{^{''}} + 4 y^{^{'}} + 13 y = cos (3 x) - sin (3 x)

- x^{^{''}} - 3 x^{^{'}} - 2 = t

y^{^{''}} - y = - 4 t + 3

x^{^{''}} + x = e^{- 2 t} sin (t)

y^{^{''}} + 6 y^{^{'}} + 9 y = 2 e^{- 2 x} + sin (2 x)