y^{''}-4y^'+4y=2t^2+4+e^{2t}+tsin(2t)

解答

y^{^{''}} - 4 y^{^{'}} + 4 y = 2 t^{2} + 4 + e^{2 t} + t sin (2 t)

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{t ^{2} + e ^{2 t} t ^{2}}{2} + t + \frac{7}{4} + \frac{t cos ( 2 t )}{8} + \frac{1}{16} cos (2 t) - \frac{1}{16} sin (2 t)

求解步骤

y^{''} (t) - 4 y^{'} (t) + 4 y = 2 t^{2} + 4 + e^{2 t} + t sin (2 t)

求解线性方程:

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{t ^{2} + e ^{2 t} t ^{2}}{2} + t + \frac{7}{4} + \frac{t cos ( 2 t )}{8} + \frac{1}{16} cos (2 t) - \frac{1}{16} sin (2 t)

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{t ^{2} + e ^{2 t} t ^{2}}{2} + t + \frac{7}{4} + \frac{t cos ( 2 t )}{8} + \frac{1}{16} cos (2 t) - \frac{1}{16} sin (2 t)

y^{^{''}} + 9 y = 4 (\frac{\frac{u ^{5}}{3} π - u ^{13}}{3 π})

y^{^{''}} + y^{^{'}} = sinh (t)

y^{^{''}} - 2 \cdot y^{^{'}} + 2 \cdot y = cos (t), y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} + 6 y = t, y (0) = - 3, y^{^{'}} (0) = 0

3 y^{^{''}} + 2 y^{^{'}} + 1 = 0