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

解

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

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{t ^{2}}{2} + t + \frac{3}{4} + \frac{2 e ^{2 t} t ^{3}}{3} + \frac{t cos ( 2 t )}{8} + \frac{1}{16} cos (2 t) - \frac{1}{16} sin (2 t)

解答ステップ

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

線形 ODE を解く:

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{t ^{2}}{2} + t + \frac{3}{4} + \frac{2 e ^{2 t} t ^{3}}{3} + \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}}{2} + t + \frac{3}{4} + \frac{2 e ^{2 t} t ^{3}}{3} + \frac{t cos ( 2 t )}{8} + \frac{1}{16} cos (2 t) - \frac{1}{16} sin (2 t)