y^{''}-4y^'+4y=t^3e^2t,y(0)=0,y^'(0)=0

解

y^{^{''}} - 4 y^{^{'}} + 4 y = t^{3} e^{2} t, y (0) = 0, y^{^{'}} (0) = 0

y = - \frac{15 e ^{2}}{8} e^{2 t} + (- 3 e^{2} + \frac{15 e ^{2}}{4}) e^{2 t} t + \frac{e ^{2} t ^{4}}{4} + e^{2} t^{3} + \frac{9 e ^{2} t ^{2}}{4} + 3 e^{2} t + \frac{15 e ^{2}}{8}

解答ステップ

y^{''} (t) - 4 y^{'} (t) + 4 y = t^{3} e^{2} t

線形 ODE を解く:

y = - \frac{15 e ^{2}}{8} e^{2 t} + (- 3 e^{2} + \frac{15 e ^{2}}{4}) e^{2 t} t + \frac{e ^{2} t ^{4}}{4} + e^{2} t^{3} + \frac{9 e ^{2} t ^{2}}{4} + 3 e^{2} t + \frac{15 e ^{2}}{8}

y = - \frac{15 e ^{2}}{8} e^{2 t} + (- 3 e^{2} + \frac{15 e ^{2}}{4}) e^{2 t} t + \frac{e ^{2} t ^{4}}{4} + e^{2} t^{3} + \frac{9 e ^{2} t ^{2}}{4} + 3 e^{2} t + \frac{15 e ^{2}}{8}