y^{''}-3y^'-4y=t^2,y(0)=1,y^'(0)=4

解

y^{^{''}} - 3 y^{^{'}} - 4 y = t^{2}, y (0) = 1, y^{^{'}} (0) = 4

y = (- \frac{2}{5} + \frac{45}{32}) e^{4 t} + \frac{2}{5} e^{- t} - \frac{t ^{2}}{4} + \frac{3 t}{8} - \frac{13}{32}

解答ステップ

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

線形 ODE を解く:

y = (- \frac{2}{5} + \frac{45}{32}) e^{4 t} + \frac{2}{5} e^{- t} - \frac{t ^{2}}{4} + \frac{3 t}{8} - \frac{13}{32}

y = (- \frac{2}{5} + \frac{45}{32}) e^{4 t} + \frac{2}{5} e^{- t} - \frac{t ^{2}}{4} + \frac{3 t}{8} - \frac{13}{32}