y^{''}+4y^'=t^2+6e^t,y(0)=0,y^'(0)=9

解

y^{^{''}} + 4 y^{^{'}} = t^{2} + 6 e^{t}, y (0) = 0, y^{^{'}} (0) = 9

y = \frac{1243}{640} - \frac{6}{5} - \frac{1243}{640} e^{- 4 t} + \frac{t ^{3}}{12} - \frac{t ^{2}}{16} + \frac{t}{32} + \frac{6}{5} e^{t}

解答ステップ

y^{''} (t) + 4 y^{'} (t) = t^{2} + 6 e^{t}

線形 ODE を解く:

y = \frac{1243}{640} - \frac{6}{5} - \frac{1243}{640} e^{- 4 t} + \frac{t ^{3}}{12} - \frac{t ^{2}}{16} + \frac{t}{32} + \frac{6}{5} e^{t}

y = \frac{1243}{640} - \frac{6}{5} - \frac{1243}{640} e^{- 4 t} + \frac{t ^{3}}{12} - \frac{t ^{2}}{16} + \frac{t}{32} + \frac{6}{5} e^{t}