ty^{''}-y^'=t^2+t,y(1)=1,y^'(1)=5

解

t y^{^{''}} - y^{^{'}} = t^{2} + t, y (1) = 1, y^{^{'}} (1) = 5

y = (\frac{13}{12} + \frac{11}{12}) t^{2} - \frac{13}{12} + \frac{4 t ^{3} + 6 t ^{2} ln ( t ) - 3 t ^{2}}{12}

解答ステップ

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

線形 ODE を解く:

y = (\frac{13}{12} + \frac{11}{12}) t^{2} - \frac{13}{12} + \frac{4 t ^{3} + 6 t ^{2} ln ( t ) - 3 t ^{2}}{12}

y = (\frac{13}{12} + \frac{11}{12}) t^{2} - \frac{13}{12} + \frac{4 t ^{3} + 6 t ^{2} ln ( t ) - 3 t ^{2}}{12}