4x^{''}+x=t^2+2cos(t)

解

4 x^{^{''}} + x = t^{2} + 2 cos (t)

x = c_{1} cos (\frac{t}{2}) + c_{2} sin (\frac{t}{2}) + t^{2} - 8 - \frac{2}{3} cos (t)

解答ステップ

4 x^{''} (t) + x = t^{2} + 2 cos (t)

線形 ODE を解く:

x = c_{1} cos (\frac{t}{2}) + c_{2} sin (\frac{t}{2}) + t^{2} - 8 - \frac{2}{3} cos (t)

x = c_{1} cos (\frac{t}{2}) + c_{2} sin (\frac{t}{2}) + t^{2} - 8 - \frac{2}{3} cos (t)