4x^{''}-4x^'+x= 8/(t^2)e^{t/2}

解

4 x^{^{''}} - 4 x^{^{'}} + x = \frac{8}{t ^{2}} e^{\frac{t}{2}}

x = c_{1} e^{\frac{t}{2}} + c_{2} e^{\frac{t}{2}} t - 2 e^{\frac{t}{2}} ln (t) - 2 e^{\frac{t}{2}}

解答ステップ

4 x^{''} (t) - 4 x^{'} (t) + x = \frac{8}{t ^{2}} e^{\frac{t}{2}}

線形 ODE を解く:

x = c_{1} e^{\frac{t}{2}} + c_{2} e^{\frac{t}{2}} t - 2 e^{\frac{t}{2}} ln (t) - 2 e^{\frac{t}{2}}

x = c_{1} e^{\frac{t}{2}} + c_{2} e^{\frac{t}{2}} t - 2 e^{\frac{t}{2}} ln (t) - 2 e^{\frac{t}{2}}