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

Lösung

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}}

Schritte zur Lösung

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

Solve linear ODE:

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}}

- y^{^{''}} + 2 y^{^{'}} + 3 y = 1

y^{^{''}} - y^{^{'}} = e^{x} (3 sin (x) + cos (x))

4 \frac{d ^{2} y}{d x ^{2}} + 9 y = 15

y^{^{''}} - y = sin (t), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 5 y^{^{'}} + 6 y = 4 x + 5