4y^{''}-4y^'+y=16e^{1/2 t}

Lösung

4 y^{^{''}} - 4 y^{^{'}} + y = 16 e^{\frac{1}{2} t}

y = c_{1} e^{\frac{t}{2}} + c_{2} e^{\frac{t}{2}} t + 2 e^{\frac{t}{2}} t^{2}

Schritte zur Lösung

4 y^{''} (t) - 4 y^{'} (t) + y = 16 e^{\frac{1}{2} t}

Solve linear ODE:

y = c_{1} e^{\frac{t}{2}} + c_{2} e^{\frac{t}{2}} t + 2 e^{\frac{t}{2}} t^{2}

y = c_{1} e^{\frac{t}{2}} + c_{2} e^{\frac{t}{2}} t + 2 e^{\frac{t}{2}} t^{2}

16 y^{^{''''}} - y = e^{\frac{x}{2}}

y^{^{''}} + 5 y^{^{'}} + 6 y = 36 e^{t}, y (0) = - 5, y^{^{'}} (0) = 21

y^{^{''}} - 6 y^{^{'}} + 8 y = 2 x - 1

y^{^{''}} = 2 y + 22 cot^{3} (x), y (0) = 11 cot (x)

2 y^{^{''}} + 18 y + 5 = 0, y (0) = 0, y^{^{'}} (0) = 0