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

Lösung

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

y = c_{1} e^{4 x} + c_{2} e^{2 x} + \frac{x}{4} + \frac{1}{16}

Schritte zur Lösung

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

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} e^{2 x} + \frac{x}{4} + \frac{1}{16}

y = c_{1} e^{4 x} + c_{2} e^{2 x} + \frac{x}{4} + \frac{1}{16}

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

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

so l v e f or y, (d^{3} + 4 d) y = 2 t + t^{2}

y^{^{''}} + y = 2 \cdot e^{x}

y^{^{''}} - y^{^{'}} - 6 y = 10 x e^{3 x} + 17 e^{3 x}