y^{''}-10y^'+24y=48x^2+10

Lösung

y^{^{''}} - 10 y^{^{'}} + 24 y = 48 x^{2} + 10

y = c_{1} e^{6 x} + c_{2} e^{4 x} + 2 x^{2} + \frac{5 x}{3} + \frac{17}{18}

Schritte zur Lösung

y^{''} (x) - 10 y^{'} (x) + 24 y = 48 x^{2} + 10

Solve linear ODE:

y = c_{1} e^{6 x} + c_{2} e^{4 x} + 2 x^{2} + \frac{5 x}{3} + \frac{17}{18}

y = c_{1} e^{6 x} + c_{2} e^{4 x} + 2 x^{2} + \frac{5 x}{3} + \frac{17}{18}

y^{^{''}} + 4 y = t^{2} + 10 e^{t}, y (0) = 0, y^{^{'}} (0) = 3

y^{^{''}} + y = 4 x \cdot sin (x)

y^{^{''}} - 10 y^{^{'}} + 25 y = \frac{1.5 e ^{5 t}}{t ^{2} + 1}

y^{^{''}} - 3 y^{^{'}} = 1 + e^{x} + cos (x) + sin (x)

y^{^{''}} + 4 y^{^{'}} + 3 y = 3 e^{x} + x