y^{''}-4y^'+20y=3x^2-1

Lösung

y^{^{''}} - 4 y^{^{'}} + 20 y = 3 x^{2} - 1

y = e^{2 x} (c_{1} cos (4 x) + c_{2} sin (4 x)) + \frac{3 x ^{2}}{20} + \frac{3 x}{50} - \frac{53}{1000}

Schritte zur Lösung

y^{''} (x) - 4 y^{'} (x) + 20 y = 3 x^{2} - 1

Solve linear ODE:

y = e^{2 x} (c_{1} cos (4 x) + c_{2} sin (4 x)) + \frac{3 x ^{2}}{20} + \frac{3 x}{50} - \frac{53}{1000}

y = e^{2 x} (c_{1} cos (4 x) + c_{2} sin (4 x)) + \frac{3 x ^{2}}{20} + \frac{3 x}{50} - \frac{53}{1000}

4 y^{^{'''}} - y^{^{''}} = 3 x

y^{^{''}} + 16 y^{^{'}} + 64 y = (36 x^{2} - 24 x + 12) e^{- 8 x}

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{t}, y (0) = 1, y^{^{'}} (0) = - 1

y^{^{''}} + 5 y^{^{'}} - 6 y = x e^{(- 5 x)}

y^{^{'''}} + 5 y^{^{''}} + 6 y^{^{'}} = 4 + 9 sin (x)