y^{''}-4y^'+8y=16x

Lösung

y^{^{''}} - 4 y^{^{'}} + 8 y = 16 x

y = e^{2 x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + 2 x + 1

Schritte zur Lösung

y^{''} (x) - 4 y^{'} (x) + 8 y = 16 x

Solve linear ODE:

y = e^{2 x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + 2 x + 1

y = e^{2 x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + 2 x + 1

(d^{2} + 1) y = x^{3}

y^{^{''}} + 4 y = 8 x^{2} + 9 x

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

y^{^{''}} - 14 y^{^{'}} + 49 y = 49 x + 4

y^{^{''}} + 3 y^{^{'}} = 7