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

Lösung

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

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

Schritte zur Lösung

y^{''} (x) - 4 y^{'} (x) + 8 y = 8 x^{2} - 2

Solve linear ODE:

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

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

y^{^{''}} - 2 y^{^{'}} + 2 y = \frac{( e ^{x} )}{sin ( x )}

(D^{2} + 1) y = sec (x) tan (x)

0.5 q^{^{''}} + 6 q^{^{'}} + \frac{1}{0.02} q = 24 sin (10 t)

y^{^{''}} - 3 y^{^{'}} + 2 y = 13 e^{3 x}

y^{^{''}} - y = e^{t} sin (t), y (0) = 0, y^{^{'}} (0) = 0