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

Lösung

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

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

Schritte zur Lösung

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

Solve linear ODE:

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

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

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