y^{''}-2y^'+2y=e^x(1/(sin(x)))

Lösung

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{x} (\frac{1}{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 = e^{x} (\frac{1}{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)

y^{^{''}} - y^{^{'}} - 6 y = 5 t, y (0) = 3, y^{^{'}} (0) = 1

so l v e f or D, D (x) = x^{8} - 4^{4}

x^{^{''}} + 2 x^{^{'}} + 3 x = t^{2} + 4 t

y^{^{''}} - 2 y^{^{'}} + y = (e^{x} - 1)^{2}, y (0) = 0, y^{^{'}} (0) = 1

2 y^{^{''}} + 3 y^{^{'}} + y = g t