y^{''}-4y^'+5y=(2e^{2x})/(sin(x))

Lösung

y^{^{''}} - 4 y^{^{'}} + 5 y = \frac{2 e ^{2 x}}{sin ( x )}

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

Schritte zur Lösung

y^{''} (x) - 4 y^{'} (x) + 5 y = \frac{2 e ^{2 x}}{sin ( x )}

Solve linear ODE:

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

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

y^{^{''}} + y^{^{'}} - 2 y = 12 cos (2 x)

y^{^{''}} + 9 y = x e^{3 x}

(D^{2} - 3 D + 2) y = 72 x e^{- x}

y^{^{''}} + 6 y^{^{'}} + 8 = (3 x - 2) e^{4 x}

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{- 4 t}, y (0) = 1, y (0) = 5