y^{''}-3y^'+2y=e^3xcos(e^x)

Lösung

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{3} x cos (e^{x})

y = c_{1} e^{2 x} + c_{2} e^{x} + e^{2 x} \cdot \int e^{- 2 x + 3} x cos (e^{x}) d x + e^{x} \cdot \int - e^{- x + 3} x cos (e^{x}) d x

Schritte zur Lösung

y^{''} (x) - 3 y^{'} (x) + 2 y = e^{3} x cos (e^{x})

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{x} + e^{2 x} \cdot \int e^{- 2 x + 3} x cos (e^{x}) d x + e^{x} \cdot \int - e^{- x + 3} x cos (e^{x}) d x

y = c_{1} e^{2 x} + c_{2} e^{x} + e^{2 x} \cdot \int e^{- 2 x + 3} x cos (e^{x}) d x + e^{x} \cdot \int - e^{- x + 3} x cos (e^{x}) d x

y^{^{''}} + 5 y^{^{'}} + y = 4 t^{2} + 8 t + 8

y^{^{''}} + 3 y^{^{'}} = x^{2} sin (x) cos (x)

y^{^{''}} + y = 4 x

2 y^{^{''}} + 6 y^{^{'}} + 1 = 95

x^{^{''}} + 2 x^{^{'}} + 41 x = 11 cos (6 t)