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

Lösung

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

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

Schritte zur Lösung

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

Solve linear ODE:

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

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

y^{^{''}} + 4 y^{^{'}} + 4 y = x^{2} + e^{- 2 x} + 1

y^{^{''}} + 8 y^{^{'}} + 16 y = \frac{2 e ^{- 4 x}}{x ^{2}}

y^{^{''}} - 2 y^{^{'}} + y = t^{- 6} e^{t}

y^{^{''''}} - 5 y^{^{''}} + 4 y = 10 cos (x)

- 9 y^{^{''}} + y = 9 t - 4 e^{t} + 9