y^{''}-4y^'=cosh(2x)

Lösung

y^{^{''}} - 4 y^{^{'}} = cosh (2 x)

y = c_{1} e^{4 x} + c_{2} + \frac{1}{8} e^{4 x} (- \frac{1}{6} e^{- 6 x} - \frac{1}{2} e^{- 2 x}) - \frac{1}{8} sinh (2 x)

Schritte zur Lösung

y^{''} (x) - 4 y^{'} (x) = cosh (2 x)

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} + \frac{1}{8} e^{4 x} (- \frac{1}{6} e^{- 6 x} - \frac{1}{2} e^{- 2 x}) - \frac{1}{8} sinh (2 x)

y = c_{1} e^{4 x} + c_{2} + \frac{1}{8} e^{4 x} (- \frac{1}{6} e^{- 6 x} - \frac{1}{2} e^{- 2 x}) - \frac{1}{8} sinh (2 x)

y^{^{''}} + 12 y^{^{'}} + 27 y = 64 e^{- 1 t}, y (0) = 5

y^{^{''}} - y = e^{t} sin (t)

y^{^{''}} + 3 y^{^{'}} - 4 y = 3 e^{2} x

y^{^{''}} - 2 y^{^{'}} - 3 y = e^{x}

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