y^{''}-2y+1=3e^x

Lösung

y^{^{''}} - 2 y + 1 = 3 e^{x}

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} - 3 e^{x} + \frac{1}{2}

Schritte zur Lösung

y^{''} (x) - 2 y + 1 = 3 e^{x}

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} - 3 e^{x} + \frac{1}{2}

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} - 3 e^{x} + \frac{1}{2}

(D^{2} + D + 1) y = x^{3}

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

y^{^{''}} + 4 y = 5 x + 16

3 y^{^{''}} + y = cos (t)

y^{^{''}} - 4 y = 3 e - 2 t