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

Lösung

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

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{e ^{2 x} x}{16} + \frac{1}{32} e^{2 x}

Schritte zur Lösung

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

Solve linear ODE:

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{e ^{2 x} x}{16} + \frac{1}{32} e^{2 x}

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{e ^{2 x} x}{16} + \frac{1}{32} e^{2 x}

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

y^{^{''}} - 8 y^{^{'}} + 16 y = 4 e^{4 x} ln (x)

y^{^{''}} - 3 y^{^{'}} + 2 y = \frac{1}{t}, y (0) = 1, y^{^{'}} (0) = 2

y^{^{''}} - 2 y^{^{'}} + y = x^{3} + 9 x

y^{^{''}} - 8 y^{^{'}} + 25 y = 4 x^{3} e^{- x} - 7 e^{- x}