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

Lösung

y^{^{''}} - 2 y^{^{'}} + y = \frac{e ^{x} ln ( x )}{x}

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

Schritte zur Lösung

y^{''} (x) - 2 y^{'} (x) + y = \frac{e ^{x} ln ( x )}{x}

Solve linear ODE:

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

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

y^{^{''}} - 2 y^{^{'}} = 5 t^{4}, y (0) = 1, y^{^{'}} (0) = 0

y^{^{'''}} - 4 y^{^{''}} + 5 y^{^{'}} - 2 y = 5 e^{- x} + sin (x)

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} = x - 1

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

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