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

Lösung

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

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

Schritte zur Lösung

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

Solve linear ODE:

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

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

(d^{2} - 13 d + 12) \cdot y = x

(d - 1)^{2} \cdot y = e^{x} \cdot sec^{2} (x) tan (x)

(d^{2} + 3 d + 2) y = sin (e^{x})

(d^{3} + 8) y = x^{4} + x + 1

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