y^{''}-2y^'+y=(3e^t)/(1+t^2)

Lösung

y^{^{''}} - 2 y^{^{'}} + y = \frac{3 e ^{t}}{1 + t ^{2}}

y = c_{1} e^{t} + c_{2} t e^{t} - \frac{3}{2} e^{t} ln (1 + t^{2}) + 3 e^{t} t arctan (t)

Schritte zur Lösung

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

Solve linear ODE:

y = c_{1} e^{t} + c_{2} t e^{t} - \frac{3}{2} e^{t} ln (1 + t^{2}) + 3 e^{t} t arctan (t)

y = c_{1} e^{t} + c_{2} t e^{t} - \frac{3}{2} e^{t} ln (1 + t^{2}) + 3 e^{t} t arctan (t)

y^{^{''}} + y = 4 x + 10 sin (x), y (π) = 0, y^{^{'}} (π) = 2

9 y^{^{''}} - 6 y^{^{'}} + y = cos (t)

y^{^{''}} - y = c^{x} + 2 c^{2 x}

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

y^{^{''}} - 11 y^{^{'}} + 10 y = 2 x^{2} - 44 x + 4