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

Lösung

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

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

Schritte zur Lösung

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

Solve linear ODE:

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

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

\int_{0}^{π} sin^{4} (3 x) d x

\int - 1 d x

\frac{\partial}{\partial x} (cos (2 x - 3 y))

\int cos (x) \cdot sin (2 x) d x

\frac{\partial}{\partial x} (cos (2 x y))