ty^'+2y= 2/(1+t^2)

Lösung

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

y = \frac{ln ( 1 + t ^{2} )}{t ^{2}} + \frac{c _{1}}{t ^{2}}

Schritte zur Lösung

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

Solve linear ODE:

y = \frac{ln ( 1 + t ^{2} )}{t ^{2}} + \frac{c _{1}}{t ^{2}}

y = \frac{ln ( 1 + t ^{2} )}{t ^{2}} + \frac{c _{1}}{t ^{2}}

\int 3 5 - 4 x^{2} (- 8 x) d x

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

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

\int \frac{( e ^{x} - e ^{- x} )}{e ^{2 x}} d x

\frac{\partial}{\partial s} (s cos (t))