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

Lösung

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

y = - \frac{cos ( 2 x )}{2 x} + \frac{sin ( 2 x )}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

Schritte zur Lösung

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

Solve linear ODE:

y = - \frac{cos ( 2 x )}{2 x} + \frac{sin ( 2 x )}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

y = - \frac{cos ( 2 x )}{2 x} + \frac{sin ( 2 x )}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

\int_{- 2}^{1} x (4 - x^{2}) - (x + 2) d x

\int \frac{1}{2 ^{x} + 3} d x

n = 1 \sum \infty \frac{1}{n ( n + 5 )}

n = 0 \sum \infty \frac{i ^{n}}{3 ^{n}}

im pl i c i t \frac{e x}{y} = 9 x - y