exo7 846

$y'+2xy=0,\ \ \ M=(0,1)$,

$y'+y \tan x = \sin x \cos x \ \ \ M=(\frac{\pi}{4},0)$,

$x(x^2-1)y'+2y=x^2$, On déterminera $a,b,c \in \Rr$ tels que $\frac{1}{x(x^2-1)}=\frac{a}{x}+\frac{b}{x-1}+\frac{c}{x+1}$.