I need help to find symmetries of 2nd order ODE $y''+y=0.$ Here is my work:
Criterion gives $\eta^{2} + \eta=0$ which I solved and got determining equations as
$1) Y': 2\eta_{xy} - \xi_{xx}+ 3y\xi_y=0$
$2) Y'^2: \eta_{yy}- 2\xi_{xy}=0$
$3) Y'^3: \xi_{yy}=0$
$4) Const: \eta_{xx}- y\eta_y+ 2y\xi_x+ \eta$
After solving this $(3)$ I got
$\xi=a(x)y+b(x)$
and solving $(2)$ yields
$\eta= a'(x)y^2+c(x)y+d(x)$
and $(1)$ gives
$a''+a=0 $ and $b''-2c'$
and $(4)$ gives
$a''+a'=0, c''+2b'=0, d''+d=0$
Where (,) denotes partial derivative w.r.t 'x'.
I solved it uptill here but here I got stuck the equations are just as same as the equation I started working with... Please if anybody can help me do it or can give a clue.. It will be highly acknowledged. (Apology for inconvenience while reading it.!)
Thanks in advance!