User:Egm6321.f11.team2.Xia/RP3.15

= R*3.15 - Second Exact condition =

Given
Given the following equations (1)p.18-3 $$ \displaystyle (15p^{4}\cos x^{2})y''+(6xy^{2})y'+[-6xy^{5}\sin x^{2}+2y^{2}]=0$$ (3.15.1) $$ \displaystyle f_{xx}+2pf_{xy}+p^{2}f_{yy}=g_{xp}+pg_{yp}-g_{y}$$

(3.15.2) $$ \displaystyle f_{xp}+pf_{yp}+2f_{y}=g_{pp}$$

(3.15.3)

Find
Verify whether (3.15.1) satisfies the 2nd exactness condition (3.15.2) and (3.15.3).

Solution
According to Prof. Loc Vu-Quoc's lecture note, we know that $$ \displaystyle \phi_{x}=3p^{5}(-\sin x^{2})(2x)+2y^{3}$$

(3.15.4)

$$ \displaystyle \phi_{y}=6xy^{2}$$

(3.15.5)

$$ \displaystyle f:=\phi_{p}=15p^{4}\cos x^{2}$$ (3.15.6)

$$ \displaystyle g:=\phi_{y}y'+\phi_{x}=(6xy^{2})y'+[-6xp^{5}\sin x^{2}+2y^{3}$$ (3.15.7)

Hence

$$ \displaystyle f_{xx}=(\phi_{p})_{xx}=-30p^{4}\sin x^{2}-60x^{2}p^{4}\cos x^{2}$$ (3.15.8) $$ \displaystyle f_{xp}=-120xp^{3}\sin x^{2},\, f_{xy}=0,\, f_{yy}=0,\, f_{y}=0,\, f_{yp}=0$$ (3.15.9) $$ \displaystyle g_{y}=(12xy)y'+6y^{2},\, g_{yp}=0$$ (3.15.10) $$ \displaystyle g_{xp}=-30p^{4}\sin x^{2}-60x^{2}p^{4}\cos x^{2},\,g_{pp}=-120xp^{3}\sin x^{2}$$ (3.15.11) Substitute (3.15.8)-(3.15.11) to (3.15.2) $$ \displaystyle LHS=-30p^{4}\sin x^{2}-60x^{2}p^{4}\cos x^{2}+2p\times0+p^{2}\times0$$ $$ \displaystyle LHS=-30p^{4}\sin x^{2}-60x^{2}p^{4}\cos x^{2}$$ (3.15.12) $$ \displaystyle RHS=-30p^{4}\sin x^{2}-60x^{2}p^{4}\cos x^{2}+p\times0-((12xy)y'+6y^{2})$$ $$ \displaystyle RHS=-30p^{4}\sin x^{2}-60x^{2}p^{4}\cos x^{2}-((12xy)y'+6y^{2})$$ (3.15.13) Thus $$ \displaystyle LHS\neq RHS$$, then (3.15.11) does not satisfy (3.15.2), but we can easily find that (3.15.1) does satisfy (3.15.2)