x^2+2xy+3y^2=1
2x + 2y + 2xy' + 6yy' = 0
(2x+6y)y' = -2(x+y)
y' = -(x+y)/(x+3y)
y" = -[(1+y')(x+3y)-(x+y)(1+3y')]/(x+3y)^2
You are correct so far. It can be simplified to
y" = -2(x^2+2xy+3y^2)/(x+3y)^3
sir steve what about these i just solved it now i didnt get it
if x^2+2xy+3y^2=1
prove that (x+3y)^3d^2y/dx^2+2(x^2+2xy+3y^2)=0
i solve with implicit derivative like these
2x+2(xdy/dx+y)+6ydy/dx=0
2x+2xdy/dx+2y+6ydy/dx=0
2xdy/dx+6ydy/dx=-2y-2x
(xdy/dx+3ydy/dx)=-(y+x)
dy/dx(x+3y)=-(y+x)
dy/dx=-(y+x)/(x+3y)
now d^2y/dx^2=-(x+3y)(dy/dx+1)-[-(y+x)(1+3dy/dx)/(x+3y)^2
is it correct b4 i plug in and see whats up
1 answer
1 answer