I got 7/2 it is dy^2/dx^2 not dx^2/dy^2
here is the oda that got me confused
find the value of the constant integration if
(x^1/2+x^1/3)dx=0
at x=1
If y=2x^2-21x^2+60x+6
what is the value of x?
dx^2/dy^2=0
2 answers
assuming you meant
y = 2x^3-21x^2+60x+6
y' = 6x^2-42x+60
y" = 12x-42
y"=0 when x = 7/2
For the other one,
(x^1/2+x^1/3)dx = 0
∫(x^1/2+x^1/3)dx = C
2/3 x^(3/2) + 3/4 x^(4/3) = C
at x=1, C = 2/3 + 3/4 = 17/12
y = 2x^3-21x^2+60x+6
y' = 6x^2-42x+60
y" = 12x-42
y"=0 when x = 7/2
For the other one,
(x^1/2+x^1/3)dx = 0
∫(x^1/2+x^1/3)dx = C
2/3 x^(3/2) + 3/4 x^(4/3) = C
at x=1, C = 2/3 + 3/4 = 17/12