Question 1 1 . x - y + 3ry = 2 (1) First, consider the term 3ry. Use the product rule. F(x) = f(x) . g(x) = F'(x) = f'(x)g(x) + f(x)g'(x) (3ry) = 3(1)y + 3ry' = 3y + 3xy' Differentiate eq.(1) w.r.t. r gives: 1 - y' + 3y + 3xy' = 0 (2) # y' (-1+ 3x) = -1 - 3y -1 - 3y y' = -1+3x 1 + 3y I signs have y = 1 -31 changed ( 3 ) We can obtain the second derivative using either eq. (2) or (3). From eq. (2): 1 - y' + 3y + 3ry = 0. why are there -y" $ 3y' + 3y - 3cy" = 0 two Bul ? 3cy" - y" + 6y' = 0 y"(3.x - 1) = -6y' signs have -6y' y" = : chunged 3r - 1 6y' 1 - 3x