Thread: Antidifferentiation: -Find equation of curve

The points (-1,3) and (0,2) are on a curve, and at any point (x,y) on the curve d^2y/dx^2 = 2 - 4x. Find an equation of the curve.
Hint: Let d^2y/dx^2 = dy'/dx' and obtain an equation involving y', x and an arbitrary constant C1. From this equation obtain another equation involving y, x, C1, and C2. Compute C1 and C2 from the conditions:

Answer from book: 3y = -2x^3 + 3x^2 +2x + 6

THanks so much!

Originally Posted by ^_^Engineer_Adam^_^

The points (-1,3) and (0,2) are on a curve, and at any point (x,y) on the curve d^2y/dx^2 = 2 - 4x. Find an equation of the curve.
Hint: Let d^2y/dx^2 = dy'/dx' and obtain an equation involving y', x and an arbitrary constant C1. From this equation obtain another equation involving y, x, C1, and C2. Compute C1 and C2 from the conditions:

Answer from book: 3y = -2x^3 + 3x^2 +2x + 6...

Hello,

dy/dx = ∫(d²y/dx²)dx therefore:

dy/dx = ∫(2-4x)dx = 2x - 2x² +C

y = ∫(dy/dx)dx therefore:

y = ∫(2x - 2x² + C)dx = x² - 2/3*x³ + Cx + D

Plug in the coordinates of the points you know:

(-1, 3): 1 + 2/3 - C + D = 3
(0, 2): 0 - 0 + 0 + D = 2 Plug in this value into the first equation:

1 + 2/3 - C + 2 = 3 and solve for C. C = 2/3

the complete equation is:

y = -2/3*x³ + x² + 2/3*x + 2

EB