Answer to Question #167393 in Calculus for Joshua

Question

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. 

Answer 1

$$\dfrac{d^2y}{dx^2} = 2-4x\\ d^2y =(2-4x)dx^2\\ d(dy) = [(2-4x)dx]dx\\ \int d(dy) = \int[(2-4x)dx]dx\\ dy = (2x - 2x^2 +c_1)dx\\ \int dy = \int(2x - 2x^2 +c_1)dx\\$$

The general solution of the curve is given as:

$${y =x^2-\frac{2x^3}{3}+c_1x + c_2}$$

at point (-1,3):

$$3 = (-1)^2 - \frac{2(-1)^3}{3} +c_1(-1) + c_2\\ 3 = 1 + \frac{2}{3} - c_1+c_2\\ 3-1-\frac{2}{3} = -c_1+c_2 \implies \frac{4}{3} = -c_1+c_2$$

At point (0,2):

$$2 = (0)^2 - \frac{2(0)^3}{3} +c_1(0) + c_2\\ \implies c_2=2$$

Substitute for $c_2$ in the previous equation:

$$\frac{4}{3} = -c_1+2\\ c_1 = 2 - \frac{4}{3}\\ c_1 = \frac{2}{3}$$

$\therefore$ the equation of the curve is:

$$y =x^2-\frac{2x^3}{3}+\frac{2x}{3} + 2 \implies y =2+\frac{2x}{3}+x^2-\frac{2x^3}{3}$$