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

To find the equation of the curve, we first find the general solution of the given differential equation.

$$\dfrac{d^2y}{dx^2} = 2-4x$$

We can rewrite the above as:

$$d^2y =(2-4x)dx^2\\ d(dy) = [(2-4x)dx]dx$$

Integrating the above:

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

Integrating again:

$$\int dy = \int(2x - 2x^2 +c_1)dx\\ \bold{y =x^2-\frac{2x^3}{3}+c_1x + c_2}$$

Which is the general solution of the curve.

At point (-1,3), where x = -1 and y = 3:

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

At point (0,2), where x = 0 and y=2;

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

Substitute eqn 2 into eqn 1:

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

Thus the equation of the curve is:

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

OR

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