Thread: problem in integral calculus(need help)

1. problem in integral calculus(need help)

The points (-1,3) and (0,2) are on a curve, and at any point (x,y) on the curve d2y = 2- 4x. Find an equation of the curve.
dx2

(HINT: let d2y = dy' , and obtain an equation involving y', x and an arbitrary
dx2 dx
constant C1. From the equation obtain another equation involving y,x, C1 and C2. Compute C1 and C2 from the conditions)

=APriL=

2. We start with the given second derivative of y, and work backwards.

$y'' = 2 - 4x$

Find an antiderivative of y'' to obtain the first derivative of y.

$y' = -2x^2 + 2x + C_1$

Find an antiderivative of y' to obtain y.

$y = -\frac{2}{3}x^3 + x^2 + C_1 \cdot x + C_2$

This equation has two unknowns for which we need to solve. We've been given the coordinates of two points which lie on the graph of y.

Substitute these values for x and y to obtain a system of two equations.

$3 = -\frac{2}{3}(-1)^3 + (-1)^2 + C_1(-1) + C_2$

$2 = -\frac{2}{3}(0)^3 + (0)^2 + C_1(0) + C_2$

Solving this system for the unknowns is simple algebra. Can you do it?

Cheers,

~ Mark

3. Originally Posted by april29
The points (-1,3) and (0,2) are on a curve, and at any point (x,y) on the curve d2y = 2- 4x. Find an equation of the curve.
dx2

(HINT: let d2y = dy' , and obtain an equation involving y', x and an arbitrary
dx2 dx
constant C1. From the equation obtain another equation involving y,x, C1 and C2. Compute C1 and C2 from the conditions)

$\frac{d^2 y}{dx^2} = 2 - 4x \Rightarrow \frac{dy}{dx} = 2x - 2x^2 + A \Rightarrow y = x^2 - \frac{2}{3} x^3 + Ax + B$.