# Calculus

For Questions 1–2, use the differential equation given by dy/dx = xy/3, y > 0.
1. Complete the table of values
x -1 -1 -1 0 0 0 1 1 1
y 1 2 3 1 2 3 1 2 3
dy/dx

2. Find the particular solution y = f(x) to the given differential equation with the initial condition f(0) = 4.

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1. I'm sure you can fill in the table by plugging in values for x and y and evaluating xy/3

As for the DE,
dy/dx = xy/3

3/y dy = x dx
3/2 y^2 = 1/2 x^2 + c
or,
3y^2 = x^2 + c
Since y(0)=4,
3*4^2 = 0^2+c
c=48

So the solution is

3y^2 = x^2+48
or, in a more standard form you see that it is a hyperbola
3y^2-x^2=48
y^2/16 - x^2/48 = 1

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posted by Steve

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