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.
I think you can complete the table, right? Just plug in the values for x and y.
As for the solution, you have
dy/dx = xy/3
3y dy = x dx
3/2 y^2 = 1/2 x^2 + c
3y^2 = x^2 + c
plug in (0,4) and you have
3*16 = 0 + c
c=48
3y^2-x^2 = 48
or, in a mor standard form, you can see that you have the hyperbola
y^2/16 - x^2/48 = 1
To complete the table of values for dy/dx, we need to substitute the given values of x and y into the differential equation dy/dx = xy/3 and calculate the corresponding values of dy/dx. Let's go through each point:
For x = -1 and y = 1:
dy/dx = (-1)(1)/3 = -1/3
For x = -1 and y = 2:
dy/dx = (-1)(2)/3 = -2/3
For x = -1 and y = 3:
dy/dx = (-1)(3)/3 = -1
For x = 0 and y = 1:
dy/dx = (0)(1)/3 = 0
For x = 0 and y = 2:
dy/dx = (0)(2)/3 = 0
For x = 0 and y = 3:
dy/dx = (0)(3)/3 = 0
For x = 1 and y = 1:
dy/dx = (1)(1)/3 = 1/3
For x = 1 and y = 2:
dy/dx = (1)(2)/3 = 2/3
For x = 1 and y = 3:
dy/dx = (1)(3)/3 = 1
Now we can complete the table of values for dy/dx:
x | y | dy/dx
-----------------------
-1 | 1 | -1/3
-1 | 2 | -2/3
-1 | 3 | -1
0 | 1 | 0
0 | 2 | 0
0 | 3 | 0
1 | 1 | 1/3
1 | 2 | 2/3
1 | 3 | 1
Now let's move on to finding the particular solution y = f(x) to the given differential equation with the initial condition f(0) = 4.
To find the particular solution, we need to solve the differential equation dy/dx = xy/3 and substitute the initial condition f(0) = 4.
Let's separate the variables and integrate both sides:
dy/y = x/3 dx
Integrating both sides gives:
ln|y| = (1/2)x^2 + C1
Applying the initial condition f(0) = 4 by substituting x = 0 and y = 4 into the equation, we can solve for the constant C1:
ln|4| = 0 + C1
ln(4) = C1
So the particular solution is:
ln|y| = (1/2)x^2 + ln(4)
We can simplify this equation as:
ln|y| = (1/2)x^2 + ln(4)
ln|y| = ln(4e^(1/2x^2))
|y| = 4e^(1/2x^2)
y = ±4e^(1/2x^2)
Since y > 0 (from the given condition), we take the positive solution:
Hence, the particular solution to the given differential equation with the initial condition f(0) = 4 is:
y = 4e^(1/2x^2)