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.
To solve question 1, we are given a differential equation dy/dx = xy/3 and we need to complete the table of values for y and find dy/dx.
We can start by plugging in the given values of x into the differential equation to find the corresponding values of y. Let's do that:
For x = -1: dy/dx = (-1)(1)/3 = -1/3
For x = 0: dy/dx = (0)(1)/3 = 0
For x = 1: dy/dx = (1)(1)/3 = 1/3
Now, for the corresponding values of y:
For x = -1: y = 1
For x = 0: y = 2
For x = 1: y = 3
We can fill in the table with these values:
x -1 -1 -1 0 0 0 1 1 1
y 1 2 3 1 2 3 1 2 3
dy/dx -1/3 -1/2 -1/3 0 0 0 1/3 1/2 1/3
Now, let's move on to question 2.
To find the particular solution y = f(x) to the given differential equation dy/dx = xy/3 with the initial condition f(0) = 4, we need to integrate the differential equation and solve for the unknown function, f(x).
First, let's rearrange the differential equation:
dy/y = (x/3) dx
Now, we can integrate both sides of the equation separately. The integral of dy/y is ln(y), and the integral of (x/3) dx is (1/6)x^2.
Integrating both sides, we get:
ln(y) = (1/6)x^2 + C
Now, let's solve for y by taking the exponential function of both sides:
y = e^((1/6)x^2 + C)
Since y > 0, we can take e^C as a positive constant, let's say A.
y = Ae^(1/6)x^2
Now, we can use the initial condition f(0) = 4 to find the value of the constant A.
When x = 0, y = 4:
4 = Ae^(1/6)(0)^2
4 = A
So, the particular solution to the differential equation with the initial condition f(0) = 4 is:
y = 4e^(1/6)x^2
To find the values of dy/dx and complete the table of values, we substitute the given values of x and y into the differential equation dy/dx = xy/3.
1. For x = -1:
- Substitute x = -1 and y = 1 into the differential equation:
dy/dx = (-1)(1)/3 = -1/3
2. For x = 0:
- Substitute x = 0 and y = 1 into the differential equation:
dy/dx = (0)(1)/3 = 0
3. For x = 1:
- Substitute x = 1 and y = 1 into the differential equation:
dy/dx = (1)(1)/3 = 1/3
Completing 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 -1/3 -2/3 -1 0 0 0 1/3 2/3 1
To find the particular solution y = f(x) with the initial condition f(0) = 4, we need to solve the differential equation with the given initial condition.
Given: dy/dx = xy/3, y > 0, f(0) = 4
To solve this, we will separate variables and integrate:
dy/y = x/3 dx
Integrating both sides:
∫ (1/y) dy = ∫ (x/3) dx
ln|y| = (x^2)/6 + C
To find the value of C, we use the initial condition f(0) = 4:
ln|4| = (0^2)/6 + C
ln(4) = C
Using the laws of logarithms, we can write this as:
C = ln(2^2) = ln(2) + ln(2) = 2ln(2)
Therefore, the particular solution is:
ln|y| = (x^2)/6 + 2ln(2)
Solving for y:
|y| = e^((x^2)/6 + 2ln(2))
Since y > 0, we can drop the absolute value:
y = e^((x^2)/6 + 2ln(2))
This is the particular solution for the given differential equation with the initial condition f(0) = 4.
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