For Questions 1–3, use the differential equation given by dx equals xy/3, y > 0.
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 ? ? ? ? ? ? ? ? ?
Find the particular solution y = f(x) to the given differential equation with the initial condition f(0) = 4.
I need help pleasee. Thank you for the help in advance! I greatly appreciate it :)
So for number 1, is it as simple as it seems. I plug in the x and y into xy/3 and get all the values of dy/dx?
Sure, I'd be happy to help! Let's start by finding the values of dy/dx for each corresponding set of x and y values in the table.
x -1 -1 -1 0 0 0 1 1 1
y 1 2 3 1 2 3 1 2 3
dy/dx ? ? ? ? ? ? ? ? ?
To find dy/dx, we can substitute the given x and y values into the differential equation dx = xy/3. Let's calculate each dy/dx:
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 = -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 = 3 / 3 = 1
Now the table of values for the derivative dy/dx looks like this:
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 can integrate the differential equation dx = xy/3:
dy/y = x dx/3
Integrating both sides:
ln|y| = (1/6) x^2 + C
Using the initial condition f(0) = 4, we can solve for C:
ln|4| = (1/6) (0)^2 + C
ln|4| = C
So the value of C is ln|4|. Plugging this back into the equation, we get:
ln|y| = (1/6) x^2 + ln|4|
To find y = f(x), we can exponentiate both sides:
|y| = 4 * e^((1/6) x^2)
Since y > 0, we can remove the absolute value:
y = 4 * e^((1/6) x^2)
Therefore, the particular solution to the given differential equation with the initial condition f(0) = 4 is y = 4 * e^((1/6) x^2).
To complete the table of values, we need to substitute each x value into the given differential equation and solve for dy/dx.
For x = -1, we have y = 1. Substituting these values into the equation dx = xy/3, we get:
(-1) * (1) / 3 = -1/3.
So, dy/dx = -1/3 when x = -1.
Similarly, we can find the values for the other x values:
For x = 0, y = 1, substituting into the equation:
(0) * (1) / 3 = 0.
So, dy/dx = 0 when x = 0.
For x = 1, y = 1, substituting into the equation:
(1) * (1) / 3 = 1/3.
So, dy/dx = 1/3 when x = 1.
Now, to find the particular solution y = f(x) with the initial condition f(0) = 4, we need to integrate the given differential equation.
The given equation is dx = xy/3. Rearranging, we get:
dx/x = y/3 * dy.
Integrating both sides with respect to x:
∫(1/x) dx = ∫(y/3) dy.
ln|x| = (y^2)/6 + C, where C is the constant of integration.
Applying the initial condition f(0) = 4, we have x = 0 and y = 4:
ln|0| = (4^2)/6 + C.
As the natural logarithm of zero is undefined, we cannot substitute x = 0 into the equation. Instead, we need to use the limit:
lim(x->0) ln|x| = (4^2)/6 + C.
Since ln|x| approaches negative infinity as x approaches 0 from the left and approaches positive infinity as x approaches 0 from the right, we have:
-ln(0) = lim(x->0^-) ln|x| = (4^2)/6 + C.
ln(0) = lim(x->0^+) ln|x| = (4^2)/6 + C.
Therefore, C = -infinity.
Hence, the particular solution y = f(x) to the given differential equation with the initial condition f(0) = 4 is:
ln|x| = (y^2)/6 - infinity.
Please note that due to the undefined nature of the logarithm at x = 0, we cannot find a particular value for y when x = 0.
To complete the table of values for dy/dx, we can use the given differential equation dx = xy/3. We'll substitute the corresponding values of x and y into the equation and simplify to find dy/dx.
For each row in the table, calculate dy/dx using the given values of x and y:
Row 1: When x = -1 and y = 1,
dx = (-1)(1)/3 = -1/3.
Row 2: When x = -1 and y = 2,
dx = (-1)(2)/3 = -2/3.
Row 3: When x = -1 and y = 3,
dx = (-1)(3)/3 = -1.
Row 4: When x = 0 and y = 1,
dx = (0)(1)/3 = 0.
Row 5: When x = 0 and y = 2,
dx = (0)(2)/3 = 0.
Row 6: When x = 0 and y = 3,
dx = (0)(3)/3 = 0.
Row 7: When x = 1 and y = 1,
dx = (1)(1)/3 = 1/3.
Row 8: When x = 1 and y = 2,
dx = (1)(2)/3 = 2/3.
Row 9: When x = 1 and y = 3,
dx = (1)(3)/3 = 1.
Now that we have the values of dy/dx for each x and y, let's find the particular solution y = f(x) with the initial condition f(0) = 4.
The given differential equation is dx = xy/3. To find the particular solution, we'll use separation of variables and integrate both sides.
dx = xy/3
dy/dx = y/3x
Separating variables,
dy/y = dx/3x
Integrating both sides,
∫(dy/y) = ∫(dx/3x)
ln|y| = (1/3)ln|x| + C
Applying the initial condition f(0) = 4,
ln|4| = (1/3)ln|0| + C
ln|4| = C
Therefore, C = ln|4|.
The particular solution is:
ln|y| = (1/3)ln|x| + ln|4|
ln|y| = ln|4x^(1/3)|
y = 4x^(1/3)
Thus, the particular solution to the differential equation with the initial condition f(0) = 4 is y = 4x^(1/3).
#1. Really?? Given x and y, you can't compute xy/3?
I'm assuming that you meant
dy/dx = xy/3
dy/y = x/3 dx
ln y = 1/6 x^2 + ln c
y = c*e^(1/6 x^2)
Since y(0) = 4, c = 4 and so
y = 4e^(1/6 x^2)
to check, just take the derivative and verify that it equals xy/3.