The slope of a curve is equal to y divided by 4 more than x2 at any point (x, y) on the curve.
A. Find a differential equation describing this curve.
B. Solve the differential equation from part A.
C. Suppose it’s known that as x goes to infinity on the curve, y goes to 1. Find the equation for the curve by using part B and determining the constant.
I got dy/dx = y/x^2+4 for A and |y| = e^c e^(arctan(x/2)/2) for B but I'm not sure
oobleck, wouldn’t it be 1/4arctan since you factor out a 1/4 to get (x/2)^2 +1
That would make c = e^(-pi/8)
A. You're on the right track! To find the differential equation describing the curve, we need to express the slope in terms of x and y.
Given: The slope of the curve is equal to y divided by 4 more than x² at any point (x, y) on the curve.
So, we have: dy/dx = y / (x² + 4)
B. Now let's solve the differential equation.
To start, we'll rewrite it slightly: dy/dx = y / (x² + 4)
First, we'll multiply both sides by (x² + 4) to get rid of the denominator: (x² + 4) dy/dx = y
Now, let's move everything involving y to one side: (x² + 4) dy = y dx
Divide both sides by y(x² + 4): dy / y = dx / (x² + 4)
Integrating both sides, we get: ln|y| = ∫(dx) / (x² + 4)
Integrating the right side using the substitution method: let u = x / 2, which gives du = (1/2) dx.
Substituting, we have: ln|y| = ∫(2 du) / (4u² + 4) = (1/2) ∫(du) / (u² + 1)
The integral on the right side is just the arctan function: ln|y| = (1/2) arctan(u) + C
Substituting back in for u: ln|y| = (1/2) arctan(x / 2) + C
C. Lastly, let's find the equation for the curve by determining the constant C. We know that as x approaches infinity on the curve, y approaches 1.
Using this information, we can substitute x = ∞ and y = 1 into the equation we found in part B:
ln|1| = (1/2) arctan(∞ / 2) + C
Since arctan(∞ / 2) approaches π/2, this simplifies to:
0 = (1/2)(π/2) + C
C = -π/4
Therefore, the equation for the curve is:
ln|y| = (1/2) arctan(x / 2) - π/4
Or, if you prefer, you can exponentiate both sides to eliminate the logarithm:
|y| = e^((1/2) arctan(x / 2) - π/4)
Note that the absolute value signs account for both positive and negative values of y.
To find the differential equation describing the curve, we can start by using the information given: the slope of the curve is equal to y divided by 4 more than x^2. The slope of a curve at any point (x, y) is represented by dy/dx. Therefore, we have:
dy/dx = y / (x^2 + 4)
This is the differential equation describing the curve.
To solve this differential equation in part B, we can start by separating the variables. The equation becomes:
dy/y = dx / (x^2 + 4)
Next, we integrate both sides of the equation.
∫(1/y) dy = ∫(1 / (x^2 + 4)) dx
Integrating the left side gives us:
ln(|y|) = ∫(1 / (x^2 + 4)) dx
To solve the integral on the right side, we can use a substitution. Let u = x/2. Then, du = (1/2) dx, and the integral becomes:
∫(1 / (u^2 + 1)) (2du)
Simplifying, we have:
2∫(1 / (u^2 + 1)) du
This integral can be solved using the arctan function. The integral now becomes:
2arctan(u) + C
Substituting back for u, we have:
2arctan(x/2) + C
Hence, the solution to the differential equation is:
ln(|y|) = 2arctan(x/2) + C
Now, for part C, we know that as x approaches infinity, y approaches 1. Substituting these values into the solution from part B, we get:
ln(|1|) = 2arctan(infinity/2) + C
ln(1) = 2arctan(0) + C
0 = 0 + C
Hence, we determine that C is equal to 0. Therefore, the final equation for the curve is:
ln(|y|) = 2arctan(x/2)
Or, exponentiating both sides:
|y| = e^(2arctan(x/2))
This is the equation for the curve.
dy/dx = y/(x^2+4)
dy/y = dx/(x^2+4)
lny = 1/2 arctan(x/2) + c
y = c*e^(1/2 arctan(x/2)) (different c ... e^c is just a constant, so call it c)
arctan(∞ ) = π/2, so
ce^(1/2 * π/2) = 1
c = e^(-π/4)
so we could say that
y = e^(1/2 arctan(x/2) - π/4)