consider dy/dx = 2-2y where y=f(x) is the particular solution to the differential equation where f(0)=2.
1. write an equation for the line tangent to f at x=0
2. is f concave up or down at (0,2)?
3. find f(x)
To solve the given differential equation dy/dx = 2 - 2y, let's follow these steps:
Step 1: Write the general solution to the differential equation:
dy/dx = 2 - 2y
Separate variables and integrate:
1/(2-2y) dy = dx
Integrating both sides:
(1/2) ln|2-2y| = x + C1
Where C1 is the constant of integration.
Step 2: Find the particular solution using the initial condition f(0) = 2.
When x = 0, y = 2:
(1/2) ln|2-2(2)| = 0 + C1
(1/2) ln|-2| = C1
-ln(2) = C1
So the particular solution is:
(1/2) ln|2-2y| = x - ln(2)
Step 3: Write an equation for the line tangent to f at x=0:
The equation for a line tangent to a curve at a specific point is given by:
y - y1 = m(x - x1)
Here, x1 = 0 and y1 = 2 (the point (0, 2)), and we need to find the slope m at x = 0.
Differentiating both sides of the particular solution equation:
(1/2) (2-2y)(-2) dy/dx = 1
Simplifying:
-(2-2y) dy/dx = 1
At x = 0, y = 2, so:
-(2-2(2)) dy/dx = 1
-(2-4) dy/dx = 1
2 dy/dx = 1
dy/dx = 1/2
Therefore, the slope m at x = 0 is 1/2.
Substituting x1, y1, and m back into the equation for the tangent line:
y - 2 = (1/2)(x - 0)
y - 2 = (1/2)x
This is the equation for the line tangent to f at x = 0.
Step 4: Determine if f is concave up or down at (0, 2):
To determine whether the function f is concave up or down at a point, we need to check the sign of the second derivative.
Differentiating both sides of the initial differential equation:
d²y/dx² = d(2 - 2y)/dx
d²y/dx² = -2(dy/dx)
Substituting x = 0 and y = 2:
d²y/dx² = -2(dy/dx) = -2(1/2) = -1
Since the second derivative is negative at (0, 2), f is concave down at that point.
Step 5: Find f(x):
Given the particular solution:
(1/2) ln|2-2y| = x - ln(2)
Let's solve for y:
ln|2-2y| = 2x - 2ln(2)
Taking the exponential of both sides:
|2-2y| = e^(2x - 2ln(2))
Since the absolute value function has two branches, we need to consider both cases:
Case 1: 2 - 2y = e^(2x - 2ln(2)) (assuming the positive branch)
2y = 2 - e^(2x - 2ln(2))
y = 1 - (1/2)e^(2x - 2ln(2))
Case 2: 2 - 2y = -e^(2x - 2ln(2)) (assuming the negative branch)
2y = 2 + e^(2x - 2ln(2))
y = 1 + (1/2)e^(2x - 2ln(2))
So the general solution for f(x) is given by the two cases:
f(x) = 1 - (1/2)e^(2x - 2ln(2)) or f(x) = 1 + (1/2)e^(2x - 2ln(2))
To solve the given differential equation dy/dx = 2 - 2y, we can use the method of separation of variables. Let's go step by step:
1. To find the equation for the line tangent to f at x=0, we need to find the derivative of f(x) and evaluate it at x=0. Given that f(0) = 2, we can start by finding dy/dx.
dy/dx = 2 - 2y
Now, rearrange the equation by moving the y term to the left side and the dy term to the right side:
dy/(2 - 2y) = dx
Next, integrate both sides of the equation with respect to their respective variables:
∫(1/(2 - 2y)) dy = ∫dx
To integrate the left side, we can use the substitution u = 2 - 2y, du = -2dy:
∫(1/u) du = ∫dx
ln|u| = x + C
Since u = 2 - 2y, we have:
ln|2 - 2y| = x + C
Now, we can use the initial condition f(0) = 2 to find the value of C. Substitute x = 0 and y = 2 into the equation:
ln|2 - 2(2)| = 0 + C
ln|2 - 4| = C
ln|-2| = C
Since the natural logarithm of a negative number is undefined, we can conclude that C does not have a real value. Therefore, the equation for the line tangent to f at x=0 is:
ln|2 - 2y| = x
2. To determine whether f is concave up or down at (0,2), we need to analyze the second derivative of f(x). Let's find it:
Taking the derivative of the given differential equation dy/dx = 2 - 2y:
d²y/dx² = d/dx(2 - 2y)
Differentiating:
d²y/dx² = -2(dy/dx)
Substituting dy/dx = 2 - 2y:
d²y/dx² = -2(2 - 2y)
Simplifying:
d²y/dx² = -4 + 4y
Now, substitute the point (0,2) into the second derivative equation:
d²y/dx² = -4 + 4(2)
d²y/dx² = -4 + 8
d²y/dx² = 4
Since the second derivative is positive, we can conclude that f is concave up at (0,2).
3. To find f(x), we need to solve the differential equation:
dy/dx = 2 - 2y
Rearrange the equation by moving the y term to the left side and the dx term to the right side:
dy / (2 - 2y) = dx
Separate the variables and integrate:
∫(1/(2 - 2y)) dy = ∫dx
Using the same substitution as before, u = 2 - 2y, du = -2dy:
∫(1/u) du = ∫dx
ln|u| = x + C
Substituting u = 2 - 2y:
ln|2 - 2y| = x + C
To solve for y, we need to eliminate the natural logarithm. Start by rewriting the equation as an exponential:
2 - 2y = e^(x + C)
Next, rearrange to solve for y:
-2y = e^(x + C) - 2
Divide by -2:
y = (2 - e^(x + C))/2
Since e^C is an arbitrary constant, we can rewrite it as another constant, A:
y = (2 - e^x)/2
Therefore, f(x) = (2 - e^x)/2.
The point (0,2) lies on the curve, so the tangent there is
y-2 = -2(x-0)
y" = -2y'
y' = -2, so y" > 0
so the graph is concave up
see what you can do with the last part.