Consider the differential equation dy/dx = 2x - y. Let y = f(x) be the particular solution to the differential equation with the initial condition f(2) = 3. Does f have a relative min, relative max, or neither at x = 2?
Since we're trying to find a max/min I don't think it's necessary to find the integral. I tried doing 2x - y = 0, but I'm not sure if I can use the intial condition here. Is the initial condition meant to be used in the antiderivative or can it be used in the derivative itself?
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Ah, the delightful world of differential equations and min/max conundrums. Quite the perplexing situation, isn't it? Well, fear not, for I shall guide you through the twisty labyrinth of this particular problem.
Now, let's start by differentiating both sides of the equation dy/dx = 2x - y, just to shake things up a bit. If we do so, we get d²y/dx² = 2 - dy/dx. Quite the funky equation, I must say.
Now, let's bring in our initial condition, f(2) = 3, just for funsies. What we can do is substitute x = 2 and y = 3 into our original differential equation and see what happens. Lo and behold, we get dy/dx = 2(2) - 3 = 1.
But hold your horses! We're not done yet. To determine if f has a relative min, relative max, or neither at x = 2, we need to consider the second derivative. Remember that d²y/dx² thing from earlier? Well, now is the time to whip it out.
If we substitute x = 2 into d²y/dx² = 2 - dy/dx, we get d²y/dx² = 2 - 1 = 1.
So, what does this mean? Well, since the second derivative is positive at x = 2, we can conclude that f has a relative min at that point. Ta-da! We did it!
Now, go forth and conquer, my brave friend. Solve more equations, find more min/max, and remember to always sprinkle a healthy dose of humor along the way.
To determine if the particular solution function f(x) has a relative minimum, maximum, or neither at x = 2, we need to analyze the behavior of the function around that point.
The differential equation given is:
dy/dx = 2x - y
Since we're given the initial condition f(2) = 3, we can use this information to find the particular solution's constant of integration. Let's substitute x = 2 and y = 3 into the differential equation:
dy/dx = 2x - y
3 = 2(2) - f(2)
3 = 4 - f(2)
f(2) = 4 - 3
f(2) = 1
Now we know the value of f(2) and can proceed to solve the differential equation. Rearranging the equation, we have:
dy = (2x - y)dx
Separating the variables, we get:
dy + ydx = 2xdx
Integrating both sides:
∫(dy + ydx) = ∫(2xdx)
Integrating, we have:
y + 1/2y^2 = x^2 + C
Now, since we're looking for the relative maximum or minimum at x = 2, we need to determine the value of f'(x) and f''(x) at x = 2.
To find f'(x), we differentiate the equation y + 1/2y^2 = x^2 + C with respect to x:
d/dx[y] + 1/2 d/dx[y^2] = d/dx[x^2] + d/dx[C]
dy/dx + y dy/dx = 2x
Substituting dy/dx = 2x - y, we have:
(2x - y) + y (∂y/∂x) = 2x
Expanding and rearranging:
2x - y + y (∂y/∂x) = 2x
y (∂y/∂x) = y
∂y/∂x = 1
Now we find ∂^2y/∂x^2 to determine if there is a potential max/min:
Taking the derivative again:
∂/∂x [∂y/∂x] = ∂/∂x[1]
∂^2y/∂x^2 = 0
Since ∂^2y/∂x^2 = 0, it means that the second derivative is zero, which implies that there is neither a relative minimum nor a relative maximum at x = 2.
In conclusion, the particular solution function f(x) does not have a relative minimum or maximum at x = 2.
To determine whether the particular solution f(x) has a relative minimum, relative maximum, or neither at x = 2, we need to analyze the behavior of the solution near that point.
First, let's find the solution to the differential equation dy/dx = 2x - y. This is a first-order linear ordinary differential equation, so we can solve it using a method called "separation of variables."
Rearrange the equation to isolate y on one side:
dy = (2x - y) dx
Now, separate the variables by moving all terms involving y to one side and all terms involving x to the other side:
dy + y dx = 2x dx
Next, integrate both sides with respect to their respective variables:
∫(dy + y dx) = ∫(2x dx)
Integrating each term separately:
∫dy + ∫y dx = ∫2x dx
y + ∫y dx = x^2 + C
Integrate y with respect to x:
y + (1/2)y^2 = x^2 + C
Now, we need to use the initial condition f(2) = 3 to find the particular solution. Substitute x = 2 and y = 3 into the equation:
3 + (1/2)(3)^2 = 4 + C
3 + 9/2 = 4 + C
Simplifying:
3/2 = C
Now, we can rewrite the equation in terms of y:
y + (1/2)y^2 = x^2 + 3/2
To determine whether f(x) has a relative minimum, relative maximum, or neither at x = 2, we need to look at the behavior of the slope of the curve at that point.
First, let's find the derivative of y with respect to x:
(dy/dx) + y(dy/dx) = 2x
Factoring out (dy/dx):
(1 + y)(dy/dx) = 2x
Substitute the particular solution y = f(x) into the equation:
(1 + f(x))(dy/dx) = 2x
Now, substitute x = 2:
(1 + f(2))(dy/dx) = 4
We already know that f(2) = 3, so we can simplify the equation further:
(1 + 3)(dy/dx) = 4
4(dy/dx) = 4
Simplifying:
(dy/dx) = 1
The derivative of y with respect to x is equal to 1 at x = 2.
Since the derivative is positive at x = 2, the function f(x) is increasing. This means it has neither a relative minimum nor a relative maximum at x = 2.