Let be a solution for all of the second order differential equation

Without attempting to solve the equation answer the following questions.

- If and has an extremum at , prove that this extremum must be a minimum.
- If has an extremum at 0 decide whether this extremum is a maximum or minimum and prove your assertion.
- If the solution satisfies the conditions
find the minimum value for the constant such that

**Incomplete.**

One way to solve part c is to assume f(x) = Ax^2. By substituting this into the differential equation, we get a quadratic with A as the variable. Use the quadratic equation and we get A as a function of x.

This equation tells us what A must be for f(x) to be equal to Ax^2 at each x, since the solutions to the differential equation must be equal if the functions are equal.

Now that we have a function for A, all we must do to find a minimum value is to find the maximum value of A for x>=0, so that Ax^2 >= f(x) for all x>= 0.

This step is a little tricky, and I cheated and used Desmos and an online limit calculator. Basically, we find that limA as x->inf = 0, and A(0) is the max value of A. We calculate the limit of A as x->0, and we get A=1/2.

a.) We know from our given information that c cannot be zero. And we know that at the point where x=c, we have a local extremum where f'(c) = 0. So our differential equation becomes

And since c is not zero, we can divide both sides, giving us

We now have two cases

1.) c is greater than zero

If c is greater than zero, then

Thus,

And by Theorem 4.9, since f”(c) is greater than 0, f'(c) must be a local minimum.

2.) c is less than zero

If c is less than zero, then

And by Theorem 4.9, since f”(c) is greater than 0, f(c) must be a local minimum.

As such, for all c not equal to zero, the point f(c) must be a local minimum.

b.) From our given information, we know that if f has an extremum at x=0, then f'(0) = 0. To determine if this is a local maximum, minimum, or neither, we must compute the value of f”(0). If we were to divide both sides by x and take the limit as x approaches zero

We would get the indeterminate form of 0/0

But if this is the case, then we can use L’Hopital’s Rule and take the derivative with respect to x of the numerator and denominator, and then take the limit as x approaches zero.

And since this value is greater than zero, the point f(0) is a local minimum.

c.) If we can assume that f(0) = f'(0) = 0, we can use the result from part (b.) to show the following:

But with our initial conditions and the result of part (b.), we know that

So the smallest value of A that fulfills the condition

And thus, the conditions

for all x greater than or equal to zero, must be

Looks like that’s all for now, thanks to RoRi for opening up the comments sections, and thanks to the readers for… reading :)

So, the latex didn’t render properly for my cases, it should be as follows

Case 1.) c > 0

Case 2.) c < 0

D’oh!

Hi. I have one question about item (c): My doubt is: why is it possible, in this case, to conclude that f'(x) <= 2Ax from f(x) <= Ax^2?

I agree it’s most probably not possible to solve c) that way, because the implication most probably doesn’t hold.

Hey, I found that you could put the upper bound 1 on f” over all x>=0, given that f'(0)=0. Express for x>0

f”(x)=(1-e^(-x))/x-3[f'(x)]^2, and we see that

f”(x)<=(1-e^(-x))/x0.

But we showed in (b) that f”(x)–>1 as x–>0, so this is the smallest possible upper bound on f”. Integrating then shows that (1/2)x^2 is larger than f.

I’m not sure if this is entirely related to what you’re asking. But I think the idea is to reverse the direction; you conclude that f(x)<=Ax^2 from the fact that f''(x)<=2A.