Question

Six plus half of my indefinite integral is the derivative of myself squared. What function am I?

Answer 1

tough. do you know the answer?

Answer 2

No, I just made it up because it sounded cute and wanted to see if anyone else would try to solve it with me for fun.

Answer 3

I'm pretty sure you could solve this as a DE

Answer 4

because we could do something like F(x) is indefinite integral of f(x)
6+F(x)/2=[f'(x)]^2
f(x)/2=2f'(x)f''(x)
lol

Answer 5

My first attempt is to just differentiate both sides to turn it into a DE but it's sort of slow going solving the thing after this.

Answer 6

yeah that's all i did

Answer 7

Doing that I'm getting: \[\Large (y')^2+yy''=\frac{1}{8}\] Now I suspect we can use either a trig or hyperbolic trig function to fulfill this.

Answer 8

i don't even see how you got that, but i'm totally lost as to where to take it from there...

Answer 9

\[\Large 6+\frac{1}{2}\int\limits y dx = \frac{d}{dx}(y^2)\]\[\Large \frac{1}{2}y = \frac{d^2}{dx^2}(y^2)*2y\]\[\Large \frac{1}{4} = \frac{d^2}{dx^2}(y^2)\]\[\Large \frac{1}{4} = \frac{d}{dx}(2yy')=2(y')^2+2yy''\]

However I think I might have a flaw in my reasoning... Wait, maybe it's ok let me think.

Answer 10

I think I ended up with 2y twice. Bleh haha.

Answer 11

\[6+\frac y2=\frac d{dx}((y')^2)\\\frac{y'}2=2y'y''\]?

Answer 12

I think when I worded it I was vague and meant (y^2)' instead of (y')^2 but I guess it doesn't matter which one we solve since they're all made up.

Answer 13

yeah but it's fun messing around with

Answer 14

do we get initial conditions?

Answer 15

oh but i think i represented it right, because i just took y to be the indefinite integral of y', though we have that constant term, so the answer depends on.... yeah^

Answer 16

either way, after taking the derivative of the square, it involves the product of the function and it's derivative on one side, and it's integral on the other. I have no clue how to solve a DE like that

Answer 17

I'm doing it with Laplace Transform right now. It isn't pretty, but it appears to be solvable

Answer 18

Oh yeah, Laplace could work...

Answer 19

@Miracrown @Mimi_x3 @ganeshie8 @ikram002p @dan815

I want to solve this lol

Answer 20

*

Answer 21

:)

Answer 22

There doesn't appear to be much literature on the subject: http://en.wikipedia.org/wiki/Integro-differential_equation

Only one text is cited: http://books.google.com.sv/books/about/Theory_of_integro_differential_equations.html?id=p1ZLP4OHp4YC

In any case, I have a feeling that any IDE containing an indefinite integral won't be solvable. We need to know the limits of integration (some constant to a variable) in order to be able to describe the equation like this.

Answer 23

@kainui - How do you know that there's even an elementary solution, if the equation is soluble at all?

Answer 24

I mean, at best, we're looking at a nonlinear ODE.

Answer 25

http://prntscr.com/52hsnq

Answer 26

I was initially under the impression that this was just a standard calculus problem, but if it's something you made up, it could turn out to be exceedingly messy.

Answer 27

But I think that if it IS soluble, it goes well beyond the
scope of calculus. It might be the case that the equation is
only numerically soluble given proper initial and boundary conditions.