Question

A weird differential equation to solve for f(x).

Answer 1

\[\frac{f'(x)-f(x)}{e^{\pi/2}}=f(x+\frac{\pi}{2})\]

Answer 2

Am I missing something? I mean all you have to do is isolate f'(x), right?

Answer 3

f is a function of x on the left side but on the right f is a function of (x+pi/2). I don't know, maybe it's easier than it looks.

Answer 4

I'm new to differential equation, but I'm sure all you need to do is isolate f'(x).

Answer 5

You can do it, right?

Answer 6

Yeah, that's not a problem just algebra.

Answer 7

Yeah.

Answer 8

But upon separation you can't integrate f(x+pi/2) with respect to x, you have to integrate it with respect to (x+pi/2) I believe.

Answer 9

\[\int\limits \frac{df}{f(x)+f(x+\pi/2)}=\int\limits e^{-\pi/2}dx\] Doesn't smell right to me.

Answer 10

no, you just integrate with respect to x. pi/2 is just constant. f(x+pi/2) is still function with one variable; x.

I don't see why you need to integrate them?

Answer 11

I guess you're too new to differential equations then if you don't know why I'm integrating...

Answer 12

hmm, I don't think you would need to integrate, I mean we don't know what f(x) is.

Answer 13

I'm not fully confident, so help? @thomaster @AccessDenied

Answer 14

For example you might be given:

y'=y

what's y?\[\frac{dy}{dx}=y\]separate it out\[\int\limits \frac{dy}{y}=\int\limits dx\] add an arbitrary constant\[\ln|y|=x+\ln(C)\]\[y=Ce^x\]

How do we know? differentiate y to get:

\[y'=Ce^x\]

So indeed it satisfies the equation, y'=y

Answer 15

"hmm, I don't think you would need to integrate, I mean we don't know what f(x) is. "

Yeah, I'm trying to find what f(x) is.

Answer 16

Sorry, I reread question. I now see that you need to solve f(x), not f'(x). sorry...

Answer 17

f(x) is really just the same as "y)

Answer 18

"doesn't smell right to me."

Answer 19

So far I got:\[f(x) = e^{\pi/2}\int\limits f\left(x+\frac{\pi}{2}\right)dx + \int\limits f(x)dx\]

I know it's not good enough. I don't know how to simplify it even more. Sorry...

Answer 20

It's fine, it's a problem I came up with that I know the answer to, but there is no obvious way that I can see that I'd be able to turn it around and find the correct answer.

Answer 21

I was just curious how people attempted to solve it so I could see how their brains work! =P

Answer 22

how about this: replace the x and what is the "e"

Answer 23

huh, replace x to what? And are you asking what e is?

Answer 24

is this some type of transformation? considering F(x+pi/2) function of cos...sin...?

Answer 25

wait I got it! This appear to be a generalization of differential equation.
Seperate the variable
\[f ' (x) = e ^{\pi/2} [f(x+\frac{ \pi }{ 2 })]+f(x)e ^{0}\]

Answer 26

I am NOT looking forward to Calculus next year.

Answer 27

@Rinru but how would you go about solving it from there?

Answer 28

So this is what my attempt is at solving it:\[f'(x)-f(x)=e^{-\pi/2}f(x+\pi/2)\] I multiply both sides by an integrating factor \[\mu f'(x)- \mu f(x)= \mu e^{-\pi/2}f(x+\pi/2)\]Then to make the left side condense by the product rule I make \[\mu '= -\mu\] which means \[\mu = e^{-x}\] So I have \[\frac{d}{dx}(e^{-x} f(x))=e^{-(x+\pi/2)}f(x+\pi/2)\]

And so the left and right sides are both functions of different variables kind of, so I'm not so sure what to do, but it looks pretty lol.

Answer 29

Actually, maybe if I say \[g(x)=e^{-x}f(x)\] then I'm just looking at: \[g'(x)=g(x+\pi/2)\] which means \[ g(x)=C(\sin(x)+\cos(x)) \] So just plugging in to that top formula, \[f(x)=Ce^x(\sin(bx)+\cos(bx))\] Hmm. I guess that works.

Answer 30

to be honest...I think leaving it in the original form
f′(x)=eπ/2[f(x+π2)]+f(x)e0 would be for the the best
are you trying to prove/simplify this equation?

Answer 31

I'm trying to solve this differential equation... Leaving it as it is really isn't the point...