Calculating Constants in indefinite integral with 2 values for f.

[tainted]

Homework Statement

Find the function f(x) such that $f''(x) = \frac{1}{x^2}$, $f(1) = 0$ and $f(e) = 0$

Homework Equations

$\int f''(x)\,dx = f'(x) + c$
$\int f'(x)\,dx = f(x) + cx + C$

The Attempt at a Solution

$f''(x)= \frac{1}{x^2}$
$f'(x)= \int \frac{1}{x^2}\,dx = \frac{-1}{x} + c$
$f(x) = -\int \frac{1}{x} + c\,dx = -ln(x) + cx + C$

My problem is that I can get the constants to satisfy one of those requirements in the original problem, but I am having trouble getting it to satisfy both.

Thanks in advance

 

[Muphrid]

Show some work? Might be able to find the issue. With only two data points and two constants, there shouldn't be any big issue setting both.

 

[tainted]

That is my work. That's what I've had for over an hour /facepalm
I've tried guessing some constants, but nothing has worked. I don't know how to get the constants to fit both those statements
$f(1) = 0$ and $f(e) = 0$

 

[Muphrid]

All right then. What are $f(1),f(e)$ in terms of these constants?

 

[tainted]

I really have no clue. In terms of what constants? I haven't seen anything like this problem before. I have seen plenty where $f'(e)=0$ and $f(1) = 0$or the likes, but I haven't seen them where both value were just for $f$

 

[Muphrid]

itex, not itext.

In terms of the constants $c, C$. You wrote $f(x)$ in the first post. Even though you don't know the values of $c,C$, you should be able to evaluate the function and get an expression in terms of these constants at $x=1,e$.

 

[tainted]

Alright thanks. I think that got me going in the right direction.

$f(x) = -ln(x) + cx + C$
$f(1) =-ln(1) + 1*c + C = 0$
$f(e) = -ln(e) + e*c + C = 0$
$-ln(1) + 1*c + C = -ln(e) + e*c + C$
$-0 + c + C = -1 + e*c + C$
$-0 + c + C = -1 + e*c + C$
$c = e*c - 1$
$c + 1 = e*c$
$1 = e*c - c$
$1 = c(e-1)$
$\frac{1}{e-1} = c$Does that look correct on evaluating for $c$ ?

 

[Muphrid]

Yeah, I think that's correct. You should be able to substitute in and find $C$ at this point. In general, as you can see, if you have an Nth order differential equation, you only need N pieces of information in order to determine all the integration constants. They can be values of derivatives or values of the function itself at any arbitrary points.

 

[tainted]

Using that value for $c$, I did the following.

$f(x) = -ln(x) + \frac{1}{e-1} * x + C$
$f(x) = -ln(x) + \frac{x}{e-1} + C$
$f(1) = -ln(1) + \frac{1}{e-1} + C = 0$
$f(1) = -0 + \frac{1}{e-1} + C = 0$
$C = \frac{-1}{e-1}$

That would make my final answer

$f(x) = -ln(x) + \frac{x}{e-1} - \frac{1}{e-1}$

I tested it, and it appeared to work

Thanks a lot man. At first I was trying to figure out what you meant, but once I did, I was able to get it. I hadn't realized that you could have only 2 values of the function to determine both the constants
Thank you

 

[Muphrid]

Yeah, you just end up with two equations, two unknowns in a linear system. It's a piece of cake once you've seen it once or twice. Just making the leap to getting there is the only hard part.