find the anti-derivative of $$f \prime \prime(x) = 2e ^{3}+3\sin t$$, f(0)=0, f(pi)=0

Question

flywalker · Answer

in my way: f'(x)= (2/3)*e^3-3cost+c
and you are doing the same thing to go to f(x)
and then (u find 1/f(x) and you find the derivative
but i dont understand what is f(pi) to be sure if i am right!
tell me to right the full explanation!

zepdrix · Answer

@Luigi0210 do you understand how to find the anti-derivative of the $2e^3$ portion?

zepdrix · Answer

Unless you made a typo, there appears to be no $t$ involved. So it is simply a constant. What is the anti-derivative of a constant?

luigi0210 · Answer

I made a mistake, sorry. It's suppose to be 2e^t

flywalker · Answer

and i thing i  made a big mistake

zepdrix · Answer

oh ok :)

zepdrix · Answer

Do you remember the derivative of $\large e^x$?
It gives us the same thing back right?
That process will be the same in reverse.

So the anti-derivative of $\large e^t$ is $\large e^t$.
That part is pretty straightforward yes?

Just ignore the 2 in front, it has no effect on the anti-differentiation process.

zepdrix · Answer

$$\large f''(x)=2e^t+3\sin t$$

The anti-derivative of $\sin t$ will be $-\cos t$.
If you're unclear about that, take the derivative of $-\cos t$ and see what it gives you. You should end up with $\sin t$.

So we get something like this when we anti-differentiate once,$$\large f'(x)=2e^t-3\cos t+c$$

luigi0210 · Answer

Yup, I got that far. But what I don't understand is finding the next anti-derivative

zepdrix · Answer

Hmm it will be a similar process, but now we have a constant that showed up. Do you remember the anti-derivative of a constant? :)

zepdrix · Answer

Or maybe an easier way to ask that would be, do you remember the anti-derivative of $1$?

luigi0210 · Answer

it'll be 2e^t-3sint+Cx+D

zepdrix · Answer

yes good c:

luigi0210 · Answer

so how do I solve? Because now I have two variables..

flywalker · Answer

it's f(x) not f(t) you don't have f(x)=2e^t-3sint+Cx+D 
but f(x)=2(x**2/2)e^t-3sint+Cx+D  actually 2e^t is a number!

zepdrix · Answer

Oh I'm sure that was a typo :) lol

zepdrix · Answer

Probably should be all t's or all x's.

luigi0210 · Answer

yea, sorry about that

zepdrix · Answer

Oh I was being sloppy earlier and writing f(x) also, lolol just noticed that XD

zepdrix · Answer

So you were able to figure this out,
$$\large f(t)=2e^t-3\sin t+Ct+D$$
Now to solve for C and D, use your initial conditions,$$f(0)=0, \qquad \qquad f(\pi)=0$$

flywalker · Answer

if u put 0 for x and 0 for t you'll find  find the one variable relative to the other!!

flywalker · Answer

pi= π???Pls tell

zepdrix · Answer

$$\large f(0)=0 \qquad \rightarrow \qquad 0=2e^0-3 \sin 0+ C \cdot 0 +D$$See how that works Luigi? c:

luigi0210 · Answer

yes, pi = $$\pi $$

zepdrix · Answer

Hint,$$\large e^0=1$$

luigi0210 · Answer

Oh right I forgot. So D=2

flywalker · Answer

ty!now you will change f(t)=y and you will have    y=2et−3sint+Ct+D
(keep in mind that you have to find C and D) after that have to so lve for t!!and that's anti f(t)!!after you find the second derivative of the new function

flywalker · Answer

but i think it's really hard to solve for t!!

flywalker · Answer

it is ok?

flywalker · Answer

(is it)*

luigi0210 · Answer

all we need to do is just find the equation

flywalker · Answer

yes if you find the anti f(t) then it's easy!the problrm is that you have sin t and $$e^{t}$$