Question

Another calculus question for a medal!
(posting the question below)

Answer 1

Find the particular solution that satisfies the initial conditions.
\[f"(x)=\frac{1}{2}(e^x+e^{-x})\]f'(0)=0 and f(0)=0

Answer 2

I know I need to take the antiderivative.

Answer 3

And I know that I take out the constant, but what do I do with what's inside the parentheses?

Answer 4

@zepdrix
Do you have an idea for this one perhaps?

Answer 5

Oh, you could be sneaky if you wanted to....
maybe they don't want you to do that though.

Answer 6

\(\large\rm \frac{1}{2}(e^x+e^{-x})=cosh x\)
have you learned about hyperbolic functions yet?

Answer 7

Hey, sorry. My connection timed out and just came back on.

Answer 8

I've seen that before, but we haven't it learned about them yet. So I'm going to assume that I can't use them.

Answer 9

\[\large\rm \int\limits y''=\frac{1}{2}\int\limits e^x+e^{-x}~dx\]so then,\[\large\rm y'=\frac{1}{2}(e^x-e^{-x})+c\]Any confusion on that first integral? :o

Answer 10

So, it's literally the same thing?

Answer 11

Mmm no. Not exactly :o

derivative of e^(-x) is -e^(-x)
So the integral of e^(-x) is also -e^(-x),
right?

Answer 12

The sign between the terms should change from addition to subtraction.

Answer 13

Ah yes, I see now.

Answer 14

So to show that in writing, could you separate the e^x and e^-x and take there derivatives separately then add them back together?

Answer 15

their*

Answer 16

their integrals separately? yes good.

Answer 17

Use the initial data to solve for your unknown constant.\[\large\rm y'(0)=0=\frac{1}{2}(e^0-e^{-0})+c\]Solve for c.

Answer 18

1-1 x .5 = 0 ...so c = 0

Answer 19

\[\large\rm y'(x)=\frac{1}{2}(e^x-e^{-x})\]k good :)
integrate again.

Answer 20

It would be the same except the - would change back to a +

Answer 21

\[\large\rm y(x)=\frac{1}{2}(e^x+e^{-x})+C\]k cool
find your constant :)\[\large\rm y(0)=0=\frac{1}{2}(e^0+e^{-0})+c\]

Answer 22

My connection is finally working now.
And thank you again!

Answer 23

c=1