Find the solution to the initial value problem in explicit form.

;

I think I screwed up somewhere because when I go to solve for C I get .

The Book's answer in explicit form is

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- September 5th 2011, 07:32 PMVitaXInitial Value Problem
Find the solution to the initial value problem in explicit form.

;

I think I screwed up somewhere because when I go to solve for C I get .

The Book's answer in explicit form is - September 5th 2011, 07:41 PMSammySRe: Initial Value Problem
- September 5th 2011, 07:55 PMVitaXRe: Initial Value Problem
Why does C go in the denominator? I thought that since you are taking the reciprocal of both sides that will just be another constant, C, so I left it alone. It's these types of things that are confusing me, the work is easy but knowing when C gets changed or not is a bit confusing here. Why exactly is C effected here if there is no variable effecting it?

- September 5th 2011, 09:11 PMSammySRe: Initial Value Problem
It's pretty basic algebra !

- September 5th 2011, 10:09 PMVitaXRe: Initial Value Problem
Yeah I screwed up on that part.

My main question is this though, after isolating does it become this: or this: .

I know this is hard to explain but I'll show a link to why I'm a bit confused with the subject of C.

Problem 7 Page 4

If you go to**page 4 and problem 7**, you can see where he raises both sides to and simply becomes C. So does become here or no? I only ask that because both and are constants, so isn't that the reason why in that problem it is written as ? The math is easy, but working with the C is confusing me with situations like this. - September 6th 2011, 06:29 PMSammySRe: Initial Value Problem
He uses C (upper case) as the original constant and c (lower case) as the 'derived' constant. Therefore, he has c = e^C, where ^ denotes the exponentiation operation.

In your case, you just need to be consistent about which way you use the constant, C, both when you plug-in the initial conditions to solve for C and then when you substitute that value of C back into your expression for y.

You should get the same answer for y either way.