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Why does C3(t) = that?

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- Jan 16th 2011, 06:32 AMadam_leedsIntegration help, how do you do this?
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Why does C3(t) = that? - Jan 16th 2011, 06:34 AMProve It
So hard to read ><

Is it ?

If so, it's first order linear, so use the Integrating Factor method. - Jan 16th 2011, 06:38 AMadam_leeds
- Jan 16th 2011, 06:42 AMProve It
I find it hard to believe that you've been given a first order linear ODE without knowing what the Integrating Factor method is.

Do you at least know about Separation of Variables? - Jan 16th 2011, 06:45 AMadam_leeds
- Jan 16th 2011, 06:54 AMadam_leeds
- Jan 16th 2011, 07:07 AMProve It
OK, then read very carefully, for I will post this only once...

To understand the Integrating Factor method, you need to have a basic idea of the Product Rule, i.e. if are functions of , then .

Now examine a general first-order linear ODE:

.

What we aim to do is to multiply both sides of the DE by some function of , call it , so that the LHS is a Product-Rule expansion, which can then be written as a derivative, and therefore be integrated. BTW is known as the Integrating Factor (hence the name of the method)

So multiplying both sides by gives

.

The LHS will be a Product Rule expansion if (Can you see why?)

This is a separable ODE, so solving for we have

(You can omit modulus signs and constants because any scalar multiple of the Integrating Factor will do)

.

So, if we multiply both sides of a first-order linear ODE by , we create a Product Rule expansion on the LHS which we can write as a single derivative and integrate both sides. Watch what happens...

which solves the ODE.

Looks complicated in theory, in practice, extremely easy. Look at your problem...

.

Here your , so the Integrating Factor is .

Multiplying both sides by the Integrating Factor gives

where is the Integration Constant

. - Jan 16th 2011, 07:11 AMadam_leeds
- Jan 16th 2011, 07:27 AMHallsofIvy
It

**doesn't**. You must have misunderstood the question. is a solution to that equation only for . Since you have NOT taken a course in differential equations, I suspect you were NOT intended to actually solve that equation. I suspect that the problem was really "Find a value of A such that satisfies the equaton ".

And you can do that just by differentiating the given function and plugging into the equation: if then so the equation becomes . That is the same as and since an exponential is never negative, we can divide both sides of the equation by to get or .