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sumpm1

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Registered: 2007-03-05

Posts: 42

Differeeenntial Eqautions Assiignment Help

Hi. I am starting an assignment  in Diffeq, the  questions and problems are in the following picture. Now I am just getting started here, but it seems that the equation (1) in question a) is separable, but what significance does alpha=1 in the question have? Can you guys see any things up front that may prove tricky? Please give me a hand here.

Ricky

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Registered: 2005-12-04

Posts: 3,791

Re: Differeeenntial Eqautions Assiignment Help

(1) in question a) is separable, but what significance does alpha=1 in the question have?

If alpha is not 1, then when you separate the variables and integrate you get a rational polynomial.  If alpha is 1, then you get ln(W).  This changes the form of the solution dramatically.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

sumpm1

Member

Registered: 2007-03-05

Posts: 42

Re: Differeeenntial Eqautions Assiignment Help

Okay, so I have found an equation for W(t), but the problem does not state the domain of alpha; I suppose it is rationals, but certainly not integers only. My concern is: if the 1-alpha term is even, then the solution is +/- the solution I have found. But I guess it wouldn't make much sense to have negative weight. Also, there are designated variables K and alpha, so what does my C variable represent? Any clues?

Edit: And do you think that the C variable is necessary here? thanks

Last edited by sumpm1 (2008-12-06 04:02:07)

sumpm1

Member

Registered: 2007-03-05

Posts: 42

Re: Differeeenntial Eqautions Assiignment Help

Now for part b I am given values for constants, but as you can see, the differential equation is much more complicated that in part a. How would I go about solving this equation for W(t)? Is it so simple that is still separable?

Ricky

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Registered: 2005-12-04

Posts: 3,791

Re: Differeeenntial Eqautions Assiignment Help

Okay, so I have found an equation for W(t), but the problem does not state the domain of alpha; I suppose it is rationals, but certainly not integers only. My concern is: if the 1-alpha term is even, then the solution is +/- the solution I have found. But I guess it wouldn't make much sense to have negative weight. Also, there are designated variables K and alpha, so what does my C variable represent? Any clues?

You are losing solutions when you take that root, you are right.  But any valid initial condition (i.e. positive) will eliminate any such solutions as well.  So in the end, you aren't losing anything.

Your C variable is indeed necessary.  Once you pick a K and alpha, there are still infinitely many solutions, depending on what your starting weight is.  If you start with 10 as opposed to 10 billion pounds, your equation will change accordingly.

And why couldn't the constants be irrationals?

b.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

sumpm1

Member

Registered: 2007-03-05

Posts: 42

Re: Differeeenntial Eqautions Assiignment Help

I really appreciate the help Ricky. So you are telling me I should be able to find the integral seen here?

Ricky

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Registered: 2005-12-04

Posts: 3,791

Re: Differeeenntial Eqautions Assiignment Help

Your equation should be:

Notice the placement of the 1-alpha.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

sumpm1

Member

Registered: 2007-03-05

Posts: 42

Re: Differeeenntial Eqautions Assiignment Help

Thanks for catching that mistake. Okay, so I have found a decent equation here, but need to know what method to use to integrate the pictured integral. My calc gives the pictured solution to that integral, but I don't know HOW to get there.

Also, I know this is skipping ahead, but my calculator can also solve the differential equation directly (handy tool for checking answers, but not a replacement for knowledge) and it gives me an implicit solution W^(1/4)= c*e^(-5t/6) +3 rather than solving for W directly. So can I just take the solution to the 4th power to solve for W directly? Since I have an initial condition W(0)=1, it would seem that I must find an explicit solution for W(t) to find C when I am done.

Finally, in part C of the assignment, it says "W(t) is as determined in part b." But W(t) is not found in equation (3), only dW/dt is. So it would seem that I would substitute

Which would keep things quite complicated. Do you have any insight into what they are looking for in part C?

Ricky

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Registered: 2005-12-04

Posts: 3,791

Re: Differeeenntial Eqautions Assiignment Help

So can I just take the solution to the 4th power to solve for W directly? Since I have an initial condition W(0)=1, it would seem that I must find an explicit solution for W(t) to find C when I am done.

You can find C without having an explicit solution for W.  It happens many times that when finding an explicit solution, you lose solutions.  It's like taking the square root without using +/-, but a whole lot worse.  In some cases you need to take the sine inverse and there you'll lose an infinite amount of solutions unless you add on an n*pi.

In this case, I don't think you'll be losing any.  But all you need to check is that once you solve for C in the implicit equation, plugging in your initial condition gives you the right value in your explicit equation.

As for the integral, work it backwards.  The solution your calculator gives you tells you exactly what u is, and that it's integrating du/u.  Find du (since you already know u), and then see if du/u is equal to the integral you want.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

sumpm1

Member

Registered: 2007-03-05

Posts: 42

Re: Differeeenntial Eqautions Assiignment Help

Hey Ricky. I do see that the calculator uses W^(1/4) - 3 as u for a substitution. But the calculator also factors the equation in a funny way so that the denominator is (w^(3/4))(w^(1/4) - 3) so that is where the calculator is getting that term. After you factor it this way, it is easier to see what the substitution for integration should be. Here is my final result (and the calculator concurs lol). I have taken the result to the  4th power and found that it does not ruin anything, plus I will need to graph W(t) and must explicitly solve for W to make this happen.

Now, is it at all apparent what I should substitute for dW/dt in part c? Or should I actually take the derivative of W(t) found in part b with respect to t to get dW/dt? Sorry if these questions seem elementary, but many of these notations are not 100% clear to me. Thanks again.

Last edited by sumpm1 (2008-12-07 21:25:16)

Ricky

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Registered: 2005-12-04

Posts: 3,791

Re: Differeeenntial Eqautions Assiignment Help

Now, is it at all apparent what I should substitute for dW/dt in part c?

Just leave it as dW/dt.  So now your equation is:

If it makes you feel more comfortable, we can set:

Now find the solution to the differential equation:

I'll give you a hint: It's really really easy.  Just remember what the integral of dW/dt is.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

sumpm1

Member

Registered: 2007-03-05

Posts: 42

Re: Differeeenntial Eqautions Assiignment Help

Thanks so much Ricky for all of your help. Since you have nearly done half of the work, it is only fair that I post my result for parts c, d, and e. I have the whole thing in word document, but this was the graphs inserted in the assignment. I found C(t) = integral(K1+K2W(t))+c = 2t/5 + W/10 + 1.