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Diff EQ's are killing me

  1. Feb 8, 2006 #1
    A teacher who mumbles proofs to himself without explaining them to the rest of the class coupled with A First Course In Differential Equations with Modeling Applications 8th edition by Dennis G. Zill are my biggest foes in conquering Diff EQ's.

    Right now I am stuck on Boundary-Value Problems.

    For example:

    Find an interval centered about x = 0 for which the given IVP has a unique solution:

    10. y" + (tanx)y = e^x, y(0) = 1, y"(0) = 0

    So naturally, I need to know what y is, right? How do I find that? All the examples in the book simply say "blah blah is a solution to blah blah blah. Verify this." without explaining how they got it in the first place... Jesus Christ! They even say that I NEED TO KNOW how to solve these things, but never show me!


    Wait... is this the thing where I go [tex]e^{ \int P(x) dx}[/tex]? I still blame the book for not being clear...

    EDIT 2:

    No... not it's not... I am clueless. Especially since the NEXT section in the book goes into reduction of order, that means that I SHOULD know how to do this...

    EDIT 3:

    Jesus Christ... I just figured out I posted this on the wrong forum, too... someone just shoot me...
    Last edited: Feb 8, 2006
  2. jcsd
  3. Feb 9, 2006 #2

    Tom Mattson

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    Gold Member

    Your post is in the correct forum. This is where we host homework/textbook questions in math from calculus on up.

    Honestly, I think your biggest foe is that you don't read either the question or the textbook very carefully.

    No, you don't need to know the solution of the IVP to answer the question. The key word here is "unique". Does that ring any bells for you? It should, because there is a "uniqueness theorem" in your book! Look it up, read the antecedent conditions of the theorem, and investigate the interval(s) on which your equation satisfies them. You can do that without solving the equation.

    It should be perfectly clear that that method is only for first order equations, and that your equation is of second order.
  4. Feb 10, 2006 #3
    I read both carefully many times. Maybe I'm just stupid, I don't know, but it's definately not that I'm lazy.

    I checked the uniqueness theorem in the book. All it says is if everything is continuous on the interval and (here it gets fuzzy...) the coefficiant of the highest order of y isn't 0, then a solution of y(x) of the IVP exists on the interval and is unique.

    Super. That's basically telling me it's possible, not how I can find it.

    Let's move on to an example then. Oops! Every example in this section starts off with "Verify that (blah blah blah) is a solution to (blah blah blah)" and then uses the solution to find the boundry. That's my entire problem. The book doesn't explain how to do it without having the solution.

    Like I said, maybe I'm brain dead, but I DO NOT understand this. As far as I can tell, only 1 student in class does understand it, and I can't always get ahold of her (there are only 11 students in class, btw).

    Another thing I would like some help on is reduction of order. I can find a second solution to a homogenious DE no problem, but then a problem in the book tells me to find a general solution to a nonhomogenous DE. Looking through the section, only a small snippet in the "remarks" at the end says "Oh yeah, you can find a general solution to nonhomogenous DE's, by the way." without going into further detail.

    Yeah, I figured out that that equation doesn't work...

    I'm just really confused by all of this.
  5. Feb 10, 2006 #4
    You're not asked to find it. You're asked for an interval on which the initial value problem in your first post you has a unique solution. For the question you asked you have absolutely no need to find a general solution.
  6. Feb 11, 2006 #5
    If [itex]a_1 \left( x \right)y'' + a_2 \left( x \right)y' + a_3 \left( x \right)y = 0[/itex] has a solution y = f(x) then the equation

    a_1 \left( x \right)y'' + a_2 \left( x \right)y' + a_3 \left( x \right)y = h\left( x \right)

    has a solution y = u(x)f(x) where u(x) can be determined if you set y = u(x)f(x), differentiate as required, and then substitute y = u(x)f(x) and its derivatives into the non homogeneous equation. This method is usually for equations with non constant coefficients but you might be asked to use reduction of order to solve an equation with constant coefficients.

    In that case you'll typically have linear combinations of sine and cosine as a solution to the homogeneous equation so something like y = Asin(wx) + Bcos(wx). In that case just ignore the constants and take either the sine or cosine and set y = u(x)sin(wx) (or u(x)cos(wx)) to find the solution to the non homogeneous equation.
  7. Feb 12, 2006 #6
    Ok, but is anybody going to actually explain how to find it? The example in the book uses a solution to the problem. It doesn't show an example without one, so I have no idea how to do this. Furthermore, the problem that I am doing does not have a solution in the back of the book so I have no idea if I'm right or not.

    Benny, thanks for the help.

    EDIT: Wait... I think I finally figured it out. It's weird how answers suddenly hit you...

    You guys were right, I did not understand the problem well, that's why I had no idea what to do. Thank you for your help.

    EDIT 2:

    Hey, I have another stupid question:

    Doing a chemical reaction equation and I got to this point:

    [tex]{{150 - x}\over{60 - x}} = ce^{90tk}[/tex]

    How do I solve for X? This is an algebra problem, I know, but I am completely stumped. I asked my younger sister, and she doesn't know either. :(
    Last edited: Feb 12, 2006
  8. Feb 13, 2006 #7
    \frac{60-x}{60-x} \times \frac{150-x}{60-x}=ce^{90tk}






    this can be simplified down to:

    you'll probably want to expand it though, so you can tell what the hell it is doing... cause that's kind of a weird function to look at without graphing.
    Last edited: Feb 13, 2006
  9. Feb 13, 2006 #8
    Wow... I kept multiplying by the denominator and ending up with 1 = blah blah.

    Nah, I just needed to find out X at time T. Thanks for the help.
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