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    the differential equation (d2y/dx2) + 6(dy/dx) = exp(-3x)

    when x = 0, y=0, dy/dx = 0

    to solve can I just put the exp(-3x) equal to zero because of the boundary condition and then solve from there, or is there something else I must do instead?
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    (Original post by danhirons)
    the differential equation (d2y/dx2) + 6(dy/dx) = exp(-3x)

    when x = 0, y=0, dy/dx = 0

    to solve can I just put the exp(-3x) equal to zero because of the boundary condition and then solve from there, or is there something else I must do instead?
    Notice how this is a disguised first order DE. Use the substitution u=\dfrac{dy}{dx}. The use the integrating factor method to get the CF and find the particular integral by plugging in a function of the form u=Ae^{-3x} and solving for A.
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    (Original post by Farhan.Hanif93)
    Notice how this is a disguised first order DE. Use the substitution u=\dfrac{dy}{dx}. The use the integrating factor method to get the CF and find the particular integral by plugging in a function of the form u=Ae^{-3x} and solving for A.

    Ahhh okay thank you, there's another question where the LHS is the same as in this question and RHS = 1 + 9x^2, do I do the same sort of thing in this question?
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    (Original post by Farhan.Hanif93)
    Notice how this is a disguised first order DE. Use the substitution u=\dfrac{dy}{dx}. The use the integrating factor method to get the CF and find the particular integral by plugging in a function of the form u=Ae^{-3x} and solving for A.
    I've often found the problem with finding solutions like that is that you then try to tend to apply them in every situation, even if you can't. Unless it's an M5 Vector differential equation, I just generally try and keep the methods used to solve normal 1st ODE and 2nd ODEs separate. In FP2 they generally make it pretty clear which is which.

    (Original post by danhirons)
    the differential equation (d2y/dx2) + 6(dy/dx) = exp(-3x)

    when x = 0, y=0, dy/dx = 0

    to solve can I just put the exp(-3x) equal to zero because of the boundary condition and then solve from there, or is there something else I must do instead?
    You were right in what you said at the start. Solve it like you normally would, with the coefficients reducing to m^2+6m=0. You can quickly spot that that's just m=0 and m=-6 and solve from there like you would any other question. You'll then need to find the PI like you always would and substitute in the boundary conditions from there.
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    (Original post by danhirons)
    Ahhh okay thank you, there's another question where the LHS is the same as in this question and RHS = 1 + 9x^2, do I do the same sort of thing in this question?
    Yes, more or less. You just need to think about what function to plug in when searching for the Particular Integral.
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    (Original post by TwilightKnight)
    I've often found the problem with finding solutions like that is that you then try to tend to apply them in every situation, even if you can't. Unless it's an M5 Vector differential equation, I just generally try and keep the methods used to solve normal 1st ODE and 2nd ODEs separate. In FP2 they generally make it pretty clear which is which.
    I disagree, I'm afraid. Solving most linear first order ODEs tends for be quite a bit less in terms of legwork than solving a second order one simply because there's less differentiation going on. So if a substitution simplifies things then you should almost definitely use it.

    EDIT: I can't see why you'd use the substitution if it was clearly not going to work. It tends to be pretty obvious if an ODE of that form is a first order one in disguise. Obviously some substitution are harder to spot than others so if you can't spot it, you may as well try your 2nd order method.
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    (Original post by Farhan.Hanif93)
    I disagree, I'm afraid. Solving most linear first order ODEs tends for be quite a bit less in terms of legwork than solving a second order one simply because there's less differentiation going on. So if a substitution simplifies things then you should almost definitely use it.
    I wasn't trying to say your method was wrong at all. In fact, it's brilliant. It makes it much quicker to solve.

    What I was trying to say, is that by this point in A Level Maths, you usually know a half dozen ways to solve certain equations. If you're doing the Further Mechanics units, then you probably know even more.

    Sometimes, you'll forget that in order to do a substitution etc, you needed very specific criteria, and the methods begin to 'bleed' together. It's happened to me in the past, and I've tried to solve questions which just didn't lend themselves to that method and gone in circles for 20 minutes on a past paper before realising actually; do that.

    All I was saying that was, in terms of making sure you do your best on the exam, sometimes it's easier to go in with a set way of doing something, rather than trying to improvise at the last minute with a method you might be unfamiliar with.
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    (Original post by TwilightKnight)
    I wasn't trying to say your method was wrong at all. In fact, it's brilliant. It makes it much quicker to solve.

    What I was trying to say, is that by this point in A Level Maths, you usually know a half dozen ways to solve certain equations. If you're doing the Further Mechanics units, then you probably know even more.

    Sometimes, you'll forget that in order to do a substitution etc, you needed very specific criteria, and the methods begin to 'bleed' together. It's happened to me in the past, and I've tried to solve questions which just didn't lend themselves to that method and gone in circles for 20 minutes on a past paper before realising actually; do that.

    All I was saying that was, in terms of making sure you do your best on the exam, sometimes it's easier to go in with a set way of doing something, rather than trying to improvise at the last minute with a method you might be unfamiliar with.
    Fair enough.

    I'm not too sure what these specific criteria are, though. There shouldn't be anything that is overly problematic when considering a substitution. It will either simplify things, or it won't. You don't need to worry about any conditions (AFAIK) so I'm not sure how it would become a problem.

    Something important about maths, which is not really pushed enough in A-Level maths IMO, is that you should be able to string all your ideas together with confidence. So by combining the knowledge of substitution with the idea of linear 1st order ODEs, you can simplify a 2nd order ODE. In A-Level exams they seem to cripple this by either asking you to solve a question with just a single idea to it OR they will completely lead you through the problem by indicating what to do at each stage through breaking up the question into parts, which completely removes the opportunity for you to think of the idea yourself.

    Exams tend to be time pressured and a method which can save time should really be considered if you're confident with it.

    Besides this, the OP is actually at University doing a non-maths degree so this idea of A-Level maths exams shouldn't really apply to him. :p:
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    (Original post by Farhan.Hanif93)
    Fair enough.

    I'm not too sure what these specific criteria are, though. There shouldn't be anything that is overly problematic when considering a substitution. It will either simplify things, or it won't. You don't need to worry about any conditions (AFAIK) so I'm not sure how it would become a problem.

    Something important about maths, which is not really pushed enough in A-Level maths IMO, is that you should be able to string all your ideas together with confidence. So by combining the knowledge of substitution with the idea of linear 1st order ODEs, you can simplify a 2nd order ODE. In A-Level exams they seem to cripple this by either asking you to solve a question with just a single idea to it OR they will completely lead you through the problem by indicating what to do at each stage through breaking up the question into parts, which completely removes the opportunity for you to think of the idea yourself.

    Exams tend to be time pressured and a method which can save time should really be considered if you're confident with it.

    Besides this, the OP is actually at University doing a non-maths degree so this idea of A-Level maths exams shouldn't really apply to him. :p:
    Lol, didn't spot that.

    Anyway. I completely agree. When I did my C4 paper, I had already recently self taught bits of FP2 and FP3, and so I used a MacLaurin series to solve a binomial question, since it seemed the most logical way. I was able to find a perpendicular vector by using the cross product rather than making two line equations and doing about 20 lines of solving.

    Unfortunately, in my case, I also tried to integrate a question directly rather than using parts because I remembered how to use Eulers relation on an integral of form e^x(trig function)[x] or using trig functions with powers more than 2-3, but the question didn't lend itself to it, and instead of getting a theta at the end, I got just a '1'. No marks. So I guess I'm an advocate of using whatever way works, but also have a cautionary tale to tell in order to make sure I don't do it in the future .
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    (Original post by TwilightKnight)
    Lol, didn't spot that.

    Anyway. I completely agree. When I did my C4 paper, I had already recently self taught bits of FP2 and FP3, and so I used a MacLaurin series to solve a binomial question, since it seemed the most logical way. I was able to find a perpendicular vector by using the cross product rather than making two line equations and doing about 20 lines of solving.

    Unfortunately, in my case, I also tried to integrate a question directly rather than using parts because I remembered how to use Eulers relation on an integral of form e^x(trig function)[x] or using trig functions with powers more than 2-3, but the question didn't lend itself to it, and instead of getting a theta at the end, I got just a '1'. No marks. So I guess I'm an advocate of using whatever way works, but also have a cautionary tale to tell in order to make sure I don't do it in the future .
    Spotting which tricks simplify things and which complicate the issue is a different matter. That's where mathematical insight comes into play.
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    I feel violated by my degree, I see an ODE and my first reaction is to use Laplace Transforms to solve it.
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    (Original post by TwilightKnight)
    Lol, didn't spot that.

    Anyway. I completely agree. When I did my C4 paper, I had already recently self taught bits of FP2 and FP3, and so I used a MacLaurin series to solve a binomial question, since it seemed the most logical way. I was able to find a perpendicular vector by using the cross product rather than making two line equations and doing about 20 lines of solving.

    Unfortunately, in my case, I also tried to integrate a question directly rather than using parts because I remembered how to use Eulers relation on an integral of form e^x(trig function)[x] or using trig functions with powers more than 2-3, but the question didn't lend itself to it, and instead of getting a theta at the end, I got just a '1'. No marks. So I guess I'm an advocate of using whatever way works, but also have a cautionary tale to tell in order to make sure I don't do it in the future .
    I seen a couple of C4 papers were they were like if Maclaurin series was used they had to refer the answer to their "senior examiner". I remember I used the cross product for a C4 past paper question and the method wasn't on the mark scheme so I had to give myself zero for that question .
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    (Original post by anshul95)
    I seen a couple of C4 papers were they were like if Maclaurin series was used they had to refer the answer to their "senior examiner". I remember I used the cross product for a C4 past paper question and the method wasn't on the mark scheme so I had to give myself zero for that question .
    Really? According to my teacher, as long as you got the right answer and used a perfectly valid method you should still get the marks, as long as they didn't explicitly state "by using the dot product" or whatever.
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    (Original post by dknt)
    Really? According to my teacher, as long as you got the right answer and used a perfectly valid method you should still get the marks, as long as they didn't explicitly state "by using the dot product" or whatever.
    well I was only practising so not much harm done, but I did ask my teachers the next day I got a mixed response.
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    (Original post by anshul95)
    well I was only practising so not much harm done, but I did ask my teachers the next day I got a mixed response.
    The main problem is, a lot of the people marking the exams may only be familiar with Maths up to A Level. Sometimes, if the exam marking isn't going as quickly as it should, then they even draft in people that don't even have that - my sister was an exam marker last year, and she predominantly does Psychology, but they were having trouble marking all of the Geography papers in time, so they gave her a few hundred to mark as well.

    90 times out of a 100, that wouldn't be a problem, exams are fairly standard these days, and most of the time, the markscheme is just like a checklist. However, with something like Maths, where you can often use methods outside of the syllabus for that paper, you really need someone who knows Maths like the back of their hand. I guess that's where the 'refer to coordinator' stuff comes from, but I'm willing to bet at least some of the time, notes like that just get ignored when the markers are tired and want to go to bed/ go out etc.

    So, I'd say generally, use the method appropriate for the paper, but, if you're absolutely certain of the method you intend to use and you know for certain you will get the right answer quicker and more efficiently, go for that. Because if you're more familiar with that method, the chances are you aren't going to slip up with it, which means that if you get a low score back, it has been mismarked and you can get it remarked etc. Chances are it won't, and you'll get the mark you deserve.
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    (Original post by Farhan.Hanif93)
    Notice how this is a disguised first order DE. Use the substitution u=\dfrac{dy}{dx}. The use the integrating factor method to get the CF and find the particular integral by plugging in a function of the form u=Ae^{-3x} and solving for A.

    Attached working.

    Is that the sort of thing you meant or have I misunderstood?

    Dan
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    (Original post by danhirons)
    Attached working.

    Is that the sort of thing you meant or have I misunderstood?

    Dan
    You've made a fair few mistakes from what I can see. I'm not at home at the moment so I can't really work through it but from what I can see:

    p(x) is not 1, it should be 6. Which gives you an integrating factor of exp(6x).

    You then multiply through both sides of the DE by that integrating factor and note that the LHS is now d/dx[m * exp(6x)]. Follow that with integration (as you did).

    Then remember that when you rearrange both sides for m, you must divide through by exp(6x) and that means that you must divide the arbitrary constant by exp(6x) too. You appear to have forgotten that.

    Then use your conditions to solve for your constant and finally integrate again for y. That's your complementary function.

    Your method for finding the PI looks fine but I haven't checked the working.
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    (Original post by Farhan.Hanif93)
    ...
    I think I've missed something.

    I thought the whole point of treating is a 1st ODE was so that, by using the substitution, you're effectively just solving 2 1st ODEs, consequently finding both the constants and the particular integral at the same time, without actually explicitly looking to find any of them.

    Usually, I wouldn't post a full solution, but I think I'm just as confused as the OP was now. I like the method, so I want to use it in the future where possible:

    Name:  Diff1.JPG
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    Name:  Diff2.JPG
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    Is it right? And if not, where did I bugger up? :/.

    EDIT: I used C twice, but they're actually two different constants, so ignore that.
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    (Original post by TwilightKnight)
    I think I've missed something.

    I thought the whole point of treating is a 1st ODE was so that, by using the substitution, you're effectively just solving 2 1st ODEs, consequently finding both the constants and the particular integral at the same time, without actually explicitly looking to find any of them.

    Usually, I wouldn't post a full solution, but I think I'm just as confused as the OP was now. I like the method, so I want to use it in the future where possible:

    Name:  Diff1.JPG
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    Name:  Diff2.JPG
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    Is it right? And if not, where did I bugger up? :/.

    EDIT: I used C twice, but they're actually two different constants, so ignore that.
    That looks fine to me. The whole point of it was that you didn't have to go through the process of dealing with a more complicated 2nd order ODE. Some could argue that it was more simple to just solve it as it was but each to their own.
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    (Original post by Farhan.Hanif93)
    That looks fine to me. The whole point of it was that you didn't have to go through the process of dealing with a more complicated 2nd order ODE. Some could argue that it was more simple to just solve it as it was but each to their own.
    After we had the back and forth I went back to doing M4 and M5 and a lot of the Damped/Forced Harmonic equations and the Vector differentials you set up are actually disguised 1st orders, so I figured that if it did come up, it'd be nice to solve a potentially easier equation and save myself time for the rest of the paper.

    Anyway, glad to see it works, I shall use it in future .
 
 
 
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