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Need help going back through test question

  1. Nov 3, 2016 #1
    1. The problem statement, all variables and given/known data
    Use variation of parameters to solve ##\frac{d^2y}{dx^2} + 4y = sec(2x)##

    2. Relevant equations
    Really just curious about comments, see picture.

    3. The attempt at a solution
    I'm having a hard time deciphering the comments I got back on an exam.

    For the first comment, is there a system of equations? Sure, they were:

    $$u_1'y_1+u_2'y_2=0$$
    $$u_1'y_1'+u_2'y_2'=g(t)$$

    I don't see the point of writing that on an exam, where there's seven other questions with a hour time limit.

    He circled the Wronskian and asked what it represented... the determinant of the Wronskian ?


    ##y_1y_2'-y_2y_1'= W##

    The note about it not being variation of parameters is the most confusing... what do you call it then?

    Then there's the note about where does this come from? I guess I could re-derive the general formula, but I didn't feel it was necessary at the time. Lesson learned.

    The next comment confuses me as well, how are:

    $$y_p = -cos(2x)\int{\frac{tan(2x)dx}{2}}+sin(2x)\int{\frac{dx}{2}}$$
    $$y_p=-cos(2x)\int{\frac{-du}{4u}} + \frac{1}{2}xsin(2x)+C_2$$
    $$y_p= \frac{1}{4}cos(2x)ln|cos(2x)| +C_1 + \frac{1}{2}xsin(2x)+C_2$$

    Not equal, just not seeing it.

    I guess I should have just wrote...

    $$y=\frac{1}{4}cos(2x)(ln|cos(2x)|+C_1) + \frac{1}{2}sin(2x)(x+C_2)$$

    And that would just give the general solution. Probably better that way instead of making ##C_1 = C_2 = 0## in my head and getting sloppy to keep it as ##y_p##

    On the bit about there being no arbitrary constants in the particular solution, sure, probably didn't handle that the best way. The book says treat them as zero. I'm thinking maybe just taking definite integrals from 0 to x would be a better way to say the same thing or just changing ##y_p## to ##y##

    The comment about how many answers? There's infinity many. The constants in front of the homogeneous part can be whatever. Not understanding how I missed 90% of the solution.

    I talked to some other classmates, they said I HAD to do it using Cramer’s . Uhhhh...? Why? It leads to the same thing, especially since the order was only two.

    Thanks for any help or advice!
     

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    Last edited: Nov 3, 2016
  2. jcsd
  3. Nov 4, 2016 #2

    Simon Bridge

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    How about you explain your reasoning for how you solved it.
     
  4. Nov 4, 2016 #3
    Err? My reasoning is already there. You solve a system of equations for ##u_1'## and ##u_2'##, take the integral of ##u_1'## and ##u_2'##, and bam ##y_p=y_1u_1+y_2u_2##, you're done. ##y_1## and ##y_2## are obtained from the homogeneous part.


    I'm really interested in interpreting the comments.
     
  5. Nov 4, 2016 #4

    Mark44

    Staff: Mentor

    @BillhB, the image you attached is too small and too faint to read.
     
  6. Nov 4, 2016 #5
    My apologies, let me know if this is better. If not, I can type it all out just worried it'll lose context that way.


    20161102_113344.jpg
     
  7. Nov 5, 2016 #6

    Simon Bridge

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    I don't think anyone can reflect, with authority, on the comments. The person you should ask about this is the person who made the comments. Anything we may say on that matter will be speculative. After all, it may be that they were just wrong. Even if they were, what authority would we have?
    However we can comment on your physics. This seems to be the more productive use of our time.

    I can see the maths you used.
    Your "reasoning" should justify the maths you used.
    What is the physics behind what you did?

    From what I can see of your working, you have failed to justify your maths ... you should comment your equations etc as you go.
     
  8. Nov 5, 2016 #7
    Hey, not really looking for authority, just wondering how to interpret the comments. I rework missed questions, but I don't really understand what I missed. I know I didn't justify everything, but like I said I didn't think there was a need to. It was a DE class, the justification for the general solution is just LA. That's something I'll do going forward with this professor. I'm a bit dumbfounded that justification was needed here, it seems trivial.

    I plan to talk to the professor about it but that's not something I normally like to do. It feels like when you talk to them about this stuff they get the instant feeling that you're grade grubbing which I couldn't bring myself to do in the first place. Too much pride. It is slightly irritating that this question was the difference between a 100 and 86 though on the exam.

    I don't think I understand the physics part. Is there any physics here? We're just studying techniques to solve DE's. The only real "physicy" stuff we've done is oscillations.

    But thanks for taking the time to reply, I appreciate it.
     
  9. Nov 5, 2016 #8

    Mark44

    Staff: Mentor

    The new image is slightly better than the original, but due to the poor lighting in the image and the instructor's handwriting, it's still hard to read. Of the part I could read, I don't understand some of the instructor's comments, such as questioning ("What is this?") for the Wronskian, which seemed pretty clear to me. Checking with the wikipedia article on variation of parameters, it seems to me that what you did matched what was in the article fairly closely.
     
  10. Nov 5, 2016 #9
    Hey Mark,

    Thanks for the input, yeah I thought what I did was variation of parameters and couldn't make out some of the comments either. I'll talk to the professor. Thanks for the input and your time.
     
  11. Nov 5, 2016 #10

    Simon Bridge

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    There are two ways to work through a variation of parameters problem - by rote or by derivation. You appear to have chosen the rote approach. If the derivation approach was expected, then that may result in comments like what you got if the paper was marked late at night or the rote approach was not explicitly taught in class or something. Part of why spelling things out as you go is a good idea.

    The derivation approach would go like this:

    v----------------------------------------v

    Solving ##y'' + 4y = \sec 2x## ...(1):
    ... this is 2nd order non-homogeneous DE of form ##py'' + qy' + ry = g##

    Method: Variation of Parameters:
    The general solution is: ##y=c_1y_1+c_2y_2+y_p##
    ... where ##y_1## and ##y_2## are independent solutions of ##y''+4y=0## ...(2)

    To find ##y_p##, find ##u(x)## and ##v(x)## so that: ##y_p = uy_1+vy_2## is a solution to (1)
    [this is the "variation of parameters" part!]

    Guessing solutions to (2) of form ##y=e^{\lambda x}##, gives (2) characteristic equation ##\lambda^2+1 =0## which suggests ##\lambda=\pm\sqrt{-1} = \pm i##
    ... so that ##y_1=e^{ix}## and ##y_2=e^{-ix}## ...(3,4)

    Therefore ##y_p = ue^{ix}+ve^{-ix}##
    ... which leads to the system of equations:
    ##u'e^{ix} + v'e^{-ix} = 0## ...(5)
    ##iu'e^{ix} - iv'e^{-ix} = \sec 2x## ...(6)

    Solving this system gives:
    $$u'=-\frac{ie^{-ix}\sec 2x}{e^{2ix} + e^{-2ix}} \qquad\qquad v'=\frac{ie^{ix}\sec 2x}{e^{2ix} + e^{-2ix}}$$
    ... 2x 1st order DEs. Solve these to get ##y_p##

    ^----------------------------------------^

    The "by rote" approach goes something like this:

    v----------------------------------------v

    Solving ##y'' + 4y = \sec 2x## ...(1):
    ... this is 2nd order non-homogeneous DE of form ##py'' + qy' + ry = g##

    Method: Variation of Parameters:
    The solution is given by ##y=c_1y_1 + _2y_2 + y_p## where ##y_1## and ##y_2## are a fundamental set of solutions for ##y'' + 4y = 0##, ##W(y_1,y_2)## is the wronskian determinant, and ##y_p## is given by:
    $$y_p(x) = y_1\int \frac{y_2\sec 2x}{W(y_1,y_2)}\text{d}x + y_2\int \frac{y_1\sec 2x}{W(y_1,y_2)}\text{d}x$$
    Proceed by finding ##y_1##, ##y_2##, and ##W##.

    ^----------------------------------------^

    Comparing these with your work and the comments, it looks like the 1st approach was expected and you used the second one. Technically the second one is not variation of parameters so much as the general result of the method ... when it works. It misses out "90%" of the working of 1st one. See where the comments may have come from?
     
    Last edited: Nov 5, 2016
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