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Math Help - Stuck on a particular integral

  1. #1
    Super Member craig's Avatar
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    Stuck on a particular integral

    Hi, I've got the following question I need a bit of help on.

    \frac{d^2y}{dx^2} + \frac{dy}{dx} = 2x + 3

    It's the particular integral that I'm not too sure on.

    Normally I would make y = px + q, differentiate that, put it into the equation to get my particular integral.

    Have tried this and it didn't work, I'm guessing because there is no y function in the original equation, only derivatives of y.

    Due to this, would I set y = px^2 + qx + r for example?

    Thanks
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  2. #2
    Super Member craig's Avatar
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    Just found an answer and they have used y = px^2 + qx + r as the particular integral.

    I'm curious however, is this case because there is no y, or is that just a coincidence?

    Thanks
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  3. #3
    AMI
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    By taking y=px^2+qx+r, you get the solution y=x^2+x+r, where r\in\mathbb{R} is a constant. (Is that right? I hope I didn't mess up the calculations..)

    However, y=x^2+x-a\exp(-x)+b, where a,b\in\mathbb{R} are constants is also a valid solution, but it doesn't fit in the form that you obtained. So, using that method, you don't arrive to all the solutions. Just thought I'd point that out, I know that's not what you asked..
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  4. #4
    Super Member craig's Avatar
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    Quote Originally Posted by AMI View Post
    By taking y=px^2+qx+r, you get the solution y=x^2+x+r, where r\in\mathbb{R} is a constant. (Is that right? I hope I didn't mess up the calculations..)

    However, y=x^2+x-a\exp(-x)+b, where a,b\in\mathbb{R} are constants is also a valid solution, but it doesn't fit in the form that you obtained. So, using that method, you don't arrive to all the solutions. Just thought I'd point that out, I know that's not what you asked..
    Here's my final general solution:

    y = A + Be^{-x} + x^2 + x, how is it I'm missing a solution, the r does not get included due to there being no y in the original question :S

    Sorry if I've misread your post.
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  5. #5
    AMI
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    Well, I guess then I am the one that misunderstood something.. I thought you said you are looking for solutions in the form y=px^2+qx+r with p,q\in\mathbb{R} constants..
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    Super Member craig's Avatar
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    Quote Originally Posted by AMI View Post
    Well, I guess then I am the one that misunderstood something
    You might not have, it could quite possibly be me being rather slow, todays not been a good day

    Quote Originally Posted by AMI View Post
    I thought you said you are looking for solutions in the form y=px^2+qx+r with p,q\in\mathbb{R} constants..
    Could you explain how I've not got all the solutions if you don't mind?

    Thanks for the reply btw
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  7. #7
    AMI
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    Well, actually you have. You wouldn't have had, if what I understood was right. Which, as it seems, is not. I simply didn't (and still don't) get what you meant when you said "set y=px^2+qx+r", but since you arrived at this:
    Quote Originally Posted by craig View Post
    Here's my final general solution:

    y = A + Be^{-x} + x^2 + x
    then my post was useless
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  8. #8
    Super Member craig's Avatar
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    Quote Originally Posted by AMI View Post
    I simply didn't (and still don't) get what you meant when you said "set y=px^2+qx+r"
    This is a particular integral, part of the general solution to a second order differential equation.

    Think we got our wires crossed somewhere, I wouldn't worry bout it
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  9. #9
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    Quote Originally Posted by craig View Post
    Hi, I've got the following question I need a bit of help on.

    \frac{d^2y}{dx^2} + \frac{dy}{dx} = 2x + 3

    It's the particular integral that I'm not too sure on.

    Normally I would make y = px + q, differentiate that, put it into the equation to get my particular integral.

    Have tried this and it didn't work, I'm guessing because there is no y function in the original equation, only derivatives of y.

    Due to this, would I set y = px^2 + qx + r for example?

    Thanks
    Normally you'd try y_p = ax + b.

    But y = b is part of the homogenous solution. When that happens the usual thing to try is multiplying your initial model for the particular solution by x:

    y_p = x(ax + b) = ax^2 + bx.
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  10. #10
    Super Member craig's Avatar
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    Of course, the auxiliary equation has a root at zero, there getting rid of the e in the equation.

    Believe it or not I do know this, I just seem to have a habit of missing the obvious at times

    Quote Originally Posted by mr fantastic View Post
    But y = b is part of the homogenous solution. When that happens the usual thing to try is multiplying your initial model for the particular solution by x:

    y_p = x(ax + b) = ax^2 + bx.
    So is it best not to use something like ax^2 + bx + c, just ax^2 + bx?

    Thanks for the reply
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  11. #11
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    I find this differential equation problem interesting because I've never seen it before. I don't understand the above explanations of the problem. Can anyone please show me the steps of solving it?

    By the way, this problem isn't in the domain of multivariable calculus, right?
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