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Math Help - Differential equation - general and parictular solution

  1. #1
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    Differential equation - general and parictular solution

    Hi!

    I have been given an equation as following:

    y^{//}-y^{/}+\frac{1}{4}y = x-4

    Which I am to find the general y^g_n and a particular solution y^p_n

    I have found that this gives a general solution:

    by
    y^g_n = r^2-r+\frac{1}{4}

    One solution = \frac{1}{2}

    Which gives:
    y^g_n = Ce^\frac{x}{2} + Dxe^\frac{x}{2}


    I am then to find the particular solution y^p_n

    How to proceed is then my question...

    By setting:

    y^p_n = an+b

    and calculate from there?

    How would that look?
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    Hi jokke22.

    Quote Originally Posted by jokke22 View Post
    I am then to find the particular solution y^p_n

    How to proceed is then my question...

    By setting:

    y^p_n = an+b

    and calculate from there?

    How would that look?
    y\ =\ ax+b

    y'\ =\ a

    y''\ =\ 0

    So y''-y'+\frac14y\ =\ -a+\frac{ax+b}4\ =\ x-4. Then just equate coefficients.
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  3. #3
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    Quote Originally Posted by TheAbstractionist View Post
    Hi jokke22.
    y\ =\ ax+b

    y'\ =\ a

    y''\ =\ 0
    So y''-y'+\frac14y\ =\ -a+\frac{ax+b}4\ =\ x-4. Then just equate coefficients.

    Hmmm... I think I misinterpret what I am to do...

    I end up with...

    \frac{-4a+ax+b}{4} = x-4

    Grateful if you could show me...
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  4. #4
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by jokke22 View Post
    I end up with...

    \frac{-4a+ax+b}{4} = x-4
    \frac a4\ =\ 1

    \frac{b-4a}4\ =\ -4
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  5. #5
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    If two polynomials are the same for all x, you can equate individual coefficients. That is what The Abstractionist did.

    You can also set x to any two values and then solve the resulting two equations for a and b. For example, since \frac{-4a+ax+b}{4} = x-4, taking x= 0 gives \frac{-4a+ b}{4}= -4 or -4a+ b= -16. Taking x= 1 gives \frac{-4a+ a+ b}{4}= 1- 4 or \frac{-3a+ b}{4}= -3 so -3a+ b= -12. Those are easy to solve.
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  6. #6
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    Okey! I think I get it know... Thank you for the help!

    However, how would this then look if you were to put a solution where,

    y_n= y^g_n + y^p_n

    = Ce^\frac{x}{2}Dxe^\frac{x}{2} +
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  7. #7
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    Maybe I get it wrong... Since \frac{a}{4}=1 and \frac{b-4a}{4}=-4

    The particular solution is:

    Ce^\frac{1}{2} + Dxe^\frac{-4}{2}

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