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Math Help - Method of Undetermined Coefficients

  1. #1
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    Method of Undetermined Coefficients

    Find yc(complimentary solution) and set up yp(particular solution) by the method of undetermined coefficients.

    y''+y'+(1/2)y = x*e^(-x/2)+x*cos(x/2)

    ok so
    yc = c1 e^(-x/2) cos(x/2) + c2 e^(-x/2) sin(x/2)

    Finding yp is confusing me. Could someone please explain how to come to the answer. What assumptions must be made?

    ***ANSWER***
    yp =(Ax + B)e^(-x/2) + (Cx+D)cos(x/2) + (Ex + F) sin(x/2)
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  2. #2
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    Quote Originally Posted by ur5pointos2slo View Post
    Find yc(complimentary solution) and set up yp(particular solution) by the method of undetermined coefficients.

    y''+y'+(1/2)y = x*e^(-x/2)+x*cos(x/2)

    ok so
    yc = c1 e^(-x/2) cos(x/2) + c2 e^(-x/2) sin(x/2)

    Finding yp is confusing me. Could someone please explain how to come to the answer. What assumptions must be made?

    ***ANSWER***
    yp =(Ax + B)e^(-x/2) + (Cx+D)cos(x/2) + (Ex + F) sin(x/2)

    Is the confusing part how the answer is setup or do you not know how to go about finding y_P?
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  3. #3
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    Quote Originally Posted by dwsmith View Post
    Is the confusing part how the answer is setup or do you not know how to go about finding y_P?
    I do not know how to find yp. I read the chapter in the book but of course it's no help.
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  4. #4
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    Do you know what annihilates x*e^{\frac{-x}{2}}+x*cos{\frac{x}{2}}?
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  5. #5
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    Quote Originally Posted by dwsmith View Post
    Do you know what annihilates x*e^{\frac{-x}{2}}+x*cos{\frac{x}{2}}?
    no idea
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  6. #6
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    \bigg(D+\frac{1}{2}\bigg)^2\bigg(D^2+\frac{1}{4}\b  igg)^2
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  7. #7
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    Quote Originally Posted by dwsmith View Post
    \bigg(D+\frac{1}{2}\bigg)^2\bigg(D^2+\frac{1}{4}\b  igg)^2
    I still do not understand what is going on. Where did that come from?
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  8. #8
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by ur5pointos2slo View Post
    Find yc(complimentary solution) and set up yp(particular solution) by the method of undetermined coefficients.

    y''+y'+(1/2)y = x*e^(-x/2)+x*cos(x/2)

    ok so
    yc = c1 e^(-x/2) cos(x/2) + c2 e^(-x/2) sin(x/2)

    Finding yp is confusing me. Could someone please explain how to come to the answer. What assumptions must be made?

    ***ANSWER***
    yp =(Ax + B)e^(-x/2) + (Cx+D)cos(x/2) + (Ex + F) sin(x/2)
    There should be a standard table in your text that outlines what you should guess given a particular L.H.S

    In this case,

    L.H.S = xe^{-x/2} + xcos(x/2)

    But let us note that we can use the prinipal of superposition here so

    L.H.S_1 = xe^{-x/2}

    L.H.S_2 = xcos( \frac{x}{2} )

    Finally,

    L.H.S= L.H.S_1 + L.H.S_2

    So lets find an appropriate Y_P for our L.H.S

    For L.H.S_1 we note that we have the form

    f(x)e^{g(x)}

    Which by standard tables suggest we try the form

    Y_{P1} = (Ax+B)e^{g(x)}

    For L.H.S_2 we note that we have the form

    f(x)cos( g(x) )

    Which, again by stanard tables suggests we try the form

    Y_{P2} = (Cx + D)cos( p(x) ) + (Ex+F) sin ( p(x) )

    We include both Sin and Cos here because both are potential solutions (remember our trig circles!). Everytime you have sin or cos as your left hand side, you guess the appropriate function for both sin and cos.

    In the end,

    Y_P= Y_{P1} + Y_{P2} =  (Ax+B)e^{g(x)}  (Cx + D)cos( p(x) ) + (Ex+F) cos ( p(x) )

     (Ax+B)e^{ \frac{-x}{2} }  (Cx + D)cos( \frac{x}{2} ) + (Ex+F) cos ( \frac{x}{2} )

    EDIT....I went through this entire thing use L.H.S when in fact we are using the R.H.S but it is late so dont castrate me lol
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  9. #9
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    Quote Originally Posted by AllanCuz View Post
    There should be a standard table in your text that outlines what you should guess given a particular L.H.S

    In this case,

    L.H.S = xe^{-x/2} + xcos(x/2)

    But let us note that we can use the prinipal of superposition here so

    L.H.S_1 = xe^{-x/2}

    L.H.S_2 = xcos( \frac{x}{2} )

    Finally,

    L.H.S= L.H.S_1 + L.H.S_2

    So lets find an appropriate Y_P for our L.H.S

    For L.H.S_1 we note that we have the form

    f(x)e^{g(x)}

    Which by standard tables suggest we try the form

    Y_{P1} = (Ax+B)e^{g(x)}

    For L.H.S_2 we note that we have the form

    f(x)cos( g(x) )

    Which, again by stanard tables suggests we try the form

    Y_{P2} = (Cx + D)cos( p(x) ) + (Ex+F) sin ( p(x) )

    We include both Sin and Cos here because both are potential solutions (remember our trig circles!). Everytime you have sin or cos as your left hand side, you guess the appropriate function for both sin and cos.

    In the end,

    Y_P= Y_{P1} + Y_{P2} = (Ax+B)e^{g(x)} (Cx + D)cos( p(x) ) + (Ex+F) cos ( p(x) )

     (Ax+B)e^{ \frac{-x}{2} } (Cx + D)cos( \frac{x}{2} ) + (Ex+F) cos ( \frac{x}{2} )

    EDIT....I went through this entire thing use L.H.S when in fact we are using the R.H.S but it is late so dont castrate me lol
    lol thanks a lot.
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