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Math Help - Set up appropriate form of solution Yp.

  1. #1
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    Question Set up appropriate form of solution Yp.

    Set up the appropriate form of a solution yp of y^(3)+y'' = 4x^2 , but DO NOT determine the values of the coefficients.

    guy this is what i have done so far

    DE : y^(3)+y'' = 4x^2
    CE: r^3 + r'' = 0
    r1 = 0, r2= 0 , r3= -1

    yc = C1 + C2x + C3 e^-x

    now here i am confuse about what to do to guess yp. should yp be Ax^2?? plz help me understand this
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  2. #2
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    Quote Originally Posted by DMDil View Post
    Set up the appropriate form of a solution yp of y^(3)+y'' = 4x^2 , but DO NOT determine the values of the coefficients.

    guy this is what i have done so far

    DE : y^(3)+y'' = 4x^2
    CE: r^3 + r'' = 0
    r1 = 0, r2= 0 , r3= -1

    yc = C1 + C2x + C3 e^-x

    now here i am confuse about what to do to guess yp. should yp be Ax^2?? plz help me understand this
    Since y = 1 and y = x are part of your complimentary solution, then you'll need to seek a particular solution of the form

     <br />
y_p = A x^4 + B x^3 + C x^2<br />
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    how u came up with Ax^4+Bx^3,, i meant how do i decided Yp in every equation?
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    Quote Originally Posted by DMDil View Post
    how u came up with Ax^4+Bx^3,, i meant how do i decided Yp in every equation?
    If the rhs is some power of x then you would need to seek a particluar solution of the form

     <br />
y = c_0 + c_1 x + c_2 x^2 + \cdots + c_n x^n<br />

    depending on the power n. Now if any of your complimentary solution contains any of these terms then you need to raise the power by the number of terms that the complimentary solution contains. In your case it contains 1 and x so you need to "bump" up the particular solutions by 2.
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    i am not sure if it is allow to ask same problem again with different values, what would Yp be if i have y^3 +9y' = x^2 exp(x)
    for Yc, i would ge Yc = C1+C2cos(3x)+C3sin(3x)
    how would i guess Yp now...is there any kind of table which can guide.
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    Cool

    This is third order linear non-homogeneous ODE which you have to seperate answers which is the Y_p and Y_h which the final solution would be the Y(x) = Y_p + Y_h , and Y_p is the solution to the left hand side of the ODE and Y_h is the solution to the right hand side of the ODE. Y_h can be solved using the Laplace transformation.
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  7. #7
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    i have not learn Laplace transformation yet. but i do know in order to solve this problem complete i have to use y(x) = Yc +Yp, Yc is easy to find since Yc is same in homogenous. but guess Yp is new thing for me. in every problem is different rule. what is the basic thing to follow here. what is it that i need to do to figure out Yp correctly?
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    Quote Originally Posted by defog171 View Post
    This is third order linear non-homogeneous ODE which you have to seperate answers which is the Y_p and Y_h which the final solution would be the Y(x) = Y_p + Y_h , and Y_p is the solution to the left hand side of the ODE and Y_h is the solution to the right hand side of the ODE. Y_h can be solved using the Laplace transformation.
    oops.. i did it again, sorry for the wrong info
    should be like this
    Y(x) = Y_h + Y_p ,
    Y_h is the solution to the left hand side of the ODE which is also same as your Y_c
    and Y_p is the solution to the right hand side of the ODE that makes your ODE non-homogeneous.

    to solve Y_p, you need to learn the (method of undetermined coefficients) or variation of parameters (if you have the basis of the solutions) which you need to find the Wronskian. But for this y'''+9y' = x^2 e^x<br />
i suggest you to learn (method of undetermined coefficients) first.
    Last edited by defog171; July 25th 2009 at 06:52 PM. Reason: sorry, i misunderstood.. not Laplace, it should be (method of undetermined coefficients)
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  9. #9
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    y_p=C(x^2e^x)
    thus,
     y_p'=C(x^2e^{x}+2xe^{x})
     y_p''=C(x^2e^{x}+2xe^{x}+2xe^{x}+2e^{x})=C(x^2e^{x  }+4xe^{x}+2e^{x})
    y_p'''=C(x^2e^x+2xe^{x}+4xe^{x}+4e^{x}+2e^{x})=C(x  ^2e^{x}+6xe^{x}+6e^{x})

    substitute this y_p and it's derivatives into the ODE, gives

    C(x^2e^x+6xe^x+6e^x) +9C(x^2e^x+2xe^x) = x^2e^x

    rearranging the equation yields

    e^x(10Cx^2+6Cx+24C) = e^x(x^2)

    by comparing coefficients for the both sides, for x^2, 10C = 1, therefore C=1/10,

    then  y_p = \frac{1}{10}x^2e^x

    can anybody verify this.. i think it's wrong
    maybe if we choose y_p=C_1x^2e^x+C_2xe^x+C_3e^x
    it will be better?
    Last edited by defog171; July 25th 2009 at 07:51 PM.
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  10. #10
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    i think i only have to solve up to Yp = A(x^2 exp(x)), since the problem says Do not Determine the value of the coefficent
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