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Math Help - variation of parameters

  1. #1
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    Question variation of parameters

    I am a bit confused on variation of parameters for solving differential equations. I understand the methods of coefficients but variation of parameters is a bit confusing. I would i use it to solve this:

    y'' - 5y' = -2x + 2

    Like how would i find independent solutions u1(x), u2(x) which would help me find derivatives v1(x) and v2(x) which then i get a general solution. I am confused on the steps and how to solve.

    Thanks
    Phil
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  2. #2
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    Quote Originally Posted by taurus View Post
    I am a bit confused on variation of parameters for solving differential equations. I understand the methods of coefficients but variation of parameters is a bit confusing. I would i use it to solve this:

    y'' - 5y' = -2x + 2

    Like how would i find independent solutions u1(x), u2(x) which would help me find derivatives v1(x) and v2(x) which then i get a general solution. I am confused on the steps and how to solve.

    Thanks
    Phil
    Have you been taught this technique or are you trying to self-learn it.

    The solutions u1 and u2 come from solving y'' - 5y' = 0.
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  3. #3
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    Briefly taught as we mainly focused on method by coefficients. But then i saw a questions like that so i thought i could get some advice on how to use the technique of variations by parameters to solve such questions I posted above.
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  4. #4
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    Quote Originally Posted by taurus View Post
    I am a bit confused on variation of parameters for solving differential equations. I understand the methods of coefficients but variation of parameters is a bit confusing. I would i use it to solve this:

    y'' - 5y' = -2x + 2

    Like how would i find independent solutions u1(x), u2(x) which would help me find derivatives v1(x) and v2(x) which then i get a general solution. I am confused on the steps and how to solve.

    Thanks
    Phil
    First we need to solve the homogeneous system

    y''-5y=0

    The auxillary equation is

    m^2-5m=0 \iff m(m-5)=0, m=0,m=5

    y_c(x)=c_1e^{0x}+c_2e^{5x}=c_1+c_2e^{5x}

    Now we can set up the wronskian

    W=\begin{vmatrix}1 && e^{5x} \\ 0 && 5e^{5x} \end{vmatrix}=5e^{5x}

    W_1=\begin{vmatrix}0 && e^{5x} \\ -2x+2 && 5e^{5x} \end{vmatrix}=(2x-2)e^{5x}

    W_2=\begin{vmatrix}1 && e^{5x} \\ 0 && -2x+2 \end{vmatrix}=-2x+2

    So finally

    v_1'(x)=\frac{W_1}{W}

    v_2'(x)=\frac{W_2}{W}

    Just integrate from here.

    I will leave the rest to you.

    Good luck. \emptyset
    Last edited by TheEmptySet; May 25th 2008 at 07:56 PM. Reason: t and x's are bad
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  5. #5
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    What exactly did you do with the wronskian? Iv never heard of that?
    why W, W1, W2?
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  6. #6
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    Quote Originally Posted by taurus View Post
    What exactly did you do with the wronskian? Iv never heard of that?
    why W, W1, W2?
    for an equation

    y''+a(x)y'+b(x)y=f(x)

    When I found the complimentry solution

    y_c=c_1+c_2e^{5x}

    This is

    y_c=c_1u_1(x)+c_2u_2(x)

    The wronskian is

    W=\begin{vmatrix}u_1(x) && u_2(x) \\ u_1'(x) && u_2'(x) \end{vmatrix}

    W_1=\begin{vmatrix}0 && u_2(x) \\ f(x) && u_2'(x) \end{vmatrix}

    W_2=\begin{vmatrix}u_1(x) && 0 \\ u_1'(x) && f(x) \end{vmatrix}

    Now the particular solution to the equation is

    y_p=u_1\cdot \int\frac{W_1}{W}dx+u_2 \cdot \int\frac{W_2}{W}dx

    I hope this clears it up.

    Good luck.
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  7. #7
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    Question

    For v1' and v2' above is it:
    v1 = ((2x-2)e^(5x))/(5e^(5x)) integrated
    which gives
    v1' = ((1/5)*(x-2)*x)

    v2 = (-2x+2)/(5e^(5x)) integrated
    gives
    v2' = ((2/125)*e^(-5*x) * (5*x-4))

    Is that correct? or am I differentiating W1/W and W2/w?

    thanks
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  8. #8
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    You can check this out. I outlined a VP problem in post #6.

    http://www.mathhelpforum.com/math-he...eous-help.html
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  9. #9
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    Question

    Quote Originally Posted by TheEmptySet View Post

    v_1'(x)=\frac{W_1}{W}

    v_2'(x)=\frac{W_2}{W}

    \emptyset
    That is what i did and doesnt seem to be right. I integrated W1/W and W2/W

    For v1' and v2' above is it:
    v1' = ((2x-2)e^(5x))/(5e^(5x)) integrated
    which gives
    v1 = ((1/5)*(x-2)*x)

    v2' = (-2x+2)/(5e^(5x)) integrated
    gives
    v2 = ((2/125)*e^(-5*x) * (5*x-4))
    What am I doing wrong?
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  10. #10
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    v_1=\int\frac{1}{5} \left( 2x-2 \right)dx=\frac{1}{5}x^2-\frac{2}{5}x

    v_2=\int -\frac{2}{5}(x+1)e^{-5x}dx=\left[\frac{2}{25}x+\frac{12}{125} \right]e^{-5x}

    So the particular solution is

    y_p=u_1v_1+u_2v_2=1\left[\frac{1}{5}x^2-\frac{2}{5}x\right]+e^{5x}\left[\frac{2}{25}x+\frac{12}{125} \right]e^{-5x}=\frac{1}{5}x^2-\frac{8}{25}x+\frac{12}{125}

    Now when you combine this with the complimentry solution we get

    y=y_c+y_p=c_1+c_2e^{5x}+\frac{1}{5}x^2-\frac{8}{25}x+\frac{12}{125}

    collecting and relabeling the constants we get the solution

    y=c_3+c_2e^{5x}+\frac{1}{5}x^2-\frac{8}{25}x
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  11. #11
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    I dont think your V1 and V2 is correct because i entered it into computer and it said its wrong?
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  12. #12
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    Quote Originally Posted by taurus View Post
    I dont think your V1 and V2 is correct because i entered it into computer and it said its wrong?
    I disagree with your computer. I did the first computations by hand first and now I have checked them in Maple. Maple has written them differently, but they are equal to what I posted before.

    Here is a screen shot. What software are you using?

    variation of parameters-capture.jpg
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  13. #13
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    Question

    hmmmm yea Im also using maple.

    The questions am trying to answer this:
    Now assume that the solution of the original d.e. is of form

    y(x) = u1v1 + u2+v2

    where u1, u2 are the functions you entered in the first part. Determine the derivatives v1'(x) and v2'(x).
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  14. #14
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    Quote Originally Posted by taurus View Post
    hmmmm yea Im also using maple.

    The questions am trying to answer this:
    Now assume that the solution of the original d.e. is of form

    y(x) = u1v1 + u2+v2

    where u1, u2 are the functions you entered in the first part. Determine the derivatives v1'(x) and v2'(x).
    As I said in post number four

    v_1'=\frac{W_1}{W}
    and
    v_2'=\frac{W_2}{W}

    Am I misunderstanding your question. Are you asking for a proof of why the method works?
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  15. #15
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    hmmm yea which is what i tried but computer says it wrong, hmm i dont know anymore
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