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B Private solution to a polynomial differential equation

  1. Oct 2, 2016 #1
    The polynomial equation and it's private solution:
    $$(1)~~ay''+by'+cy=f(x)=kx^n,~~y=A_0x^n+A_1x^{n-1}+...+A$$
    If i, for example, take ##f(x)=kx^3## i get, after substituting into (1), an expression like ##Ax^3+Bx^2+Cx+D## , but that doesn't equal ##kx^3##
     
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  3. Oct 2, 2016 #2

    Simon Bridge

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    Please show your working.
     
  4. Oct 2, 2016 #3

    pasmith

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    The general solution of [tex]
    ay'' + by' + cy = ky[/tex] is [itex]Ae^{r_1x} + Be^{r_2x}[/itex] where [itex]r_1 \neq r_2[/itex] are the roots of [itex]ar^2 + br + (c - k) = 0[/itex]. If the roots are equal then the general solution is instead [itex]Ae^{r_1x} + Bxe^{r_1x}[/itex].

    You can therefore obtain [itex]y(x) = x[/itex] as a solution by taking [itex]b = 0[/itex] and [itex]k = c[/itex].
     
  5. Oct 2, 2016 #4

    Simon Bridge

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    But that is that the DE under consideration in post #1?

    Compare:
    ##ay'' + by' + cy = ky## with
    ##ay'' + by' + cy = kx^3## ...
     
  6. Oct 2, 2016 #5
    $$y=A_0x^3+A_1x^2+A_2x+A,~~y'=3A_0x^2+2A_1x+A_2,~~y''=6A_0x+2A_1$$
    $$ay''+by'+cy=6aA_0x+2aA_1+3bA_0x_2+6bA_1x+bA_2+cA_1x^2+cA_2x+cA=(cA_0)x^3+(3bA_0+cA_1)x^2+(6aA_0+6bA_1+cA_2)x+(2aA_1+bA_2+cA)$$
    Is there a phrase about a polynomial with many members to equal only one with the highest power?
     
  7. Oct 2, 2016 #6

    Simon Bridge

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    OK ... is that as far as you got?
    You need an ##=kx^3## in there someplace?
    Check the algebra on the others too ... then solve for the unknown coefficients.
    (Probably easier to write the coefficients as A,B,C,D like you did in post #1)
     
  8. Oct 2, 2016 #7
    $$(cA_0)x^3+(3bA_0+cA_1)x^2+(6aA_0+6bA_1+cA_2)x+(2aA_1+bA_2+cA)=kx^3$$
    I
    $$\left\{ \begin{array}{l} cA_0=k \\ 3bA_0+cA_1=0 \\ 6aA_0+6bA_1+cA_2=0 \\ 2aA_1+bA_2+cA=0 \end{array}\right.$$
    $$\rightarrow~~A_0=\frac{k}{c},~~A_1=-\frac{3bk}{c^2},~~A_2=\left( \frac{3b_2c-a}{c^3} \right)6k,~~A=\frac{6abk}{c^3}-\left( \frac{3b_2c-a}{c^3} \right)6bk$$
    Is that what you meant?
     
    Last edited: Oct 2, 2016
  9. Oct 2, 2016 #8

    Simon Bridge

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    That's the idea... test for a=b=c=d=k=1 (say).
     
  10. Oct 3, 2016 #9
    Thank you pasmith and Simon
     
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