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Linear homogeneous D.E. with constant coefficients - known solutions

  1. Aug 12, 2014 #1

    etf

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    My task is to find Linear homogeneous D.E. with constant coefficients which has solutions:
    $$\\\varphi 1(x)=x^2,\varphi 2(x)=e^{-3x},\varphi 3(x)=cos(5x)$$ Any idea?
     
    Last edited: Aug 12, 2014
  2. jcsd
  3. Aug 12, 2014 #2

    etf

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    I think that here we have root $$\\\lambda 1=0$$ with multiplicity 3 (from $$\\\varphi 1(x)$$), root $$\\\lambda 2=-3$$ with multiplicity 1 (from $$\\\varphi 2(x)$$).
     
  4. Aug 12, 2014 #3

    Zondrina

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    I assume you mean second order.

    The solution ##x^2## should come from a Cauchy Euler equation I believe. So I think you need to find an equation of that form by multiplying the roots together to "work backwards" to the indicial equation.
     
  5. Aug 12, 2014 #4
    It is possible to have [itex]\varphi_1(x) = x^2[/itex] as a solution to a linear homogeneous equation with constant coefficients. In addition, a linear homogeneous differential equation has as many linearly independent solutions as the order of the equation. Since the 3 functions you listed are linearly independent, the equation we are looking for must be at least of order 3.
    As a hint, note that any linear combination of the solutions must also be a solution. Thus, the 3-parameter family of functions [itex]\varphi(x) = C_1x^2 + C_2e^{-3x} + C_3\cos(5x)[/itex] is a solution to the unknown differential equation. As a further hint, try taking this solution's derivatives. Can you find a combination using constant coefficients that is homogeneous?
     
  6. Aug 12, 2014 #5

    pasmith

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    You're missing [itex]\cos(5x) = (e^{5ix} + e^{-5ix})/2[/itex].

    Thus the auxiliary equation must be a polynomial with roots 0 (with multiplicity 3), -3, 5i and -5i.
     
  7. Aug 12, 2014 #6

    Zondrina

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    Wow, I totally forgot equations higher than second order existed ^_^.

    My bad.
     
  8. Aug 12, 2014 #7

    etf

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    I calculated characteristic polynomial as $$\\(\lambda -\lambda1)^{3}(\lambda-\lambda2)(\lambda-\lambda3)(\lambda-\lambda4)=(\lambda-0)^{3}(\lambda+3)(\lambda-(0-5i))(\lambda-(0+5i))=...$$
    $$=\lambda^{6}+3\lambda^{5}+25\lambda^{4}+75\lambda^{3}$$ so D.E. is $$\\y^{(6)}+3y^{(5)}+25y^{(4)}+75y^{(3)}$$
     
    Last edited by a moderator: Aug 13, 2014
  9. Aug 12, 2014 #8

    ehild

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    It looks correct. The result is the same using Slider142's hint.

    ehild
     
  10. Aug 12, 2014 #9

    LCKurtz

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    Except that that isn't an equation.:frown:
     
    Last edited by a moderator: Aug 13, 2014
  11. Aug 12, 2014 #10

    ehild

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    Oh, =0 is missing! I did not notice.:blushing:

    ehild
     
  12. Aug 13, 2014 #11

    etf

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    Forgot to write = 0 :smile:
     
  13. Aug 13, 2014 #12

    etf

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    Here is another one (this one is not with constan coefficients):
    Find linear homogeneous second order D.E. which has general solution $$\\\varphi (x)=C1\frac{sinx}{\sqrt{x}}+C2\frac{\cos{x}}{\sqrt{x}}, x>0$$
    Any idea?
     
  14. Aug 13, 2014 #13

    ehild

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    What is the differential equation for y(x)=√(x) φ(x)?

    ehild
     
  15. Aug 13, 2014 #14

    etf

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    I found it. $$\\-\frac{1}{x}y''-\frac{1}{x^{2}}y'+(\frac{1}{4x^{3}}-\frac{1}{x})y=0$$
     
  16. Aug 13, 2014 #15

    ehild

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    Congratulation!
    Have you used my hint? It is very easy with that.

    ehild
     
  17. Aug 13, 2014 #16

    etf

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    To be honest, I didn't understand your hint :(
    I solved it using Wronskian:

    $$\\y1(x)=\frac{\sin{x}}{\sqrt{x}},$$
    $$\\y2(x)=\frac{\cos{x}}{\sqrt{x}},$$
    $$\\W(y1,y2)=\begin{vmatrix}
    y1(x) & y2(x) & y \\
    y1'(x) & y2'(x) & y'(x) \\
    y1''(x)& y2''(x) & y''(x)
    \end{vmatrix}=0$$

    However, this is a little bit complicated.
     
  18. Aug 13, 2014 #17

    ehild

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    [tex]\varphi (x)=C1\frac{sinx}{\sqrt{x}}+C2\frac{\cos{x}}{\sqrt{x}}[/tex]

    Multiply by √x:

    [tex]\sqrt{x}\varphi (x)=C1\sin(x)+C2\cos(x)[/tex]

    Define a new function ##y(x)=\sqrt{x}\varphi (x)##. The general solution of a linear homogeneous differential equation for y(x) is ##y(x)= C1\sin(x)+C2\cos(x)##. But that is the solution of the well known differential equation ##y''+ y=0 ##.

    Find the second derivative of ##y(x)=\sqrt{x}\varphi (x)## in terms of x and φ and substitute into the equation ##y''+ y=0 ##...

    ehild
     
  19. Aug 13, 2014 #18

    etf

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    Wow :smile:
    Your method is definitely much better :)
    Thanks!
     
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