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anonymous

  • one year ago

Question regarding one step in this problem

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  1. anonymous
    • one year ago
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    \[x^2 y'' - 7xy' + 16; y_1 = x^4\]I have to set it in standard form following this rule\[y''+P(x)y'+Q(x)=0\]which results in\[y''-\frac{ 7 }{ x } y' + \frac{ 16 }{ x^2 } y = 0\] How did the x and x^2 moved on the denominator?

  2. dan815
    • one year ago
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    divide by x^2

  3. anonymous
    • one year ago
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    why is it that we have to divide it by x^2 after changing it to standard form?

  4. dan815
    • one year ago
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    you dicide to change it to standard form

  5. thomas5267
    • one year ago
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    That looks like it, unless the equation is inhomogeneous and you have to figure out the inhomogeneous part by the given solution.

  6. dan815
    • one year ago
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    divide*, you cannot have any f(x) multiypling your highest degree

  7. thomas5267
    • one year ago
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    \(x^2 y'' - 7xy' + 16=0\) is not in standard form because there is \(x^2\) in front of \(y''\) where in standard form there is nothing.

  8. anonymous
    • one year ago
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    ohh I see!

  9. dan815
    • one year ago
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    first solve the homogenous equation then use variation of parameters to get general solution to your non homogenous equation

  10. anonymous
    • one year ago
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    so to set it to standard form whatever is in front of y'' has to be divided to just have y'' alone

  11. thomas5267
    • one year ago
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    \[ x^2 y'' - 7xy' + 16\\ y_1 = x^4 \] But plugging \(y_1\) into the above expression I get \(-16x^4+16\).

  12. anonymous
    • one year ago
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    oh no you have to do reduction of sides

  13. thomas5267
    • one year ago
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    Explain?

  14. anonymous
    • one year ago
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    \[y_2 = y_1 \int\limits_{}^{}\frac{ e ^{-\int\limits_{}^{}P(x)dx} }{y_1^2 }\]

  15. thomas5267
    • one year ago
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    I am assuming \(y_1=x^4\) is a solution to the incomplete differential equation \(x^2 y'' - 7xy' + 16\). What does this equation equal to?

  16. thomas5267
    • one year ago
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    Or is \(y_1\) not a solution?

  17. anonymous
    • one year ago
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    The final answer is \[Y=C_1x^4 + C_2 x^4\ln(x)\]or simplified \[Y = x^4(C_1 + C_2 \ln(x))\]

  18. anonymous
    • one year ago
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    using reduction of sides you find y_2

  19. thomas5267
    • one year ago
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    In other words, \(y_1=x^4\) should satisfy the equation \(x^2 y'' - 7xy' + 16=0\)? \[ x^2(x^4)''-7x(x^4)'+16\\ 12x^4-28x^4+16\\ -16x^4+16\neq0 \]

  20. anonymous
    • one year ago
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    \[y'' \neq y\]

  21. Michele_Laino
    • one year ago
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    try to search for solution of this type: \[y = k{x^s},\quad s \in \mathbb{Z}\]

  22. Michele_Laino
    • one year ago
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    \[\Large y = k{x^s},\quad s \in \mathbb{Z}\]

  23. Michele_Laino
    • one year ago
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    after the substitution, you should get a quadratic equation for \(s\)

  24. thomas5267
    • one year ago
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    \[ x^2 y'' - 7xy' + 16=0 \text{ not }x^2 y'' - 7xy' + 16y=0 \] I got\[ r(r-1)x^r-7rx^r+16=0 \] after assuming \(y=x^r\).

  25. Michele_Laino
    • one year ago
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    \(+16x^r\)

  26. thomas5267
    • one year ago
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    16 does not have y behind it in the differential equation. Not sure whether it is a typo or not.

  27. Michele_Laino
    • one year ago
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    from first reply I see 16y/x^2

  28. thomas5267
    • one year ago
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    This is horrible!

  29. Michele_Laino
    • one year ago
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    if we ahve only \(16\) and not \(16y\) then we have to apply my substitution above to the homogeneous equation only

  30. Michele_Laino
    • one year ago
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    have*

  31. thomas5267
    • one year ago
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    Exactly the same equation on this website. http://tutorial.math.lamar.edu/Classes/DE/EulerEquations.aspx

  32. thomas5267
    • one year ago
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    If the equation is \[ x^2y''-7xy'+16y=0 \] Divide the whole thing by \(x^2\) to get the standard form. Or just read the website.

  33. anonymous
    • one year ago
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    oh yeahI ate the y after the 16, sorry. I do undeerstand this though, just had doubt in the beginning

  34. thomas5267
    • one year ago
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    Y U ATE THE Y? YYYYY? Sorry.....

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