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Thread: Getting 0 for a particular solution: r^2*dy/dr = K, r>0, K constant

  1. #1

    Getting 0 for a particular solution: r^2*dy/dr = K, r>0, K constant

    I have this question that I am struggling to see if it is correct or not.

    A quantity y(r) satisfies the first-order fifferential equation r^2*dy/dr = K where r > 0 and K is a constant.

    The general solution I got from this FO differential equation is: y = K(-1/r + F)


    The second part to this question is to find a particular solution for y(R) = 0, where R is a constant.

    I seem to get 0 as R = 1/F which cancels out everything.


    Much appreciated for anyone that can help.

  2. #2
    Elite Member
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    Why is F inside the parentheses?
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

  3. #3
    Quote Originally Posted by tkhunny View Post
    Why is F inside the parentheses?
    This is the process I went through:


    integrate both sides: 1/r^2 * dr = 1/K*dy


    = -1/r + C = Y/K + D
    = -1/r = Y/K + (D-C)
    = -1/r = Y/K + F where F = D-c
    Y = K(-1/r - F)

  4. #4
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    Okay, that's fine, but:

    1) There is no need to produce TWO constants of integration. One will do.
    2) Is K(Ar + B) really different from KAr + KB? KB is still just come constant of itegration.
    3) You are VERY close.

    We'll do it your way.

    [tex]K\cdot\left(-\dfrac{1}{R}+F\right)=0[/tex] has two solutions. One of those is K = 0. You are not interested in this solution. The other solution is F = 1/R.

    You are done. What is this cancelling out of which you speak?

    Substitute: [tex]Y(r) = K\cdot\left(-\dfrac{1}{r} + \dfrac{1}{R}\right)[/tex] -- Perhaps you are confusing "R" with "r"?
    Last edited by tkhunny; 11-16-2017 at 03:45 PM.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

  5. #5
    Quote Originally Posted by tkhunny View Post
    Okay, that's fine, but:

    1) There is no need to produce TWO constants of integration. One will do.
    2) Is K(Ar + B) really different from KAr + KB? KB is still just come constant of itegration.
    3) You are VERY close.

    We'll do it your way.

    [tex]K\cdot\left(-\dfrac{1}{R}+F\right)=0[/tex] has two solutions. One of those is K = 0. You are not interested in this solution. Please find the other solution.
    OK.

    -1/R + F = 0

    You are left with R = 1/F. Yet, if we subsitute this back into the general solution, we get K*0, as the two F's cancel each other out.

  6. #6
    Quote Originally Posted by tkhunny View Post
    Okay, that's fine, but:

    1) There is no need to produce TWO constants of integration. One will do.
    2) Is K(Ar + B) really different from KAr + KB? KB is still just come constant of itegration.
    3) You are VERY close.

    We'll do it your way.

    [tex]K\cdot\left(-\dfrac{1}{R}+F\right)=0[/tex] has two solutions. One of those is K = 0. You are not interested in this solution. The other solution is F = 1/R.

    You are done. What is this cancelling out of which you speak?

    Substitute: [tex]Y(r) = K\cdot\left(-\dfrac{1}{r} + \dfrac{1}{R}\right)[/tex] -- Perhaps you are confusing "R" with "r"?
    You're right. I am confusing "r" with "R".

    Shouldn't the answer be y(r) = K(-1/r + 1/R) ? <== Good catch! Typos have been repaired, above..
    Last edited by tkhunny; 11-16-2017 at 03:46 PM.

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