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

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. Why is F inside the parentheses?

3. Originally Posted by tkhunny
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. 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.

$K\cdot\left(-\dfrac{1}{R}+F\right)=0$ 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: $Y(r) = K\cdot\left(-\dfrac{1}{r} + \dfrac{1}{R}\right)$ -- Perhaps you are confusing "R" with "r"?

5. Originally Posted by tkhunny
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.

$K\cdot\left(-\dfrac{1}{R}+F\right)=0$ 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. Originally Posted by tkhunny
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.

$K\cdot\left(-\dfrac{1}{R}+F\right)=0$ 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: $Y(r) = K\cdot\left(-\dfrac{1}{r} + \dfrac{1}{R}\right)$ -- 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..

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