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Product Rule?

  • Thread starter juice34
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  • #1
juice34
My question is how do they transform equation A. into B.. I know they are using the product rule but dont know what is going on.

EQ A.)D(1/r^2)d/dr(r^2(dC/dr))-kC=0
now how do they get Eq B.

EQ B.) (d^2C/dr^2)+(2/r)(dC/dr)-(kC)/D=0
 

Answers and Replies

  • #2
HallsofIvy
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My question is how do they transform equation A. into B.. I know they are using the product rule but dont know what is going on.

EQ A.)D(1/r^2)d/dr(r^2(dC/dr))-kC=0
now how do they get Eq B.

EQ B.) (d^2C/dr^2)+(2/r)(dC/dr)-(kC)/D=0
The derivative of r^2 dC/dr is, by the product rule, [r^2 d(dC/dr)/dr]+ [d(r^2)/dr] dC/dr= r^2 d^2C/dr+ 2r dC/dr. Multiplying that by 1/r^2 gives d^2C/dr^2+ 2/r dC/dr so the equation is the same as D(d^2C/dr^2+ 2/r dC/dr)- kC= 0.

Dividing the entire equation by D, then, gives eq. B.
 

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