The transformation from equation A to equation B involves the application of the product rule in calculus. Starting with equation A, D(1/r^2)d/dr(r^2(dC/dr))-kC=0, the product rule is utilized to differentiate r^2(dC/dr), resulting in r^2(d^2C/dr^2) + 2r(dC/dr). This leads to the simplified form d^2C/dr^2 + (2/r)(dC/dr) - (kC)/D = 0, which is equation B. The final step involves dividing the entire equation by D to achieve the desired format.
PREREQUISITESStudents and professionals in mathematics, physics, and engineering who are working with differential equations and require a deeper understanding of transformation techniques in mathematical modeling.
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.juice34 said:My question is how do they transform equation A. into B.. I know they are using the product rule but don't 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