Another Calc2/diff eq integral problem

kdinser

it's a diff eq problem

dN/ds=k(250-s)

The way I'm doing it. (S=integral sign)

SdN/ds=k[S250 ds - Ss ds]

N = k[250s-(s^2)/2]+C

Not sure where I'm going wrong here, but it looks nothing like the answer in the solution manual which just jumps from Sk(250-s)ds to -k/2(250-s^2)^2+C

Thanks for any help.

marlon

seems to me your answer is correct, only you gotta loose the ds on the right hand side. There should only be the integral of dN which indeed is equal to N.

How did you come to that other answer ???
Were there initial conditions given ???
µ

regards
marlon

kdinser

The original problem exactly as it appears in the book.

Write and solve the differential equation that models the verbal statement.

The rate of change of N with respect to s is proportional to 250-s

I took that to mean that
the derivative of N, N', or dN/ds is equal to k(250-s)

So to find N, integrate both sides with respect to s, I came up with;

N = k[250s-(s^2)/2]+C

The solution manual goes through these steps.
(S=integral sign)

dN/ds=k(250-s)

S(dN/ds)ds=Sk(250-s) ds (shouldn't that 1/ds on the left side have moved to the right before integrating?)

SdN=-(k/2)(250-s)^2 + C

N=-(k/2)(250-s)^2 + C

Does this make any sense? I don't see where the 1/2 is coming from or the squaring of the (250-s). Integrating k(250-s) looks about as straight forward as it gets unless I'm missing something that is happening on the left side before doing the integrations. Thanks

marlon

ahh, ok , got it. In the integration of k(250-s) they used a substitution. Set u =250-s. Then du = -ds and the integral yields kS(u)(-du) = -kSudu = -k*u²/2 = -k/2 * (250-s)²

This is it...
regards
marlon

kdinser

Just got an email about this one from my prof, she said my answer is fine. The book did it as a u substitution. I just tried it and got the same answer as the book. Thanks for the input Marlon.

Haha, whenever I come up with du=ds in a substitution, it usually means I'm doing it the hard way.