Another Calc2/diff eq integral problem

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The discussion revolves around solving the differential equation dN/ds = k(250-s). The original poster initially integrates both sides and arrives at N = k[250s - (s^2)/2] + C but struggles to match the solution in the manual. A key point raised is the need for a substitution, u = 250 - s, which simplifies the integration process. After applying this substitution, the correct solution is found to be N = -k/2(250-s)^2 + C. Ultimately, the original poster confirms that their initial answer is valid, and the book's solution aligns with the substitution method.
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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.
 
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kdinser said:
SdN/ds=k[S250 ds - Ss ds]
Thanks for any help.


seems to me your answer is correct, only you got to 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
 
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
 
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
 
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.
 
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