Another Calc2/diff eq integral problem

  • Thread starter kdinser
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In summary: Thanks again.In summary, the conversation is about solving a differential equation that models a verbal statement about the rate of change of N with respect to s being proportional to 250-s. The person originally integrated both sides with respect to s and came up with an answer that does not match the solution manual. They then ask for help and Marlon explains that the book used a substitution to get the correct answer. The person follows up with an email from their professor confirming that their answer is correct.
  • #1
kdinser
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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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  • #2
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
 
  • #3
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
 
  • #4
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
 
  • #5
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.
 

1. What is the purpose of solving "Another Calc2/diff eq integral problem"?

The purpose of solving this type of problem is to gain a deeper understanding of integration and its applications in calculus and differential equations. It also helps to improve problem-solving skills and critical thinking abilities.

2. What are the common methods used to solve "Another Calc2/diff eq integral problem"?

Some common methods used to solve this type of problem include substitution, integration by parts, trigonometric substitution, and partial fractions. The choice of method depends on the complexity and structure of the integral.

3. Is it necessary to memorize integration formulas to solve "Another Calc2/diff eq integral problem"?

No, it is not necessary to memorize integration formulas. However, having a good understanding of basic integration rules and techniques can help in solving these problems more efficiently.

4. How can I check if my solution to "Another Calc2/diff eq integral problem" is correct?

You can check your solution by differentiating it and comparing it to the original integrand. If the results match, then your solution is correct. You can also use online integral calculators or ask a teacher or tutor for help.

5. Are there any tips or tricks for solving "Another Calc2/diff eq integral problem" more easily?

Practicing regularly and understanding the concepts is the best way to improve your problem-solving skills for this type of problem. Additionally, breaking down the integral into smaller parts, using symmetry and substitution, and being familiar with common integration patterns can also make the process easier.

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