Problem

In Exercise 9.2.28 we discussed a differential eq…

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Problem 37 Hard Difficulty

Solve the initial-value problem in Exercise 9.2.27 to find an expression for the charge at time $ t. $ Find the limiting value of the change.

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Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 9

Differential Equations

Section 3

Separable Equations

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Video Transcript

we have our multiplied by Dick, you boy d t plus one Oversee on the pipe by q equals e off t we have values off RNC. And if you have article five and see equal opened or five and e off the equal 60 we have then five months applied by Dick you by DT plus one over, Over off ice multiplied by Cube A quite 60 is in after simplification we have five, but about ridiculed by DT plus 20 que equals 60. We will divide off all the question over five. Then we have dick you by DT Plus for two equal 12 is in. We may have Dick you by DT equal 12 minus 40. Then when we take negative four common factor, we have the few by DT equals before Q minus three. We will divide the question over humanist sees and we get dick you over Q minus three equal Negative four DT Then we will integrate poor sites. We get dick you over. Q minus three equals integral negative for DT. We will make integrations and we get Len a trade value offer Q minus three equal negative four T plus Len See this constant, Nancy. We will now take expansion for both sites. We get you power, Lynn que Babson Value Q minus three equal E power Negative 40 plus Lindsay. Remember that E Borland equally. This is our rule. So we have now Q minus three equal C but by E bar negative 40 then we get the Q equals three plus C e power. Negative 40. Now we have initial conditions that Q equals zero when t equals zero, we will substitute was Q equals zero and equals zero. In our question, we get zero equestrian lost, see? But they put by e How are zero now we have see equal industry. Then we get to equal three minus ST E power. Negative 40. So we have Q equal Common factor three one minus e Born in the 40 we will know calculates a limiting value for our function. Win t approaches infinity, so limit the approaches. Infinity three minus three He boarding the 14 equal slee minus three. Limit 40 across. Approaches Infinity E power Negative 40. This equals V minus three. I e. Power negative Infinity people negative. Infinity equals heroes and we get three minus lee that apply by zero. It's four city

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Problem

In Exercise 9.2.28 we discussed a differential eq…

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Related Courses

Calculus: Early Transcendentals

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Discussion

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Grace He

Catherine Ross

Samuel Hannah

Joseph Lentino

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Recommended Videos

Watch More Solved Questions in Chapter 9

Video Transcript

James Stewart

Calculus: Early Transcendentals

Grace He

Catherine Ross

Samuel Hannah

Joseph Lentino

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Grace He

Catherine Ross

Samuel Hannah

Joseph Lentino

James StewartCalculus: Early Transcendentals

Grace He

Catherine Ross

Samuel Hannah

Joseph Lentino

James Stewart

Calculus: Early Transcendentals