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Problem 18 Hard Difficulty

Find the solution of the differential equation that satisfies the given initial condition.
$ \frac {dL}{dt} = kL^2\ln t, L(1) = -1 $

Answer

$L=\frac{-1}{k t \ln t-k t+k+1}$

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Video Transcript

this question is asking us to find the solution of the differential equation that satisfies the given initial condition. We know we have d l over. DT is Kate Times all squared natural log of tea. We know that what we're gonna have to do is write this in notation given differential parts, and then we must integrate. So what this means is that we want to put all the l terms on the left hand side and all the tea and then originally, our constant integration on the right hand side. So Thean Toral of D of L over l squared is equivalent to the intro of CR constant integration times natural log of T d t. Now we know we can integrate the left hand side B negative one over l plus C is see, Time's the integral of natural log of t de tete. Remember, the constant can be pulled out and put in front of the integral like I've done over here not to integrate. We know we have negative one over L on on the right hand side. C t natural log of T minus K t plus seek. Now remember l of one's negative one, which means our X is one in our wise negative one. Obviously, in this context we have l and T instead of why an ex. But it's the same idea. So in other words, we're gonna be plugging in from El of one's negative one. Okay, now that we have this, we know we're going to be writing this just in terms of are negative one over L. Because then we can write this singularly, as in terms of l. So in other words, we have negative one over Els. We're going back to Original Equation. But now we're plugging in because now we know what C S. C is K plus one. So we should no longer have are seeing our equation. We should just talk in terms of K l and T. So this is over here plus capers one. Now we know we want this just in terms of l, right. That's the endgame. Why equals r l equals whatever the variables, we don't want it. Negative one over l. The final answer should just be in terms of l. So divide everything. So this part now all goes on the denominator