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1 month, 3 weeks ago
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Please answer the PART II C and D of the uploaded picture for me
All right on this problem we've got a differential equation that we would like to solve. I'm just going to be able to do part C. Here. You will have to make another request for part D. We've got seven over Y squared plus one. Dy the odds. Does he put a one over at And they were told why is zero when X. This one? Mhm. This is what is called a separable differential equation. In other words you can get all of your Y. Values on one side and all of your X. Values on the other and make that happen for this particular problem. All we need to do first is just multiply both sides my ideas. That's what I do here. I just multiply both sides by my Ds. And so here on the left This leaves me with seven over Y squared post one the why April is one over X. Yeah. Now that I have that what I need to do is integrate both sides. And so that's what I'm gonna do next year. I'm going to integrate both sides of this equation. Now here on the left this is a very good integral to have memorized anti derivative of one over Y squared is the tangent inverse of y. And so we have a seven. So this is seven tangent inverse of why. And then on the other side We have the integral of one over X. And so that's the natural on our backs. Well perhaps ELISA or C. Now to find what C. Is. Again I know that why was zero? I have seven tangent inverse of wine. But why is 0? Mhm. Whatever access one changing universe of 00. And so this is seven times zero Equals the natural log of one is 0. I see. Mhm. And so it tells us CF0. I'm coming back up here to this question, right? I'm just going to get rid of the sea because it was just zero. Actually going back up there to the equation there in green. I'm also we're gonna get rid of it there. Uh huh. This tells me that the tangent inverse of why is equal to 1/7 times the natural automat yourselves me that? Why is the tangent on 1/7 comes the natural alone event? Oh.
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University of Kansas
University of California, Santa Cruz
Ans: 1.35 kgm ^-3
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