<p>given the equation: dy/dx=-xy/lny,y>0
Find the general solution of the differential equation?</p>
<p>Can anybody help me with this question. Thank you</p>
<p>dy/dx = -xy/lny</p>
<p>Rearrange:</p>
<p>lny/y dy = -x dx</p>
<p>Let's do a sub: u = lny
du = 1/y dy</p>
<p>Therefore it can be rewritten as:
u du = - x dx</p>
<p>Integrate:
u^2/2 = -x^2/2</p>
<p>Resub u = lny
ln^2(y) = -x^2</p>
<p>:)</p>
<p>Unless there's a trick to this, I don't think it's that difficult...</p>
<p>First, separate the variables, so:</p>
<p>lnydy/y = -xdx</p>
<p>Integrate each side and you should get:</p>
<p>ln^2 y / 2 = -x^2 / 2</p>
<p>Which then gives:</p>
<p>ln^2 y = - x^2</p>
<p>;)</p>
<p>Edit: Damn! Got beaten to it :p And I was so excited, too, here I thought it would be disgustingly difficult and it really wasn't.</p>
<p>Unless you want to solve for y, in which case you will get something + 1 since u get e^c, and c=0 for a general solution</p>
<p>the real question is... why in god's creation is there a monkey climbing up a frictionless pulley with a block attached to the other end sliding with coefficient of friction u on a flat surface?!</p>
<p>I miss my Diffeq class. :(</p>
<p>Pebbles, why WOULDN'T there be a monkey climbing up a frictionless pulley with a block attached to the other end sliding with a coefficient of friction u on a flat surface? :)</p>
<p>because I'd kick it in the neck.</p>
<p>But then, who would pull the block along a horizontal surface with coefficient of friction u? :(</p>
<p>the block can then go to hell.</p>
<p>:P</p>
<p>Remember the +c at the end of the equation.
There are no specified bounds so you need to add a constant.</p>
<p>Pebbles, I have always wondered about those sorts of things.</p>
<p>One of the '09's I know was recently complaining about this 8.01 gem: "A 500-kg partridge is suspended from a pear tree by an ideal spring of negligible mass."</p>
<p>What? Who did that to that poor bird? And why is the bird so fat?</p>
<p>question</p>
<p>why would a class cram an entirely different topic into the last two weeks of class and then 'focus' the exam on those two weeks even though the class never had a problem set covering it?</p>
<p>Oh yeahhhhhh </p>
<p>lol sorry, venting pre-exam frustrations :P</p>
<p>If you were in 6.012, you'd know lol</p>
<p>Open circuit time constants anyone???</p>
<p>ok now what if we had to solve in terms of y?
Is that possible?</p>
<p>I guess our path toward scientific enlightenment just happens to plough through some animal rights (better the partridge than Fido).</p>
<p>But one of these days, when all the air molecules flee our perfectly spherical earth and we all turn into featureless ideal point masses wielding rods of negligible width and pulleys of negligible mass, we'll turn off the frictional forces that we don't like as much and we'll know exactly what to do.</p>
<p>ln^2(y) = -x^2</p>
<p>Take the square root of both sides.
ln(y) = -x
Raise e to both powers.
e^(lny) = e^-x.
y = e^-x.</p>
<p>Blah blah some domain things come into play, but that's the gist of it.</p>
<p>Addendum:
Don't forget to stop the earth's rotation. And honestly, for one of our pop quizzes, we had the question "Calculate your own moment of inertia about your vertical axis with all your limbs conracted. You may assume you have perfect bilaterial symmetry."</p>
<p>:/. He let us sit there for five minutes before bursting into laughter going, "okay, okay, here's your REAL quiz."</p>
<p>Here's the one I've always wanted to have solved: If a poor tired undergrad with a mass of 54 kg is carrying a 5-kg bookbag and walking due north with a velocity of 4.5 km/h, and her forward motion is entirely stopped, how fast is the wind blowing right now in Cambridge, MA?</p>
<p>Make any simplifying assumptions necessary, I suppose, although I'm better modeled by a rod of negligible width than a sphere. ;)</p>
<p>WHAT?@! NUMBERS/?!11 But we don't work with numbers. What is this km/h you speak of?</p>
<p>dL/dt = I<greek letter=""><greek letter="">!!</greek></greek></p>
<p>Said undergrad should catch a Safe Ride.</p>
<p>lmao mollie!</p>
<p>Olo--Wouldn't it be ln(y) = ix (or -ix)?</p>
<p>In which case I guess it would be y = e^(ix)...</p>
<p>And then factoring in constants...</p>
<p>ln^2 (y) = -x^2 + c1</p>
<p>so</p>
<p>ln(y) = sqrt(-x^2 + c1)</p>
<p>could you assume that this is equal to, with some other constant c2</p>
<p>ln(y) = ix + c2</p>
<p>? And then it should be:</p>
<p>y = Ce^(ix)</p>
<p>Maybe? lol</p>
<p>EDIT: Yeah, it's wrong to assume you can take the square root with the constant, the way I did...hmmm...</p>