| Author | Message | ||
| James Bennett |
Please show the working (and final answer) to the following 2 questions on this topic; I have got the wrong answers and would therefore like to compare MY working with your own CORRECT working, so that I can see where I went wrong... Thanks very much! 1) Find the general solution of the differential equation: xy(dy/dx) = ln x 2) Find the particular solution of: ex(dy/dx) = Öy given that y=9 when x=0 ~ Thanks again. | ||
| Marcos |
Hi James... For (1), we have: y(dy/dx) = (ln x)/x Integrating wrt x (and using the chain rule), òy dy = ò (ln x)/x dx Now notice that we've got ln x and it's derivative being multiplied on the RHS, and use the substitution u = ln x. Thus, ò y dy = ò u du (Can you see why?) Thus, y2/2 = u2/2 + C ie. y2/2 = (ln x)2/2 + C [If you're picky you can change it to the equivalent and neater form y2 = (ln x)2 + A, where A is an arbitrary constant] You should now be able to do (2) yourself as it follows along similar lines. By all means ask if you get stuck or if you didn't understand any of my post... Marcos | ||
| James Bennett |
Yes, think I understand but I STILL don't get the right answer for q.2!! Please go through it and then I can check EXACTLY where I went wrong! | ||
| Marcos |
Okay, first thing: we need to "separate the variables"... So, in this case we need to get rid of the ex on the LHS and the Öy on the RHS. ex(dy/dx) = Öy So we get: y-1/2dy/dx = e-x (Right?) Integrating wrt x, ò y-1/2 dy = ò e-x dx That is, 2y1/2 = - e-x + C Can you finish it off now using the conditions you're given? (I get C = 7) Marcos | ||
| James Bennett |
Yes - I can do it - just a stupid little error by me along the way. LOL. Thanks for your help. One further question though - I'm a little confused as to why only 'lnx' is substituted for 'u' in the very first question (i.e. why is '(lnx)/x' not being substituted instead? - Please explain the reasoning!) | ||
| Mark Durkee |
If you substitute u = (ln x)/x then du/dx = (1 - ln x)/x2 Therefore the integral becomes ò ux2/(1-lnx)du which is far more difficult to integrate than the original function. | ||
| James Bennett |
Still confused with above point - please elaborate! | ||
| Marcos |
James, let's take a look at a simpler example (this can be done in a couple of ways but we'll do it using substitutions): ò x ex^2 dx We can use the substitution u = x2 Thus dx/du = 1/(du/dx) = 1/(2x) We can now rewrite our integral: ò x ex2 dx = ò x*eu*1/(2x) du (Make sure you understand this line) = ò eu/2 du (You can then finish this off and substitute u = x2 back into the equation) The important thing is that it is simpler to integrate ò eu/2 du rather than ò x ex^2 dx, that was the whole point of the substitution... Now, in your example we had: ò (ln x)/x dx Say we substitute u = (ln x)/x.. Well, we need to figure out what dx/du is (as we are going to multiply by it when changing the integration variable) Now, it's quite difficult (I think impossible - someone may care to correct me) to express dx/du only in terms of u (which would be ideal as we're going to integrate wrt u) but we can do what we did in the previous example (ie. express it in terms of x and hopefully the terms in "x" will cancel). Now, as Mark has already said, dx/du = 1/(du/dx) = 1/[(1 - ln x)/x2] = x2/(1 - ln x) Thus in our integral: ò (ln x)/x dx = ò u*x2/(1 - ln x) du This has definitely not simplified as we've got both u and x (which isn't a constant) to integrate wrt u and this is harder to integrate than the original function (ln x)/x, contrary to what we're trying to do... That's why in this example it's easier to use u = ln x: If you notice that 1/x is the derivative of ln x, you can immediately substitute u for ln x and du/dx for 1/x: ò (ln x)/x dx = ò u*du/dx dx = ò u du I hope things are clearer now. If you have any questions or don't agree with some point I've made please ask/say so  ... Marcos Minor note: (1) wrt = with respect to (2) a^b = ab (3) a*b = a multiplied by b | ||
| James Bennett |
Thank you! | ||
| Marcos |
No problem |