Differential Equations Ordinary and Partial Differential Equations Math Forum

 March 10th, 2011, 12:06 PM #11 Senior Member   Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 522 Math Focus: Calculus/ODEs Re: Exact form, differential equations When you divide through by x, the (1/4)x^4 becomes (1/4)x^3 March 10th, 2011, 12:51 PM #12 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 Re: Exact form, differential equations Yes ofcourse. Show that x^2 dy/dx + 3xy = x^3 can be made exact and hence obtain its general solution, please can you check my answer Multiply by integrating factor x, x^3 dy/dx + 3x^2 y = x^4 d/dx (x^3 y) = x^4 Exact form Now integrate wrt x, x^3 y = (1/5)x^5 + A y = (1/5)x^2 + A/x^3 General solution March 10th, 2011, 12:53 PM #13 Senior Member   Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 522 Math Focus: Calculus/ODEs Re: Exact form, differential equations Flawless!  March 10th, 2011, 12:56 PM #14 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 Re: Exact form, differential equations There's hope yet my friend  March 10th, 2011, 02:06 PM #15 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 Re: Exact form, differential equations Similar question, Find the integrating factor for the following equation and find the general solution x^2 dy/dx + xy =1 Rewrite in std form.... divide by x^2 dy/dx + 1/x^2 y =1/x^2 Obtain integrating factor, I got e^-1/x this may be incorrect as I have a feeling the std form may be wrong..! Before I go any further can you advise me on the integrating factor. March 10th, 2011, 02:17 PM #16 Senior Member   Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 522 Math Focus: Calculus/ODEs Re: Exact form, differential equations When you divide through by x�. you should have: Now we find the integrating factor is...? March 10th, 2011, 02:34 PM #17 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 Re: Exact form, differential equations integrating factor, ?(x) = e^?1/x = e^ln(x) = x integrating factor = x yes? March 10th, 2011, 02:37 PM #18 Senior Member   Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 522 Math Focus: Calculus/ODEs Re: Exact form, differential equations Yes. The same result could be achieved by dividing through the original equation by x. However, it's not always so easy to spot what you need, and the formula you are using is a great help in those cases. March 10th, 2011, 02:40 PM #19 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 Re: Exact form, differential equations Yes, not so easy to see from the start, best to work through it to make sure. so, multiply by x x^3 dy/dx +x^2 y = x d/dx (x^2 y) = x exact form integrate wrt to x x^2 y =(1/2)x^2 + A y= (1/2)x + A/x^2 March 10th, 2011, 02:52 PM #20 Senior Member   Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 522 Math Focus: Calculus/ODEs Re: Exact form, differential equations No, you must multiply the equation in standard form by the integrating factor x: Now integrate and solve for y. Notice you have: which is not true. Tags differential, equations, exact, form Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post summsies Differential Equations 2 July 24th, 2013 06:24 PM helloprajna Differential Equations 12 July 19th, 2013 10:21 PM Survivornic Differential Equations 3 October 1st, 2012 08:02 PM Survivornic Differential Equations 2 September 30th, 2012 03:40 PM conjecture Calculus 0 July 27th, 2008 11:25 AM

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