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 Differential Equations Ordinary and Partial Differential Equations Math Forum

 March 9th, 2011, 12:46 AM #1 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 Exact form, differential equations hey, Ive had a go at this question but im not too sure. Solve the following differential equations given in exact form Qa) d/dx (yx^2)=x^3 Qb)d/dx (ye^x)=e^2x Aa) integrate wrt x. ?d/dx (yx^2)dx=?x^3 dx yx^2=x^4/4+A divide by x^2 to get y---> y=x^2/4 +A Ab) integrate wrt x. ?d/dx(ye^x)dx=?e^2x ye^x=e^2 x^2 /2 divide by e^x to get y---> y=( 1/2(e^2 x^2) )/ e^x Jake March 9th, 2011, 01:03 AM #2 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Exact form, differential equations Look after the constants, as they help you find general solutions. a) b) I take it that it's definitely e�x and not e^(2x)... March 9th, 2011, 03:03 AM #3 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 Re: Exact form, differential equations thanks for that! for b) it is d/dx (ye^x)=e^(2x) sorry about that. March 9th, 2011, 03:18 AM #4 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Exact form, differential equations In that case you get , hence , a family of curves which includes cosh x (for C=�). March 9th, 2011, 05:59 AM #5 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 Re: Exact form, differential equations brilliant. March 9th, 2011, 07:56 AM #6 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 Re: Exact form, differential equations Ok so, By expressing the following differential equations in exact form, find their general solutions. a) e^2x dt/dx + 2e^2x y = x^2 b) x dy/dx + y = x^3 The first one when writing in standard form, dividing by the exponential, do they cancel out leaving e^2x dx as the integrating factor? and for the second one I end up with e^ln(x) .... = x being the integrating factor..? March 9th, 2011, 08:53 AM #7 Senior Member   Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Exact form, differential equations I am assuming a) is: The left side is the differentiation of the product with respect to x: b) That's how I would solve those, however, you are instructed to express the equations in exact form: a) Test for exactness: Since the equation is exact. Thus, we have and integrating with respect to x, we have: Now we take the partial derivative with respect to y: Since we have thus giving: The solution is given implicitly by: thus we have: I'll leave this method for b) to you!  March 10th, 2011, 01:58 AM #8 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 Re: Exact form, differential equations Thanks Mark, testing for exactness the process just confirms that the equation is exact, producing the same general solution, when testing for exactness on b) xdy/dx +y=x^3 (y-x^3) dx + x dy =0 Test for exactness d/dy(y-x^3) = 1/2 x^2 d/dx(x) = 1/2 x^2 Is this correct so far March 10th, 2011, 06:49 AM #9 Senior Member   Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Exact form, differential equations No, you would have: When taking the partial derivative of a function of more than one variable, you treat the other other variables (those you are not differentiating with respect to) as a constant. March 10th, 2011, 10:52 AM #10 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 Re: Exact form, differential equations ok then, so I followed your method but I could only get y - 1/4x^4 +C => y=1/4x^4+Cx^-1 , but it does not match the solution as previously stated y=1/4 x^3 +Cx^-1 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 05:24 PM helloprajna Differential Equations 12 July 19th, 2013 09:21 PM Survivornic Differential Equations 3 October 1st, 2012 07:02 PM Survivornic Differential Equations 2 September 30th, 2012 02:40 PM conjecture Calculus 0 July 27th, 2008 10:25 AM

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