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 March 10th, 2011, 11:06 AM #11 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 When you divide through by x, the (1/4)x^4 becomes (1/4)x^3
 March 10th, 2011, 11:51 AM #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, 11:53 AM #13 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 Flawless!
 March 10th, 2011, 11:56 AM #14 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 Re: Exact form, differential equations There's hope yet my friend
 March 10th, 2011, 01:06 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, 01:17 PM #16 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 When you divide through by x². you should have: $\frac{dy}{dx}+\frac{1}{x}y=\frac{1}{x^2}$ Now we find the integrating factor is...?
 March 10th, 2011, 01:34 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, 01:37 PM #18 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 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, 01:40 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, 01:52 PM #20 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 must multiply the equation in standard form by the integrating factor x: $x\frac{dy}{dx}+y=\frac{1}{x}$ $\frac{d}{dx}$$xy$$=\frac{1}{x}$ Now integrate and solve for y. Notice you have: $x^3\frac{dy}{dx}+x^2y=\frac{d}{dx}\(x^2y)$ which is not true.

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