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 February 9th, 2012, 02:34 AM #1 Member   Joined: Jan 2012 Posts: 82 Thanks: 0 Integrating Factor Method I am trying to use the Integrating Factor Method to find the general solution to the differential equation x^3 dy/dx = y+x^2e^4x After dividing through by x and rearranging I think I get dy/dy - 1/x y = xe^4x Not sure this is correct Can anyone help? February 9th, 2012, 05:30 AM #2 Senior Member   Joined: Dec 2011 Posts: 277 Thanks: 2 Re: Intergating Factor Method To solve any first order linear ordinary differential equations by using integrating factor, we need to rewrite the given ODE into the standard form of . Notice that we need to have 1 as the coefficient of the term . Hence, from the given ODE , I find we need to divide the left and right sides of the given ODE by : February 9th, 2012, 10:09 AM #3 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: Integrating Factor Method Method for solving linear first order ODEs: a) Write the equation in the standard form b) Calculate the integrating factor by the formula: c) Multiply the equation in standard form by and, recalling that the left-hand side is just , obtain d) Integrate the last equation and solve for y by dividing by . The given ODE in standard form is: The integrating factor is then: and multiplying through, we have: The integral on the right cannot be expressed in elementary terms. February 9th, 2012, 06:50 PM #4 Senior Member   Joined: Dec 2011 Posts: 277 Thanks: 2 Re: Intergating Factor Method Argh...By re-reading this post, to leave off at this step is what's killing me, but never mind, I'll blame my weary head in the middle of last night, . I really should write it in the form , i.e. Sorry and Mark's complete workout is great! February 10th, 2012, 06:09 AM #5 Member   Joined: Jan 2012 Posts: 82 Thanks: 0 Re: Integrating Factor Method Thanks for your help guys but I made a mistake, I should have written x dy/dx = y +x^2e^4x After dividing through by x we get dy/dx = y/x + x^2e^4x /x rearranging give dy/dx - y/x = x^2e^4x /x not sure about the rest. Can anyone help? February 10th, 2012, 06:20 AM #6 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: Integrating Factor Method Writing the ODE in standard linear form, we have: Now, we calculate the integrating factor: and multiplying through: February 11th, 2012, 01:20 AM #7 Member   Joined: Jan 2012 Posts: 82 Thanks: 0 Re: Integrating Factor Method Thank you so much for your help. I now need to find particular solution of the differential equation that satisfies the initial condition y(1)=0. So far I have For y(1)=0 0=1(1/4e^4*1 + C) Not sure about the algebra here to find out what C is. Can anyone help? February 11th, 2012, 01:26 AM #8 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: Integrating Factor Method You would simply have, upon solving for C, by subtracting from both sides: February 11th, 2012, 08:19 AM #9 Member   Joined: Jan 2012 Posts: 82 Thanks: 0 Re: Integrating Factor Method Would this then leave y = x (1/4^4x - 1/4e^4) as the particular solution. February 11th, 2012, 08:45 AM #10 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: Integrating Factor Method No, it would be: Tags factor, integrating, method ,

### when should i use the integration factor method to solve an ordinary differential equation

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