My Math Forum Integrating Factor Method

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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 $\frac{dy}{dx}+P(x)y= Qx$. Notice that we need to have 1 as the coefficient of the term $\frac{dy}{dx$. Hence, from the given ODE $x^3 \frac{dy}{dx}= y+x^2e^{4x}$, I find we need to divide the left and right sides of the given ODE by $x^3$: $\frac{x^3}{x^3} \frac{dy}{dx}= \frac{y}{x^3} +\frac{x^2e^{4x}}{x^3}$ $\frac{dy}{dx}= \frac{y}{x^3} +\frac{e^{4x}}{x}$ $\frac{dy}{dx} - \frac{y}{x^3}=\frac{e^{4x}}{x}$ $\therefore \frac{dy}{dx} - (\frac{1}{x^3})y=\frac{e^{4x}}{x}$
 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 $\frac{dy}{dx}+P(x)y=Q(x)$ b) Calculate the integrating factor $\mu(x)$ by the formula: $\mu(x)=e^{\int P(x)\,dx}$ c) Multiply the equation in standard form by $\mu(x)$ and, recalling that the left-hand side is just $\frac{d}{dx}$$\mu(x)y$$$, obtain $\mu(x)\frac{dy}{dx}+P(x)\mu(x)=\mu(x)Q(x)$ $\frac{d}{dx}$$\mu(x)y$$=\mu(x)Q(x)$ d) Integrate the last equation and solve for y by dividing by $\mu(x)$. The given ODE in standard form is: $\frac{dy}{dx}+$$-x^{-3}$$y=x^{-1}e^{4x}$ The integrating factor is then: $\mu(x)=e^{\int -x^{-3}\,dx}=e^{\frac{1}{2}x^{-2}}$ and multiplying through, we have: $e^{\frac{1}{2}x^{-2}}\frac{dy}{dx}+$$-x^{-3}$$e^{\frac{1}{2}x^{-2}}y=x^{-1}e^{4x}e^{\frac{1}{2}x^{-2}}$ $\frac{d}{dx}$$e^{\frac{1}{2}x^{-2}}y$$=x^{-1}e^{\frac{8x^3+1}{2x^2}}$ $y=e^{-\frac{1}{2}x^{-2}}\int x^{-1}e^{\frac{8x^3+1}{2x^2}}\,dx$ 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 $\therefore \frac{dy}{dx} - (\frac{1}{x^3})y=\frac{e^{4x}}{x}$ 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 $\frac{dy}{dx}+P(x)y=Q(x)$, i.e. $\frac{dy}{dx} - (\frac{1}{x^3})y=\frac{e^{4x}}{x}$ $=\frac{dy}{dx} +(-\frac{1}{x^3})y =\frac{e^{4x}}{x}$ $=\frac{dy}{dx} +(-x^{-3})y =\frac{e^{4x}}{x}$ 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: $\frac{dy}{dx}+$$-\frac{1}{x}$$y=xe^{4x}$ Now, we calculate the integrating factor: $\mu(x)=e^{\int-\frac{1}{x}\,dx}=\frac{1}{x}$ and multiplying through: $\frac{1}{x}\cdot\frac{dy}{dx}+$$-\frac{1}{x}$$\frac{1}{x}\cdot y=\frac{1}{x}\cdot xe^{4x}$ $\frac{1}{x}\cdot\frac{dy}{dx}+$$-\frac{1}{x^2}$$ y=e^{4x}$ $\frac{d}{dx}$$\frac{y}{x}$$=e^{4x}$ $\frac{y}{x}=\frac{1}{4}e^{4x}+C$ $y=x$$\frac{1}{4}e^{4x}+C$$$
 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 $\frac{1}{4}e^4$ from both sides: $C=-\frac{1}{4}e^4$
 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: $y=x$$\frac{1}{4}e^{4x}-\frac{1}{4}e^4$$$

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

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