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 March 11th, 2011, 10:00 AM #1 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 Linear differential equations Problem: Use the integrating factor method to solve the following linear differential equation x dy/dx - y = x I used the same method as I did last night with the other equations and found the integrating factor to be ?(x)=e^?-1/x dx = e^-ln(x) = -x (or in this case would it be just (x) ? ) Then Multiplying the standard form equation by (-x) , -x(dy/dx)-y= -x Integrating to get y=1/2 x + A / x Im a bit unsure on it
 March 11th, 2011, 10:12 AM #2 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: Linear differential equations $e^{-\ln x}=e^{\ln$$\frac{1}{x}$$}=\frac{1}{x}$ Now try it.
 March 11th, 2011, 10:26 AM #3 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 Re: Linear differential equations y= - (ln(x) +A / x)
 March 11th, 2011, 10:36 AM #4 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: Linear differential equations We have in standard form: $\frac{dy}{dx}-\frac{1}{x}y=1$ Multiplying through by $\mu(x)=\frac{1}{x}$ gives: $\frac{1}{x}\frac{dy}{dx}-\frac{1}{x^2}y=\frac{1}{x}$ $\frac{d}{dx}$$\frac{y}{x}$$=\frac{1}{x}$ Integrate w.r.t. x: $\frac{y}{x}=\ln|x|+C$ $y=x\ln|x|+Cx$
 March 11th, 2011, 11:38 AM #5 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 Re: Linear differential equations oh yes, i made the mistake on multiplying by the integrating factor. my next equation cos(t) dx/dt + x sin(t) = cos^2 (t) Re write in Std.form, divide by cos(t) dx/dt + x (sin(t)/cos(t)) = cos(t) Obtain Integrating factor, ?(x)=e^?sin(t)/cos(t) = e^-ln(cos(t)) = -cos(t) or is it 1/cos(t)
 March 11th, 2011, 12:37 PM #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: Linear differential equations It is $\frac{1}{\cos t}=\sec t$, since by the properties of logarithms, we have $-\ln(x)=\ln$$\frac{1}{x}$$$. This transforms your equation in standard form to: $\sec t\frac{dx}{dt}+(\sec t\tan t)x=1$ Then you have: $\frac{d}{dt}$$x\sec t$$=1$
 March 11th, 2011, 01:09 PM #7 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 Re: Linear differential equations oh ok, yes i thought the laws of logs would be involved. so then i integrate d/dx (x sect)=1 ?
 March 11th, 2011, 01:19 PM #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: Linear differential equations You integrate: $\frac{d}{dt}$$x\sec t$$=1$ w.r.t. t (Notice the slight difference in what I wrote and what you wrote. )
 March 11th, 2011, 01:32 PM #9 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 Re: Linear differential equations oh yes it was suposed to say t. x sec(t) = x + C , do i divide by sec(t)
 March 11th, 2011, 02:21 PM #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: Linear differential equations Or multiply through by cos(t), same thing. Although your right side should be t + C (recall we integrated with respect to t).

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