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March 12th, 2011, 06:11 AM  #11 
Senior Member Joined: Feb 2011 Posts: 150 Thanks: 0  Re: Linear differential equations
x = 1/sec(t) + c

March 12th, 2011, 07:21 AM  #12 
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 need to multiply both t and C by cos(t) = 1/sec(t). 
March 12th, 2011, 11:33 AM  #13 
Senior Member Joined: Feb 2011 Posts: 150 Thanks: 0  Re: Linear differential equations
Oh i see, i made the mistake on integrating swopping from one sin to another, more practice on these sort of equations is needed I think I have a tricky question to solve, well it seems it. The temperature T(measured in degrees) of a body immersed in an atmosphere of varying temperature is given by : dT/dt + 0.1T = 5  2.5t , find the formula for the temperature at time t if T=60degrees when t=0 Would the use of the usual method be suitable for solving this problem, and given the boundaries would allow me to evaluate at the end if the solution is worked correctly. 
March 12th, 2011, 11:40 AM  #14 
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
Yes, your equation is already given in standard linear form, so compute your integrating factor, then integrate to get the general solution (you'll have to use integration by parts on the righthand side), then use your initial conditions to find the particular solution. I got: If you did not get something equivalent, post your work, and we'll figure out where the error is. 
March 12th, 2011, 01:06 PM  #15 
Senior Member Joined: Feb 2011 Posts: 150 Thanks: 0  Re: Linear differential equations
right, im not confident with this one at all but heres my shot at it . obtain integrating factor = e^0.1t multiply std form equation by integrating factor e^0.1t dT/dt e^(0.1t) T = ^0.1t (52.5t) d/dt T/e^0.1t = e^0.1t (52.5t) integrate wrt t T/e^0.1t = e^0.1t (30025t) + C T=e^0.1t (300+25t)+Ce^0.1t 
March 12th, 2011, 02:05 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: Linear differential equations
You obtained the right integrating factor, but from there it went south! Multiplying the equation in standard form by the integrating factor gives: Use integration by parts on remaining integral, where: giving: thus we have: thus giving: 
March 13th, 2011, 12:11 AM  #17 
Senior Member Joined: Feb 2011 Posts: 150 Thanks: 0  Re: Linear differential equations
I tried, but thank you for going over that. Going to study that today to make sure it goes in 

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