 My Math Forum problem in solution of differential equation
 User Name Remember Me? Password

 Differential Equations Ordinary and Partial Differential Equations Math Forum

 June 5th, 2008, 05:55 AM #1 Newbie   Joined: May 2008 Posts: 29 Thanks: 0 problem in solution of differential equation y' + (1/t)y = 3cos2t t>0 the integrating factor u(t) = e now i multiply both sides of equation by e and get: ey' + (1/t)*y*e = 3e*cos2t normally i would apply this rule to the left side of above equation: (uv)' = u'v + v'u but its not possible cuz i have 1/t there... any ideas? thanks June 5th, 2008, 08:25 AM #2 Newbie   Joined: May 2008 Posts: 19 Thanks: 0 You have to use a better integrating factor Let's take u(t)=e^(ln t)=t (fortunately t>0). Now the reason why I chose ln(t) as the exponent is that d/dt ln(t)=1/t, so that by the chain rule u'(t)=1/t*e^(ln t). In this case it happens that u(t)=t, and u'(t)=1. Now after multiplying both sides with u we have: y't + y*1/t*t = 3*cos(2t)*t <=> y't+y=3t*cos(2t) which can be written as y'u+yu'=3t*cos(2t) You can do the rest Generally, to solve an equation of the form y'+hy=g, where h and g are (known) functions, one can take an integrating factor of the form u(t)=e^(some integral function of h), so that u'(t)=he^(some integral function of h)=hu. June 10th, 2008, 05:15 AM #3 Newbie   Joined: May 2008 Posts: 29 Thanks: 0 Re: problem in solution of differential equation thank you very much! i did understand your idea, but i am stuck again i can see now that (ty)' = 3t*cos2t integral of cos2t is sin2t + c, but how to integrate 3t*cos2t? thanks! i tried also to show cos2t as 1 - sin^2 t... but it doesnt help much June 10th, 2008, 08:46 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,916 Thanks: 2199 June 11th, 2008, 12:30 AM #5 Newbie   Joined: May 2008 Posts: 29 Thanks: 0 Re: problem in solution of differential equation ok, then it will be like that: we choose values for integration by parts: u = 3t, du = 3dt and dv = cos2t, v = sin2t then we write: Int[3t*cos2t] = 3t*sin2t - Int[3sin2t dt] = 3t*sin2t + 3cos2t + c so we have: ty = 3t*sin2t + 3cos2t + c, and: y = 3sin2t + (3/t)cos2t + c/t but the answer is: [color=#FF0000](c) y = (c/t) + (3 cos2t)/4t + (3 sin 2t)/2;[/color] June 11th, 2008, 12:32 AM #6 Newbie   Joined: May 2008 Posts: 29 Thanks: 0 Re: problem in solution of differential equation could u show me my error plz? June 11th, 2008, 02:04 AM #7 Newbie   Joined: May 2008 Posts: 19 Thanks: 0 Re: problem in solution of differential equation sin(2x) is not an integral funcion of cos(2x). (By the chain rule, d/dx cos(2x)=sin(2x)*(d/dx 2x)=2sin(2x)?sin(2x).) Generally, if F is an integral function of f, then: ?f(kx)dx=1/k*F(kx)+C, since by the chain rule: d/dx (1/k*F(kx)+C)=1/k*f(kx)*d/dt (kx)=k*1/k*f(kx)=f(kx). By this formula, ?cos(2t)dt=1/2*sin(2t)+C and your v should be 1/2sin(2t). (Also remember the constant 1/2 when computing ?vdu.) June 11th, 2008, 02:11 AM   #8
Newbie

Joined: May 2008

Posts: 19
Thanks: 0

Re: problem in solution of differential equation

Where is the edit button? Anyways, this:
Quote:
 Originally Posted by tumppi (By the chain rule, d/dx cos(2x)=sin(2x)*(d/dx 2x)=2sin(2x)?sin(2x).)
should of course have been (By the chain rule, d/dx sin(2x)=cos(2x)*(d/dx 2x)=2cos(2x)?cos(2x).)  June 11th, 2008, 03:02 AM #9 Newbie   Joined: May 2008 Posts: 29 Thanks: 0 Re: problem in solution of differential equation thanks guys, and thank u tumppi for help, everything is very clear now! Tags differential, equation, problem, solution Search tags for this page
,

,

,

,

,

,

,

,

# how to solve the equation for sin2t - 3cos2t

Click on a term to search for related topics.
 Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post jazz_1969 Differential Equations 2 May 5th, 2013 08:52 PM Zilee Differential Equations 3 May 19th, 2012 11:03 AM bluetrain Differential Equations 6 February 26th, 2010 04:42 AM cosette Differential Equations 0 October 18th, 2009 11:36 AM cosette Differential Equations 0 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2019 My Math Forum. All rights reserved.      