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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 
Use integration by parts.

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:
 
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!


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