My Math Forum problem in solution of differential equation

 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
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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!

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# how to solve the equation for sin2t - 3cos2t

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