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June 5th, 2008, 05:55 AM   #1
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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
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June 5th, 2008, 08:25 AM   #2
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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.
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June 10th, 2008, 05:15 AM   #3
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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
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June 10th, 2008, 08:46 AM   #4
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Use integration by parts.
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June 11th, 2008, 12:30 AM   #5
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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]
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June 11th, 2008, 12:32 AM   #6
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Re: problem in solution of differential equation

could u show me my error plz?
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June 11th, 2008, 02:04 AM   #7
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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.)
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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).)
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June 11th, 2008, 03:02 AM   #9
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Re: problem in solution of differential equation

thanks guys, and thank u tumppi for help, everything is very clear now!
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