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Old October 31st, 2009, 07:20 PM
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Default Second Order DE

I'm having trouble trying to solve the differential equation y^{(2)} + (\frac{1}{4t^2})y=f cos(t), t>0 given that y_{1}=\sqrt{t} is a solution of the homogeneous equation. I have been trying to solve this by reduction of order and am having no luck. I beleive the correct answer to be y(t)=\sqrt{t}[c_{1}+c_{2}ln(t)+\int_0^t f \sqrt{s} cos(s)[ln(t)-ln(s)]ds] but can't seem to figure it out. Any help would be appreciated.
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Old October 31st, 2009, 10:15 PM
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Reduction of order certainly should work. Exactly what have you tried?


If y(t)= u(t)t^{1/2} then y'= u't^{1/2}+ (1/2)ut^{-1/2} and y"= u"t^{1/2}+ u't^{-1/2}- (1/4)ut^{-3/2}.


Putting that into the equation, y"+ (1/(4t^2)y= u"t^{1/2}+ u't^{-1/2}- (1/4)ut^{-3/2}+ (1/4)ut^{-3/2}[math]= u"t^{1/2}+ u't^{-1/2}= f cos(t)[math]. Let v= u' and that is ][math]v't^{1/2}+ vt^{-1/2}= f cos(t)[math], a linear first order equation. You might have to leave part of the solution in terms of an integral.
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Old November 1st, 2009, 08:58 AM
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I have everything you have up to that point. That was the easy part. Solving for u seems to be the problem. If I solve using y_{2}=y_{1}(t)\int  \frac{exp[-\int P(t)dt]}{y_1^2(t)}dt I get a solution, but not the given solution. If I try to solve for u use using brute force method I get hung up and am unable to produce the given solution.

Last edited by AnDiesel; November 1st, 2009 at 12:45 PM.
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Old November 1st, 2009, 11:50 AM
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Why are you integrating it from zero? Looks to me it should be solved for t_0>0. Letting y=u\sqrt{t} as HallsofIvy suggested, suppose I do it from t_0 and not zero, I get the expression:

\int_{t_0}^{t}du=\int_{t_0}^{t}\left[\frac{c_1}{t}+\frac{1}{t}\int_{t_0}^t s^{1/2}f\cos(s)ds\right]dt

Doing the left side and changing the variable names on the right so that we can follow it better:

u(t)-u(t_0)=\int_{t_0}^{t}\left[\frac{c_1}{w}+\frac{1}{w}\int_{t_0}^w s^{1/2}f\cos(s)ds\right]dw

u(t)-u(t_0)=c_1(\ln(t)-\ln(t_0))+\int_{t_0}^{t}\int_{t_0}^{w} \frac{1}{w}s^{1/2}f\cos(s)ds dw

Now, can you justify switching the order of integration in order to arrive at a solution which looks like yours but starts at t_0?
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Last edited by shawsend; November 1st, 2009 at 12:13 PM.
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Old November 1st, 2009, 01:59 PM
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I see what you did there. Thank you for your help, both of you.
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