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Given \begin{array}{l} y^{\prime} +\dfrac{1}{3}\,\sec\left(\dfrac{t}{3}\right) y= 4\, \cos\left(\dfrac{t}{3}\right) \\ y(0)=3 \end{array} where $ 0<\dfrac{t}{3}<\dfrac{\pi}{2}$, I must find the general solution $y(t)$. The problem I keep running into is when I am trying to find the integrating factor. I'm wondering if there is a way for me to simplify the integrating factor in such a way that I am able to get the differential equation into the form $(\mu(t)y(t))'=\mu(t)g(t)$ so that I can integrate both sides? Thanks!

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  • $\begingroup$ the solution looks terrible, from where do you get this? $\endgroup$ – Dr. Sonnhard Graubner Oct 1 '17 at 14:52
  • $\begingroup$ It was a question I found on a worksheet I am studying from. I have no solutions for this worksheet and I have tried putting it into Wolfram Alpha to compute, but yes, the solution looks terrible, which leads me to believe that there must be a way I can simplify this question using the given conditions. $\endgroup$ – grizzly.bear Oct 1 '17 at 14:55
  • $\begingroup$ ok i have posted a hint for you $\endgroup$ – Dr. Sonnhard Graubner Oct 1 '17 at 14:57
  • $\begingroup$ @Dr.SonnhardGraubner I have simplified it a bit $$y=12 \sin \left(\frac{t}{3}\right)-4 t \tan \left(\frac{t}{3}\right)-15 \tan \left(\frac{t}{3}\right)+(4 t+15) \sec \left(\frac{t}{3}\right)-12$$ $\endgroup$ – Raffaele Oct 1 '17 at 15:08
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Hint:

For first order equation $y'+py=q$ find integrating factor: $$I=e^{\int p(t)dt}=e^{\displaystyle \int \frac13\sec\frac{t}{3}dt}=\sec\frac{t}{3}+\tan\frac{t}{3}$$ because $$\int \frac13\sec\frac{t}{3}dt=\ln(\sec\frac{t}{3}+\tan\frac{t}{3})$$

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  • $\begingroup$ In this case, would the left side simplify down to just $(sec(\frac{t}{3})+tan(\frac{t}{3})y)'$ ? $\endgroup$ – grizzly.bear Oct 1 '17 at 15:29
  • $\begingroup$ definitely!.... $\endgroup$ – Nosrati Oct 1 '17 at 15:33
  • $\begingroup$ Now when I integrate both sides, would I integrate using the boundaries given, ie. from $0$ to $\frac{\pi}{2}$? $\endgroup$ – grizzly.bear Oct 1 '17 at 15:41
  • $\begingroup$ I think no. but you can integrate from $0$ (initial point) to $x\leq\dfrac{\pi}{2}$. $\endgroup$ – Nosrati Oct 1 '17 at 15:44
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compute $$\mu(t)=e^{\int\frac{1}{3}\sec(t/3)dt}=\frac{\sin(t/6)+\cos(t/6)}{\cos(t/6)-\sin(t/6)}$$ and multiply both sides with $$\mu(t)$$

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Integration factor is simply $e^{f(t)}$ where $$f(t) = \frac{1}{3}\int \sec\left(\dfrac{t}{3}\right) dt \\ = \ln\left(\sec\left(\dfrac{t}{3}\right) + \tan\left(\dfrac{t}{3}\right)\right)$$

So integrating factor is: $$\sec\left(\dfrac{t}{3}\right) + \tan\left(\dfrac{t}{3}\right)$$

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