Solving linear differential equation $y'+\frac{1}{3}sec(\frac{t}{3})y=4cos(\frac{t}{3})$ using integrating factor

Question

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!

Answer 1

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 $$\int \frac13\sec\frac{t}{3}dt=\ln(\sec\frac{t}{3}+\tan\frac{t}{3})$$

Answer 2

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 $$\mu(t)$$

Answer 3

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: $$\sec\left(\dfrac{t}{3}\right) + \tan\left(\dfrac{t}{3}\right)$$