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How to solve this differential equation with the initial condition $$\begin{cases} x(0)=1\\ x'(t) \cos t - 2x \sin t = 1 \end{cases}$$ for $t \in (\frac{-\pi}{2}, \frac{\pi}{2})$?

I tried to use $z(t)=x(t)\cos(t), z'(t)=x'(t)\cos(t) -x(t) \sin(t)$, but then I have $z(t)-x(t)\sin t =1$, so there is still a problem to be solved.

john1235
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1 Answers1

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Hint: $x' - (2\tan t)x = \sec t$. Can you take it from here as you have a linear $1$st order DE.

Wang YeFei
  • 6,390