Author Topic: TT2-P1  (Read 1942 times)

Victor Ivrii

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TT2-P1
« on: November 20, 2018, 05:43:50 AM »
(a) Find the general solution of
\begin{equation*}
y''+4y=2\tan (t),\qquad -\frac{\pi}{2}<t<\frac{\pi}{2}.
\end{equation*}

(b) Find solution, such that $y(0)=0$, $y'(0)=0$.

Samarth Agarwal

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Re: TT2-P1
« Reply #1 on: November 20, 2018, 09:35:29 AM »
first we change to characteristic equation
$$ r^2 + 4 = 0 $$
$$ \mbox{therefore, } r = \pm 2i $$
$$ \mbox{Therefore, the general solution = } y(t) = c_1 \cos2t + c_2 \sin2t $$
The particular solution, upon integration is
$$ y_p(t) = -\cos2t(t - \sin(t) \cos(t)) + \sin2t(\log(\cos(t)) - 1/2\cos2t) $$
$$ \mbox{Therefore, the general solution is} y(t) = c_1 \cos2t + c_2 \sin2t -\cos2t(t - \sin(t) \cos(t)) + \sin2t(\log(\cos(t)) - 1/2\cos2t) $$
$$ y(0) = c_1 = 0 $$
upon differentiating y(t) and plugging in t = 0, we get c_2 = 1/2
$$ \mbox{Therefore, the general solution is} y(t) = 1/2 \sin2t -\cos2t(t - \sin(t) \cos(t)) + \sin2t(\log(\cos(t)) - 1/2\cos2t) $$

Victor Ivrii

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Re: TT2-P1
« Reply #2 on: November 25, 2018, 11:08:01 AM »
You should leave text out of mathjax and avoid \mbox{ (which is unappropriate anyway, \text{ would be better.

Also after cancellations solution becomes
$$
y=-t\cos(2t)+\ln(\cos(t))\sin(2t)+\frac{1}{2}\sin(2t).
$$