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MAT244--2018F => MAT244--Tests => Term Test 2 => Topic started by: Victor Ivrii on November 20, 2018, 05:43:50 AM

Title: TT2-P1
Post by: Victor Ivrii on November 20, 2018, 05:43:50 AM
(a) Find the general solution of
\begin{equation*}
\end{equation*}

(b) Find solution, such that $y(0)=0$, $y'(0)=0$.
Title: Re: TT2-P1
Post by: Samarth Agarwal 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
$$\mbox{Therefore, the general solution is} y(t) = 1/2 \sin2t -\cos2t(t - \sin(t) \cos(t)) + \sin2t(\log(\cos(t)) - 1/2\cos2t)$$
Title: Re: TT2-P1
Post by: Victor Ivrii 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
$$y=-t\cos(2t)+\ln(\cos(t))\sin(2t)+\frac{1}{2}\sin(2t).$$