Toronto Math Forum
MAT2442018F => MAT244Tests => Term Test 2 => Topic started by: Victor Ivrii 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 $y(0)=0$, $y'(0)=0$.

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
$$ \mbox{Therefore, the general solution is} y(t) = 1/2 \sin2t \cos2t(t  \sin(t) \cos(t)) + \sin2t(\log(\cos(t))  1/2\cos2t) $$

You should leave text out of mathjax and avoid \mbox{ (which is unappropriate anyway, \text{ would be better.
Also after cancellations solution becomes
$$
y=t\cos(2t)+\ln(\cos(t))\sin(2t)+\frac{1}{2}\sin(2t).
$$