Author Topic: Q1: TUT 0501  (Read 2186 times)

Victor Ivrii

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Q1: TUT 0501
« on: September 28, 2018, 03:35:09 PM »
Find the value of $y_0$ for which the solution of the initial value problem
\begin{equation*}
y' - y = 1 + 3 \sin (t),\qquad y(0) = y_0.
\end{equation*}
remains finite as $t\to \infty$.

Yifei Gu

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Re: Q1: TUT 0501
« Reply #1 on: September 28, 2018, 07:50:46 PM »
here is the solution to the question.

Yiting Zhang

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Re: Q1: TUT 0501
« Reply #2 on: September 28, 2018, 08:18:05 PM »
$p(t) = -1, q(t) = 1+3 \sin(t)$

$\mu(t) = e^{\int p(t)dt} = e^{\int (-1)dt}=e^{-t}$

$\mu y' - \mu y = \mu (1+3\sin(t))$

$e^{-t} y' - e^{-t}y = e^{-t} (1+3\sin(t))$

$\frac{d}{dt}(e^{-t} y) = e^{-t} +3e^{-t}\sin(t)$

$\int \frac{d}{dt}(e^{-t} y) dt= \int e^{-t} +3e^{-t}\sin(t) dt$

$e^{-t} y = -e^{-t} - \frac{3}{2}\sin te^{-t} - \frac{3}{2}\cos te^{-t} + c$

$y(t) = -1 - \frac{3}{2}\sin t - \frac{3}{2}\cos t + ce^t$

$y(0) = y_0 = -1 - \frac{3}{2}\sin(0) - \frac{3}{2}\cos(0) + ce^0 = -1 - \frac{3}{2} + c$

$c = y_0 + \frac{5}{2}$

$y(t) = -1 - \frac{3}{2}\sin t - \frac{3}{2}\cos t + (y_0 + \frac{5}{2}) e^t$

If $y_0 < - \frac{5}{2}, y_0 + \frac{5}{2} < 0$,

$$\lim_{t\to\infty} y(t) = \lim_{t\to\infty} (y_0 + \frac{5}{2}) e^t = -\infty$$
If $y_0 > - \frac{5}{2}, y_0 + \frac{5}{2} > 0$,
$$\lim_{t\to\infty} y(t) = \lim_{t\to\infty} (y_0 + \frac{5}{2}) e^t = +\infty$$
If $y_0 = - \frac{5}{2}, y_0 + \frac{5}{2} = 0$,
$$\lim_{t\to\infty} y(t) = \lim_{t\to\infty} (-1 - \frac{3}{2}\sin t - \frac{3}{2}\cos t)$$
is the only value of $y_0$ that makes the solution finite
$y_0 = - \frac{5}{2}$
« Last Edit: September 29, 2018, 03:08:15 PM by Victor Ivrii »

Victor Ivrii

Yiting, please correct $x\to \infty$ should be replaced by $t\to \infty$