MAT244--2018F > Quiz-1

Q1: TUT 0201, TUT 5101 and TUT 5102

(1/2) > >>

Victor Ivrii:
Find the general solution of the given differential equation, and use it to determine how solutions behave as $t\to \infty$:
ty' + 2y = \sin (t)

Qin Wang:
Solution in the following file.

Victor Ivrii:
Waiting for a typed solution.

Tzu-Ching Yen:
Rephrase equation
$y' + \frac{2}{t}y = \frac{sin(t)}{t}$
Find integrating factor
$u(t) = e^{\int \frac{2}{t}} = t^2$
the constant from integration is chosen to be zero. Now
$y = \frac{1}{u(t)}\int u(t)\frac{sin(t)}{t}$
Use int by parts,
$y = -\frac{cos(t)}{t} + \frac{sin(t)}{t^2} + \frac{c_1}{t^2}$
Since $t$ is in denominator and the nominators are bounded, as $t \to \infty$, $y \to 0$

Wei Cui:
Question: $ty^{'}+2y=sin(t)$,   $t>0$

standard equation form: $y^{'}+\frac{2}{t}y=\frac{sin(t)}{t}$

$p(t) = \frac{2}{t}$,  $g(t) = \frac{sin(t)}{t}$

$u = e^{\int p(t)}dt = e^{2\int \frac{1}{t}dt} = t^2$, then we multiply both sides with $u$, and we get:

$t^2y^{'} + 2ty = tsin(t)$


$d(t^2y) = tsin(t)dt$

$t^2y = \int tsin(t) dt$

(integrating by parts, $u=t \implies du =dt$ and $dv=sin(t)$ and $v=-cos(t)$

$\int tsin(t)dt = uv - \int vdu$

                   $=-tcos(t)-\int (-cos(t))dt$
                   $=-tcos(t) +\int cos(t)dt$

                   $=-tcos(t) +sin(t) + C$)

Therefore, $t^2y=-tcos(t)+sin(t) + C$


Since $t>0$, and when $t \rightarrow \infty, y \rightarrow 0$


[0] Message Index

[#] Next page

Go to full version