Toronto Math Forum
MAT2442018F => MAT244Tests => Quiz1 => Topic started by: Victor Ivrii on September 28, 2018, 03:28:20 PM

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

Solution in the following file.

Waiting for a typed solution.

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 $t \to \infty$, $y \to 0$

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:
$t^2y^{'} + 2ty = tsin(t)$
$(t^2y)^{'}=tsin(t)$
$d(t^2y) = tsin(t)dt$
$t^2y = \int tsin(t) dt$
(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, $t^2y=tcos(t)+sin(t) + C$
$y=\frac{tcos(t)+sin(t)+C}{t^2}$.
Since $t>0$, and when $t \rightarrow \infty, y \rightarrow 0$

Solution for TUT5101
Question: Find the solution of the given initial value problem y'2y = e^2t , y(0)=2
let p(t)=2 and set u=e^(integral p(t)dt) then you get u = e^(2t)
then you multiply u on each side of the standard form equation and you get
e^(2t)y'2e^(2t)y = 1
then you can find the LHS is equal to (e^(2t)y)' = 1
integral each side you get (e^2t)y = t +C
rearrange the equation you get y=e^2t(t+C)
y(t)=(t+C)e^2t
plug in y(0)=2 and you get C=2
y can be written as y=e^2t(t+2)

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