### Author Topic: Q1: TUT 0201, TUT 5101 and TUT 5102  (Read 5243 times)

#### Victor Ivrii

• Administrator
• Elder Member
• Posts: 2607
• Karma: 0
##### Q1: TUT 0201, TUT 5101 and TUT 5102
« 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*}
« Last Edit: September 28, 2018, 03:32:44 PM by Victor Ivrii »

#### Qin Wang

• Newbie
• Posts: 2
• Karma: 1
##### Re: Q1: TUT 0201, TUT 5101 and TUT 5102
« Reply #1 on: September 28, 2018, 04:33:53 PM »
Solution in the following file.

#### Victor Ivrii

• Administrator
• Elder Member
• Posts: 2607
• Karma: 0
##### Re: Q1: TUT 0201, TUT 5101 and TUT 5102
« Reply #2 on: September 29, 2018, 03:00:26 PM »
Waiting for a typed solution.

#### Tzu-Ching Yen

• Full Member
• Posts: 31
• Karma: 22
##### Re: Q1: TUT 0201, TUT 5101 and TUT 5102
« Reply #3 on: September 29, 2018, 09:23:56 PM »
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

• Full Member
• Posts: 16
• Karma: 11
##### Re: Q1: TUT 0201, TUT 5101 and TUT 5102
« Reply #4 on: September 29, 2018, 09:51:28 PM »
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 $t \rightarrow \infty, y \rightarrow 0$
« Last Edit: September 29, 2018, 09:56:03 PM by Wei Cui »

#### Boyu Zheng

• Jr. Member
• Posts: 12
• Karma: 8
##### Re: Q1: TUT 0201, TUT 5101 and TUT 5102
« Reply #5 on: September 30, 2018, 12:10:17 AM »
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)

#### Victor Ivrii

• Administrator
• Elder Member
• Posts: 2607
• Karma: 0
##### Re: Q1: TUT 0201, TUT 5101 and TUT 5102
« Reply #6 on: September 30, 2018, 12:23:04 AM »
All "mathoperators" should be typed with escape character \: \sin \cos \ln \tan , and after it should not be  alphanumerical character (letter or digit)