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### Messages - Coollight

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1
##### Quiz-4 / Re: 0501 quiz4
« on: October 18, 2019, 02:31:34 PM »
Correct me, if I am wrong, but I think the solution should be following:

Solution:

$\text{According to Problem 34, if we have an Euler equation in the form of } t^2 y''+aty' + by = 0 \\ \text{Then by changing variables } x = \ln(t) \text{, we can transform the equation into } \frac{d^2y}{dx^2} + (a-1)\frac{dy}{dx} + by = 0$
\begin{align*}
\text{In our case, } t^2 y''+3ty' + 1.25y = 0 \text{ is a Euler eqaution where a=3, b=1.25 } \\
\text{Then by changing variables } x = \ln(t), \\
\text{we can transform the equation into } y(x)'' + (3-1)y(x)' + 1.25y = 0\\
Then \ y(x)'' +2y(x)'+ 1.25y = 0 \\
\text{To solve }y(x)'' +2y(x)'+ 1.25y = 0, \text{assume } y =e^{rx} \text{ is a solution,}\\
\text{and we will have } y'=re^{rx}, y''&=r^2e^{rx}\\
Then \ y(x)'' +2y(x)'+ 1.25y = 0 &\Rightarrow r^2e^{rx}+2re^{rx} +1.25e^{rx}=0\\
e^{rx}(r^2+2r+1.25) &= 0\\
\text{Since }e^{rx} \neq 0, \text{then } r^2+2r+1.25 = 0\\
r &= \frac{-2 \pm \sqrt{4-(4 \times 1.25)}}{2} = -1 \pm \frac{1}{2}i  \text{ where }\lambda =-1 \ and \ \mu =\frac{1}{2} \\
y(x) &= c_{1} e^{\lambda x} cos(\mu x) + c_{2} e^{\lambda x} sin(\mu x) \\
&= c_{1} e^{-x} cos(\frac{x}{2}) + c_{2} e^{-x} sin(\frac{x}{2})\\
&=  e^{-x}(c_{1}cos(\frac{x}{2}) + c_{2} sin(\frac{x}{2}) \\
\text{Since we let }x = \ln(t) \text{ at the beginning, }\\
\text{Then we will have, } \\
y(t) &= e^{- \ln(t)}(c_{1}cos(\frac{ \ln(t)}{2}) + c_{2} sin(\frac{ \ln(t)}{2}) \\
&= t^{-1}(c_{1}cos(\frac{ \ln(t)}{2}) + c_{2} sin(\frac{ \ln(t)}{2})\\
&\text{as the general solution.} \\
\end{align*}

2
##### Quiz-4 / Q4: TUT0801
« on: October 18, 2019, 02:00:13 PM »
Solve the equation. Use the method of Problem 34 to solve the given equation for t > 0. No need to do an actual change of variables, just use the result the Problem 34.
\begin{align*}
\  t^2 y''+ty' + y = 0,\\
\end{align*}
Solution:

$\text{According to Problem 34, if we have an Euler equation in the form of } t^2 y''+aty' + by = 0\\ \text{Then by changing variables } x = \ln(t) \text{, we can transform the equation into } \frac{d^2y}{dx^2} + (a-1)\frac{dy}{dx} + by = 0$
\begin{align*}
\text{In our case, } t^2 y''+ty' + y = 0 \text{ is an Euler eqaution where a = 1, b = 1 } \\
\text{Then by changing variables } x = \ln(t), \\
\text{we can transform the equation into } y(x)'' + (1-1)y(x)' + y = 0\\
Then \ y(x)'' + y = 0 \\
\text{To solve }y(x)'' + y = 0, \text{assume } y =e^{rx} \text{ is a solution,}\\
\text{and we will have } y'=re^{rx}, y''&=r^2e^{rx}\\
Then \ y(x)'' + y = 0 &\Rightarrow r^2e^{rx}+re^{rx}=0\\
e^{rx}(r^2+1) &= 0\\
\text{Since }e^{rx} \neq 0, \text{then } r^2+1 = 0\\
r^2 &= \sqrt{-1}\\
r &= \pm i  \text{ where }\lambda =0 \ and \ \mu =1 \\
y(x) &= c_{1} e^{\lambda x} cos(\mu x) + c_{2} e^{\lambda x} sin(\mu x) \\
&= c_{1} e^{0} cos(x) + c_{2} e^{0} sin(x)\\
&= c_{1}cos(x) + c_{2}sin(x) \\
\text{Since we let }x = \ln(t) \text{ at the beginning, }\\
\text{Then we will have, } \\
y(t) &= c_{1}cos(\ln t) + c_{2}sin(\ln t) \\
&\text{as the general solution.} \\
\end{align*}

3
##### Quiz-3 / Re: quiz 3 tut 0401
« on: October 11, 2019, 02:37:00 PM »
if I am wrong, please correct me.

\begin{align*}
\ r^{2} + 3r = 0 \ should \ be \\
\ r(r+3) &= 0 \\
\ r_{1} = 0 ,& \ r_{2} = -3 \\
\ so \ the \ general \ solution \ should \ be \\
\ y &= c_{1} + c_{2} e^{-3t} \\
\ since \ y(0) = -2 \\
\ c_{1}+ c_{2} e^{0} &= -2 \\
\ c_{1}+ c_{2}  &= -2 \\
\ since \ y'(0) = 3 \\
\ y' &= -3c_{2} e^{-3t} \\
\ y' &= -3c_{2} e^{0} \\
\ -3c_{2} &= 3 \\
\ so \ c_{1} =-1 ,& \ c_{2} = -1 \\
\ so \ the \  solution \ will \ be \\
\ y &= -1 - e^{-3t}
\end{align*}

4
##### Quiz-3 / Re: TUT0102 Quiz3
« on: October 11, 2019, 02:13:44 PM »
Hi, I believe the final answer should be

\begin{align*}
\ g(t)= 3te^{2t} + ce^{2t} \\
\end{align*}

5
##### Quiz-3 / Q3: TUT0801
« on: October 11, 2019, 02:05:13 PM »
Find the solution of the initial value problem
\begin{align*}
\ 2y”- 3y’+y=0, \ y(0)=2, \ y’(0)=\frac{1}{2} \\
\end{align*}
Solution:
\begin{align*}
\ Assume \ y=e^{rt} \ a \ solution \ of \ the \ equation. \\
\ Then \ we \ will \ have \\
\ y&=e^{rt} \\
\ y’&=re^{rt} \\
\ y”&=r^{2}e^{rt} \\
\ Then \ 2y”-3y’+y=0 \ will \ be \\
\ 2(r^{2}e^{rt})-3(re^{rt})+e^{rt}&=0 \\
\ e^{re}(2r^{2}-3r+1)&=0 \ where \ e^{rt} \neq 0 \\
\ 2r^2-3r+1 &=0 \\
\ (2r-1)(r-1)&=0\\
\ r_{1}=\frac{1}{2}&,\ r_{2}=1 \\
\ Then \ we \ will \ have \ the \ general \ solution \\
\ y=c_{1}e^{r_{1}t}+c_{2}e^{r_{2}t}&=c_{1}e^{\frac{1}{2}t}+c_{2}e^{t} \\
\ Since \ y(0)=2 \\
\ c_{1}e^0 + c_{2}e^0 &= 2 \\
\ c_{1} + c_{2} &=2 \\
\ Since \ y’(0)=\frac{1}{2} \\
\ y’&=\frac{1}{2}c_{1}e^{\frac{1}{2}t}+c_{2}e^t \\
\ \frac{1}{2}c_{1}e^0 + c_{2}e^0 &= \frac{1}{2}\\
\ \frac{1}{2}c_{1}+c_{2}&= \frac{1}{2}\\
\ So \ solve \ for \ c_{1}, \ c_{2}\ to \ get: \\
\ c_{1}= 3, \ c_{2}=-1\\
\ So \ the \ solution \ will \ be\ y&=3e^{\frac{1}{2}t}-e^{t} \\
\end{align*}

6
##### Quiz-2 / Q2: TUT0801
« on: October 04, 2019, 04:40:30 PM »
Show the given equation is not exact but becomes exact when multiplied by the given integrating factor, then solve the equation

\begin{align*}
\  x^2 y^3 + x(1+y^2)y' = 0, \ \mu(x, y) = \frac{1}{xy^3}\\
\end{align*}
Solution:
\begin{align*}
\ (x^2 y^3)dx + x(1+y^2)dy &= 0 \\
\ M = x^2 y^3 \ &and \ N = x(1+y^2) \\
\ M_{y} = 3x^2y^2 \ &and \ N_{x} = 1+y^2\\
\ Since \ M_{y} \neq N_{x}, \ the \ given \ equation \ is \ not \ exact. \\
\\
When \ multiplied \ by \ \mu(x, y) = \frac{1}{xy^3} \ on \ both \ sides, \ then \\
\ \frac{1}{xy^3}(x^2 y^3)dx + \frac{1}{xy^3}x(1+y^2)dy &= 0 \\
\ x dx + (y^{-3} + y^{-1})dy &= 0  \ is \ exact. \\
\ Then \ there \ exists \ a \ function \ \phi(x,y) \ s.t. \ \phi_{x} = M = x \ &and \ \phi_{y} = N = y^{-3} + y^{-1} \\
\ \phi = \int M dx = \int x dx &= \frac{1}{2}x^2 + h(y) \\
\ \phi_{y} = h(y)' \equiv N &= y^{-3} + y^{-1} \\
\ h(y)' &= y^{-3} + y^{-1} \\
\ h(y) &= -\frac{1}{2}y^{-2} + \ln|y| \\
\ \phi &= \frac{1}{2}x^2 + -\frac{1}{2}y^{-2} + \ln|y| \\
So \ we \ will \ have, \\
\ \frac{1}{2}x^2 -\frac{1}{2}y^{-2} + \ln|y| &= C \ as \ the \ general \ solution
\end{align*}

7
##### Quiz-1 / Q1: TUT0801
« on: September 27, 2019, 02:07:43 PM »
Find the general solution of the given equation by using variation of parameter:

\begin{align*}
\  y' + \frac{1}{t} y = 3 cos(2t), t > 0 \\
\end{align*}
Solution:
\begin{align*}
\ y = u(t) e^ {-\int\frac{1}{t}dt} = u(t)e^ {-\ln(t)} &= u(t)t^{-1} \\
\\
\ Sub \ y = u(t)t^{-1} \ into \ the \ equation \ to \ get: \\
\ (u(t)t^{-1})' + \frac{1}{t} u(t)t^{-1} &= 3 cos(2t) \\
\ -t^{-2}u(t) + t^{-1}u(t)' + t^{-2}u(t) &= 3 cos(2t) \\
\ t^{-1}u(t)' &= 3 cos(2t) \\
\ u(t)' &= 3tcos(2t) \\
\ u(t) &= \int 3t\ cos(2t)dt \\
\ Using \ Integral \ by \ parts, & \\
\ n = 3t & \ \ \ \ m = \frac{1}{2} sin(2t) \\
\ dn = 3dt & \ \ \ \ dm = cos(2t)dt \\
\ \int 3t \ cos(2t)dt & = nm - \int mdn \\
\ &= \frac{3tsin(2t)}{2} - \int \frac{3}{2} sin(2t)dt \\
\ &= \frac{3tsin(2t)}{2} + \frac{3cos(2t)}{4} + C \\
\ Then \ we \ will \ get: \\
\ u(t) &= \frac{3tsin(2t)}{2} + \frac{3cos(2t)}{4} + C \\
\ y &= u(t)t^{-1} \\
\ &= \frac{3sin(2t)}{2} + \frac{3cos(2t)}{4t} + \frac{C}{t} \ as \ the \ general \ solution. \\
\end{align*}

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