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### Messages - Nan Yang

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1
##### Term Test 1 / Re: Problem 2 (noon)
« on: October 23, 2019, 12:08:59 PM »
Thanks for your advise. For showing a solution, I need write more details.  BUT I think it is very clear to see that $y_2$ is $sin(x)$  and it can save time for us in writing a test.

2
##### Term Test 1 / Re: Problem 2 (noon)
« on: October 23, 2019, 08:51:47 AM »
$$\text{emm just parts of solution？？}$$
Thanks for waiting. Now I finish all solutions

3
##### Term Test 1 / Re: Problem 2 (noon)
« on: October 23, 2019, 08:05:41 AM »
Solution
a)
$y'' + \frac{xsin(x)}{xcos(x)-sin(x)}y'- \frac{sin(x)}{xcos(x)-sin(x)}y = 0$

$p(x) = \frac{xsin(x)}{xcos(x)-sin(x)}$

$w = ce^{-\int p(x)dx} =ce^{-\int \frac{xsin(x)}{xcos(x)-sin(x)}}$

let $u = xcos(x)-sin(x), du =-xsin(x)$

$w = ce^{- \int \frac{-1}{u} du} = ce^{ \int \frac{1}{u} du} = ce^{lnu} = ce^{ln(xcos(x)-sin(x))} = c(xcos(x)-sin(x))$

let $c = 1 , w = xcos(x)-sin(x)$

b)
check $y_1 =x$ is a solution.

$y_1' = 1, y_2'' = 0$

substitute them into equation,

we get

$xsin(x)- sin(x)x = 0$

so x is a solution

w = $\begin{vmatrix} x& y_2 \\ 1 & y_2' \end{vmatrix}$

$xy_2' - y_2 = xcos(x)-sin(x)$

so $y_2 = sinx$  OK. V.I.

c)

$y(t) = c_1x + c_2 sinx$

since $y(π) = π, y'(π) = 0$

$π = c_1 π + c_2sin(π)$

$π = c_1π$

so $c_1 = 1$

$y'(t) = c_1 + c_2 cos(x)$

$π = 1 - c2$

$c_2 = 1 -π$ Wrong.

$y(t) = x + (1-π) sinx$

4
##### Term Test 1 / Re: Problem 3 (afternoon)
« on: October 23, 2019, 08:04:06 AM »
Question: y'' - 2y' - 3y = 16coshx
y(0) = 0 y'(0) = 0
Solution
Since $coshy = \frac{e^x + e^{-x}}{2}$
Then $y'' - 2y' - 3y = 16\frac{e^x - e^{-x}}{2}$ = $8e^x + 8 e^{-x}$
$y'' - 2y' - 3y = 0$
$r^3 - 2r - 3 = 0$
$r = 3 , -1$
$y_c = c_1 e^{3x} + c_2 e^{-x}$

$y'' -2y' -3y = 8e^x$.
Let $Y_1= Ae^x$
$Y_1'= Ae^x$
$Y_1''= Ae^x$
$A -2A -3A = 8$
$A = -2$
$Y_1= -2e^x$

$y'' -2y' -3y = 8e^{-x}$.
Let $Y_1= Ae^{-x} \cdot x$
Then $Y_1'= Ae^{-x} - Ae^{-x} \cdot x$
Then $Y_1''= -Ae^{-x} - Ae^{-x} + Ae^{-x}x$
$-2A -2A = 8$   $A = -2$
$Y_2= -2e^{-x}x$

$Y = -2e^x -2e^{-x}x$

y(t) = $c_1 e^{3x} + c_2 e^{-x} -2e^x -2e^{-x}x$
since $y(0) = 0, y'(0) = 0$
$c_1 + c_2 -2 = 0$
$3c_1 -c_2 -2 -2= 0$
Then $c_2 =\frac{1}{2}$ $c_1 = \frac{3}{2}$
$y(t) = \frac{3}{2} e^{3x} + \frac{1}{2} e^{-x} -2e^x -2e^{-x}x$

5
##### Quiz-3 / Re: TUT 0501 Quiz 3
« on: October 11, 2019, 04:11:12 PM »
we got the same question and I type the solution of this question.
Question: find the general solution of the given differential equation
y'' $-$ 2y' $-$ 2y = 0

Solution:
Let y'' $-$ 2y' $-$ 2y = 0 be equation (1)
we assume that $y= e^{rt}$ is solution of (1)
Then we have:
$y= e^{rt}$
$y'= re^{rt}$
$y''= r^2e^{rt}$
we substitute them into equation (1),
we have $r^2e^{rt}$ $-$ 2$re^{rt}$ $-$ 2$e^{rt}$ = 0 ,
$e^{rt}$$(r^2 - 2r -2) = 0 since e^{rt} is not zero so (r^2 - 2r -2) = 0 we can get r = \frac{2 \pm \sqrt{4 +8}}{2} simplify it we get r = \frac{2 \pm 2\sqrt{3}}{2} Then r = 1+ \sqrt{3} r = 1- \sqrt{3} Then two roots of equation (1) is e^{(1+ \sqrt{3})t} and e^{(1- \sqrt{3})t} Therefore, the general solution is y =c_1 e^{(1+ \sqrt{3})t} + c_2 e^{(1- \sqrt{3})t} 6 ##### Quiz-3 / TUT0201 Quiz3 « on: October 11, 2019, 04:09:55 PM » Question: find the general solution of the given differential equation y'' - 2y' - 2y = 0 Solution: Let y'' - 2y' - 2y = 0 be equation (1) we assume that y= e^{rt} is solution of (1) Then we have: y= e^{rt} y'= re^{rt} y''= r^2e^{rt} we substitute them into equation (1), we have r^2e^{rt} - 2re^{rt} - 2e^{rt} = 0 , e^{rt}$$(r^2 - 2r -2) = 0$
since $e^{rt}$ is not zero
so $(r^2 - 2r -2) = 0$
we can get r = $\frac{2 \pm \sqrt{4 +8}}{2}$
simplify it we get  r = $\frac{2 \pm 2\sqrt{3}}{2}$
Then r = $1+ \sqrt{3}$ r = $1- \sqrt{3}$
Then two roots of equation (1) is  $e^{(1+ \sqrt{3})t}$ and $e^{(1- \sqrt{3})t}$
Therefore, the general solution is  y =$c_1$ $e^{(1+ \sqrt{3})t}$ + $c_2$ $e^{(1- \sqrt{3})t}$

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