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### Messages - Xinyu Jing

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1
##### Term Test 2 / Re: Problem 2 (main sitting)
« on: November 19, 2019, 06:30:34 AM »
(𝑎)
𝑊=$𝑐𝑒^{−∫𝑝(𝑡)𝑑𝑡}$=$𝑐𝑒^{−∫4𝑑𝑡}$=$𝑐𝑒^{−4𝑡}$
(𝑏)
$𝑟^{3}+4𝑟^{2}+𝑟−6$=0
(𝑟−1)($𝑟^{2}$+5𝑟+6)=0
𝑟=1,𝑟=−2,𝑟=−3
∴𝑦(𝑡)=$𝑐_{1}𝑒^{𝑡}+𝑐_{2}𝑒^{−2𝑡}+𝑐_{3}𝑒^{−3𝑡}$

W=\begin{vmatrix}
e^{t} & e^{-2t} &  e^{-3t} \\
e^{t} & -2e^{-2t} & -3e^{-3t} \\
e^{t} & 4e^{-2t} & 9e^{-3t} \\
\end{vmatrix}=−12𝑒−4𝑡
∴𝑐=−12 compare with (a)
(𝑐)
𝑦𝑝(𝑡)=$𝐴𝑡𝑒^{𝑡}$
𝑦′𝑝(𝑡)=$𝐴𝑒^{𝑡}+𝐴𝑡𝑒^{𝑡}$
𝑦′′𝑝(𝑡)=$2𝐴𝑒^{𝑡}+𝐴𝑡𝑒^{𝑡}$
𝑦′′′𝑝(𝑡)=$3𝐴𝑒^{𝑡}+𝐴𝑡𝑒^{𝑡}$
$3𝐴𝑒^{𝑡}+𝐴𝑡𝑒^{𝑡}+8𝐴𝑒^{𝑡}+4𝐴𝑡𝑒^{𝑡}+𝐴𝑒^{𝑡}+𝐴𝑡𝑒^{𝑡}−6𝐴𝑡𝑒^{𝑡}=12𝐴𝑒^{𝑡}$
12𝐴=24
𝐴=2
∴𝑦(𝑡)=$𝑐_{1}𝑒^{𝑡}+𝑐_{2}𝑒^{−2𝑡}+𝑐_{3}𝑒^{−3𝑡}+2𝑡𝑒^{𝑡}$

2
##### Quiz-5 / LEC0101 QUIZ5
« on: October 31, 2019, 08:35:16 PM »
Given
$x^{2}y''+xy'+(x^{2}-0.25)y=3x^{3/2}sinx$,x>0;
$y_{1}(x)=x^{-1/2}sinx$, $y_{2}(x)=x^{-1/2}cosx$
Step 1
The equation is written in standard form as:
$y''+\frac{1}{x}y'+\frac{(x^{2}-0.25)}{x}y=3x^{-1/2}sinx$
$g(x)=3x^{-1/2}sinx$
Now further, Wronskian is evaluated as:
W($x^{-1/2}sinx$,$x^{-1/2}cosx$)=$\left | {x^{-1/2}sinx \qquad \qquad \qquad \qquad \qquad \qquad x^{-1/2}cosx} \right |$
$\left | \frac{-1}{2}x^{-1/2}sinx+x^{-1/2}cosx \qquad \frac{-1}{2}x^{-3/2}cosx-x^{-1/2}sinx \right |$
=$\frac{-1}{2}x^{-2}sinxcosx-x^{-1}sin^{2}x+\frac{1}{2}x^{-2}sinxcos-x^{-1}cos^{2}x$
=$-x^{-1}(sin^{2}x+cos^{2}x)$=$-x^{-1}$
Step 2
The parameters U1 and U2 are evaluated as:
$u_{t}=-\int{\frac{y_{2}g(x)}{W(y_{1},y_{2})}}dx$
$u_{t}=-\int{\frac{x^{-1/2}cosx*3x^{-1/2}sinx}{-x^{-1}}dx}$
$u_{t}=\int{3cosxsinxdx}$
$u_{t}=\frac{-3}{2}cos^{2}x$
$u_{2}=-\int{\frac{y_{1}g(x)}{W(y_{1},y_{2})}dx}$
$u_{2}=-\int{\frac{x^{-1/2}sinx*3x^{-1/2}sinx}{-x^{-1}dx}}dx$
$u_{2}=\int3sinxcosx-\frac{3}{2}x$
STEP 3
Futher,
Y(x)=$y_{1}u_{1}$+$y_{2}u_{2}$
Y(x)=$x^{-1/2}sinx*\frac{-3}{2}cos^{2}x+x^{-1/2}cosx(\frac{3}{2}sinxcosx-\frac{3}{2}x)$
Y(x)=$\frac{-3}{2}x^{-1/2}cos^{2}xsinx+\frac{3}{2}x^{-1/2}cos^{2}xsinx-\frac{3}{2}x^{1/2}cosx$
Y(x)=$\frac{-3}{2}x^{1/2}cosx$
Hence, the solution is $Y(x)=\frac{-3}{2}x^{1/2}cosx$

3
##### Term Test 1 / Re: Problem 1 (main sitting)
« on: October 24, 2019, 01:08:06 PM »
𝑀𝑦=$1+6𝑦𝑒^{3𝑥}$

𝑁𝑥=$4𝑦𝑒^{2𝑥}$

𝑀𝑦≠𝑁𝑥,it is not exact

$𝑅_{2}$=(𝑀𝑦−𝑁𝑥)/𝑁=$\frac{1+2𝑦𝑒^{2𝑥}}{1+2𝑦𝑒^{2𝑥}}$=1

μ=$𝑒∫𝑅_{2}𝑑𝑥$=𝑒∫1𝑑𝑥=$𝑒^{𝑥}$

Multiplying both sides by 𝜇, we get

$𝑦𝑒^{𝑥}+3𝑦_{2}𝑒^{3𝑥}+(𝑒𝑥+2𝑦𝑒^{3𝑥})𝑦′=0$

$𝑀′𝑦=𝑒𝑥+6𝑦𝑒^{3𝑥}$

$𝑁′𝑥=𝑒𝑥+6𝑦𝑒^{3𝑥}$

𝑀′𝑦=𝑁′𝑥,it is exact

∃φ(𝑥,𝑦)𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 φ𝑥=𝑀′,φ𝑦=𝑁′

φ(𝑥,𝑦)=∫𝑀′𝑑𝑥=$∫𝑦𝑒^{𝑥}+3𝑦_{2}𝑒^{3𝑥}𝑑𝑥=𝑦𝑒^{𝑥}+𝑦_{2}𝑒^{3𝑥}+ℎ(𝑦)$

$φ𝑦=𝑒^{𝑥}+2𝑦𝑒^{3𝑥}+ℎ(𝑦)′=𝑒^{𝑥}+2𝑦𝑒^{3𝑥}$

Then ℎ(𝑦)′=0

Hence h(y)=c

$φ(𝑥,𝑦)=𝑦𝑒^{𝑥}+𝑦_{2}𝑒^{3𝑥}=𝑐$

Since y(0)=1

$1⋅𝑒^{0}+12⋅𝑒^{0}=2=𝑐$

$φ(𝑥,𝑦)=𝑦𝑒^{𝑥}+𝑦_{2}𝑒^{3𝑥}=2$

4
##### Term Test 1 / Re: Problem 3 (morning)
« on: October 24, 2019, 12:45:24 PM »
𝑦′′−6𝑦′+8𝑦=48sinh(2𝑥)

𝑦′′−6𝑦′+8𝑦=$24𝑒^{2𝑥}−24𝑒^{−2𝑥}$

Let $𝑦=𝑒^{𝑟𝑥(2)}$

𝑦′=sec(𝑟𝑥) 𝑦′′=$𝑟^{2}𝑒^{𝑟𝑥(3)}$

$r^{2}−r𝑥+8=0$

$𝑟_{1}=2, 𝑟_{2}=4$

$𝑦=𝑐_{1}𝑒^{4𝑥}+𝑐_{2}𝑒^{2x}$

let 𝑦=$𝐴𝑥𝑒^{2𝑥}$

𝑦′=$𝐴𝑒^{2𝑥}+2𝐴𝑥𝑒^{2𝑥}$

𝑦″=$4𝐴𝑒^{2𝑥}𝑥+2𝐴𝑒^{2𝑥}+2𝐴𝑒^{2𝑥}=4𝐴𝑒^{2𝑥}𝑥+4𝐴𝑒^{2𝑥}$

$(8𝐴−12𝐴+4𝐴)𝑡⋅𝑒^{2𝑥}+(−6𝐴+4𝐴)𝑒^{2𝑥}=24𝑒^{2𝑥}−2𝐴𝑒^{2𝑥}=24𝑒^{2𝑥}$

𝐴=−12 $𝑦=−12𝑥𝑒^{2𝑥}$

Let 𝑦=$B𝑒^{−2𝑥}$

𝑦′=$−2B𝑒^{−2𝑥}$

𝑦′′=$4B𝑒^{−2𝑥}$

$4𝐵𝑒^{−2𝑥}+12𝐵𝑒^{−2𝑥}+8𝐵𝑒^{−2𝑥}=−24𝑒^{−2𝑥}$

$24𝐵𝑒^{−2𝑥}=−24𝑒^{−2𝑥}$

𝐵=−1

∴$𝑦𝑝(𝑥)=−𝑒^{−2𝑥}$

so
$𝑦=𝑎𝑒^{2}+𝑐_{2}𝑒^{4𝑥}−12𝑒^{2𝑥}−𝑒^{−2𝑥}$

$𝑦′=2𝑐_{1}𝑒^{2𝑡}+4𝑐_{2}𝑒^{4𝑡}−24𝑟𝑒^{2𝑥}+2𝑒^{−2𝑥}−12𝑒^{2𝑥}$

𝑦(0)=𝑦′(0)=0

$𝑦=−3𝑒^{2𝑥}+4𝑒^{4𝑥}−12𝑥𝑒^{2𝑥}−𝑒^{−2𝑥}$

5
##### Quiz-4 / QUIZ4 TUT 0502
« on: October 19, 2019, 10:47:18 AM »
Question: 1 + [x/y - sin(y)]y’ = 0

We firstly simplify the equation into
dx + [x/y - sin(y)] dy= 0

Then we have M and N
M(x,y) = 1
N(x,y) = [x/y - sin(y)]

Then, we find the derivative of M with respect to y and N with respect to x
My = 0
Nx = 1/y

Since My is not equal to Nx, it is not exact.
Thus, we need to multiply a factor 𝓾 that satisfies the equation

R1 = [ (My - Nx)/ M] = [(0-1/y)] = -1/y
𝓾 = e-∫R1dy = e-∫(-1/y)dy = elny = y

We multiply the 𝓾 on both sides of the equation to find an exact equation
𝓾 dx + 𝓾[x/y - sin(y)] dy= 0
y dx + y [x/y - sin(y)] dy= 0

Then we have our new M’ and N’

M’(x,y) = y
N’(x,y) = y[x/y - sin(y)] = x - ysin(y)

Thus, there exist a function 𝒞(x,y) such that
𝒞x = M’
𝒞y = N’
By Integrating M’ with respect to x
𝒞x = M’
𝒞 = ∫ M’ dx  =  ∫ y dx = xy + h(y)

By differentiating with respect to y and equating to 𝒞y = N’
We get x + h’(y) = x - ysin(y)
Therefore, h’(y) = - ysin(y)

By integrating on both sides
h(y) =∫ - ysin(y) dy = ycos(y) - sin(y)
Now, we have
𝒞 = xy + ycos(y) - sin(y)
Thus, the solutions of differential equation are given implicitly by
xy + ycos(y) - sin(y) = C

6
##### Quiz-4 / QUIZ4 TUT 0502
« on: October 19, 2019, 01:23:47 AM »
Find the general solution of the differential equation
𝑦″+2𝑦′+2𝑦=0

The characteristic equation of the given equation is:

$𝑟^{2}$+2𝑟+2=0

𝑟=$\frac{−𝑏±\sqrt{𝑏^{2}−4𝑎𝑐}}{2𝑎}$=$\frac{−2±\sqrt{-4}}{2}$=−1±𝑖

Then,
$𝑟_{1}$=−1+𝑖 $𝑟_{2}$=−1−𝑖

Therefore, the general solution of the given differential equation is:
𝑦=$𝑐_{1}𝑒^{−𝑡𝑐𝑜𝑠𝑡}+𝑐_{2}𝑒^{−𝑡𝑠𝑖𝑛𝑡}$

7
##### Quiz-3 / QUIZ3 TUT 0502
« on: October 12, 2019, 12:20:07 AM »
Question: 𝑐𝑜𝑠(𝑡)𝑦″+𝑠𝑖𝑛(𝑡)𝑦′−𝑡𝑦=0
Find the Wronskian of two solutions of the given differential equation without solving the equation.

Solution:
Divide both sides by 𝑐𝑜𝑠(𝑡)
𝑦″+𝑡𝑎𝑛(𝑡)𝑦′−𝑡𝑐𝑜𝑠(𝑡)𝑦=0
𝑊(𝑦1,𝑦2)(𝑡)=𝑐𝑒−∫𝑝(𝑡)𝑑𝑡
𝑊(𝑦1,𝑦2)(𝑡)=𝑐𝑒−∫𝑡𝑎𝑛(𝑡)𝑑𝑡=𝑐𝑒−(−𝑙𝑛|𝑐𝑜𝑠(𝑡)|)
𝑊(𝑦1,𝑦2)(𝑡)=𝑐𝑒𝑙𝑛|𝑐𝑜𝑠(𝑡)|=𝑐𝑐𝑜𝑠(𝑡)

Therefore, the Wronskian of any pair of solutions of the given equation is 𝑊(𝑦1,𝑦2)(𝑡)=𝑐𝑐𝑜𝑠(𝑡)

8
##### Quiz-3 / QUIZ3 TUT 0502
« on: October 11, 2019, 08:06:25 PM »
Question:Find the general solution of the given differential equation.
𝑦″−2𝑦′−2𝑦=0

Solution:
𝑦=$𝑒^{𝑟𝑡}$
and it follows that r must be a root of characteristic equation
$𝑟^{2}$−2𝑟−2=0

𝑟=$\frac{−𝑏±\sqrt{𝑏^{2}−4𝑎𝑐}}{2𝑎}$

$𝑟_{1}$=1+\sqrt{3} $𝑟_{2}$=1−\sqrt{3}

Therefore, the general solution of the given differential equation is:
𝑦=$C_{1}𝑒^{(1+\sqrt{3})𝑡}+C_{2}𝑒^{(1−\sqrt{3})𝑡}$

9
##### Quiz-2 / QUIZ2 TUT 0204
« on: October 07, 2019, 09:26:17 AM »
Question: (𝑥+2)𝑠𝑖𝑛(𝑦)+𝑥𝑐𝑜𝑠(𝑦)𝑦′=0   𝑢=𝑥𝑒𝑥
Solution:
𝑀=(𝑥+2)𝑠𝑖𝑛(𝑦)    𝑁=𝑥𝑐𝑜𝑠(𝑦)
𝑀𝑦=(𝑥+2)𝑐𝑜𝑠(𝑦)     𝑁𝑥=𝑐𝑜𝑠(𝑦)
therefore 𝑀𝑦≠𝑁𝑥 , therefore the equation is not exact.
multiplies the given integrating factor
(𝑥+2)𝑥𝑒𝑥𝑠𝑖𝑛(𝑦)+𝑥2𝑒𝑥𝑐𝑜𝑠(𝑦)𝑦′=0
𝑀=(𝑥+2)𝑥𝑒𝑥𝑠𝑖𝑛(𝑦)  𝑁=𝑥2𝑒𝑥𝑐𝑜𝑠(𝑦)
𝑀𝑦=(𝑥+2)𝑥𝑒𝑥𝑐𝑜𝑠(𝑦)  𝑁𝑥=𝑥2𝑒𝑥𝑐𝑜𝑠(𝑦) + 2𝑥𝑒𝑥𝑐𝑜𝑠(𝑦)
then 𝑀𝑦=𝑁𝑥,  therefore it becomes exact.
𝜙(𝑥,𝑦) s.t 𝜙𝑥=𝑀   𝜙𝑦=𝑁
𝜙=∫𝑥2𝑒𝑥𝑐𝑜𝑠(𝑦)𝑑𝑦
𝜙=𝑥2𝑒𝑥𝑠𝑖𝑛(𝑦)+ℎ(𝑥)
𝜙𝑥=𝑥2𝑒𝑥𝑠𝑖𝑛(𝑦) + 2𝑥𝑒𝑥𝑠𝑖𝑛(𝑦)+ℎ′(𝑥)=𝑀=(𝑥+2)𝑥𝑒𝑥𝑠𝑖𝑛(𝑦)+0
ℎ′(𝑥)=0
ℎ(𝑥)=𝑐
𝜙(𝑥,𝑦)=𝑥2𝑒𝑥𝑠𝑖𝑛(𝑦)+ℎ(𝑥)=𝑐

10
##### Quiz-2 / QUIZ2 TUT 0502
« on: October 07, 2019, 12:12:42 AM »
Question: (3𝑥+6𝑦)+(𝑥2𝑦+3𝑦𝑥)𝑑𝑦𝑑𝑥=0

Solution: We want to find an integrating factor 𝜇 as a function of 𝑥𝑦 such that
(𝜇𝑀)𝑦=(𝜇𝑁)𝑥, Let 𝑧=𝑥𝑦. Thus, 𝜇(𝑥𝑦)=𝜇(𝑧(𝑥,𝑦)) Then

𝜇𝑥(𝑥𝑦)=𝑑𝜇𝑑𝑧∂𝑧∂𝑥=𝑦𝑑𝜇𝑑𝑧
𝜇𝑦(𝑥𝑦)=𝑑𝜇𝑑𝑧∂𝑧∂𝑦=𝑥𝑑𝜇𝑑𝑧

Therefore,
(𝜇𝑀)𝑦=(𝜇𝑁)𝑥

𝜇𝑀𝑦+𝑥𝑀𝑑𝜇𝑑𝑧=𝜇𝑁𝑥+𝑦𝑁𝑑𝜇𝑑𝑧

𝜇(𝑀𝑦−𝑁𝑥)=𝑑𝜇𝑑𝑧(𝑦𝑁−𝑥𝑀)

d𝜇d𝑧=𝜇(𝑁𝑥−𝑀𝑦𝑥𝑀−𝑦𝑁)

Therefore,

𝜇(𝑧)=exp(∫𝑅(𝑧)d𝑧)
\quad where 𝑅(𝑧)=𝑅(𝑥𝑦)=𝑁𝑥−𝑀𝑦𝑥𝑀−𝑦𝑁

𝑀(𝑥,𝑦)=3𝑥+𝑦 \quad and \quad 𝑁(𝑥,𝑦)=𝑥2𝑦+3𝑦𝑥=0

Then

∂∂𝑦𝑀(𝑥,𝑦)=−6𝑦2 \quad and \quad ∂∂𝑥𝑁(𝑥,𝑦)=2𝑥𝑦−3𝑦𝑥2

𝑁𝑥−𝑀𝑦𝑥𝑀−𝑦𝑁=2𝑥𝑦−3𝑦𝑥2+6𝑦2𝑥(3𝑥+6𝑦)−𝑦(𝑥2𝑦+3𝑦𝑥)
=2𝑥𝑦−3𝑦𝑥2+6𝑦22𝑥2+6𝑥𝑦−3𝑦2𝑥
=2𝑥𝑦−3𝑦𝑥2+6𝑦2𝑥𝑦(2𝑥𝑦−3𝑦𝑥2+6𝑦2)=1𝑥𝑦

Let 𝑥𝑦=𝑧

𝜇(𝑥𝑦)=exp(∫1𝑧d𝑧)=𝑒log|𝑧|=𝑧=𝑥𝑦

(3𝑥2𝑦+6𝑥)+(𝑥3+3𝑦2)𝑑𝑦𝑑𝑥=0

∂∂𝑦(3𝑥2𝑦+6𝑥)=3𝑥2=∂∂𝑥(𝑥3+3𝑦2)

𝜓𝑥(𝑥,𝑦)=3𝑥2𝑦+6𝑥(1)

𝜓𝑦(𝑥,𝑦)=𝑥3+3𝑦2(2)

𝜓(𝑥,𝑦)=𝑥3𝑦+3𝑥2+ℎ(𝑦)

𝜓𝑦(𝑥,𝑦)=𝑥3+ℎ′(𝑦)

Therefore,
ℎ′(𝑦)=3𝑦2

ℎ(𝑦)=𝑦3

𝜓(𝑥,𝑦)=𝑥3𝑦+3𝑥2+𝑦3

𝑥3𝑦+3𝑥2+𝑦3=𝐶

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