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### Topics - Xinyu Jing

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1
##### 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 $Y(x)=\frac{-3}{2}x^{1/2}cosx$

2
##### 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

3
##### 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=$𝑐_{1}𝑒^{−𝑡𝑐𝑜𝑠𝑡}+𝑐_{2}𝑒^{−𝑡𝑠𝑖𝑛𝑡}$

4
##### 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)

5
##### 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=$C_{1}𝑒^{(1+\sqrt{3})𝑡}+C_{2}𝑒^{(1−\sqrt{3})𝑡}$

6
##### 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

7
##### 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

8
##### Quiz-1 / TUT 0502 Quiz1
« on: September 27, 2019, 11:19:51 PM »
The answer to my quiz question is attached.

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