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### Messages - taojinwe

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##### Quiz-5 / TUT0801 QUIZ5
« on: November 01, 2019, 01:17:17 PM »
Question: Find the solution of  y''+9y=9sec2(3t) 0<t< 𝞹/6

We first find the homogenous solution of y''+9y=0
r2+9=0
Then r1=3i and r2=-3i

Yc(t)=C1y1(t)+C2y2(t)
=C1cos3t+C2sin3t

W=y1 × y2' - y2 × y1' = cos3t×cos3t-sin3t×(-sin3t)=3

Let Yp(t)= 𝝁1y1+ 𝝁2y2

𝝁1=-∫ (sin3t × 9sec2(3t))/3 dt
=- ∫3sin3t × [1/cos2(3t)] dt
=-3 ∫(sec3t × tan3t)dt
= -sec3t
Therefore 𝝁1 = -sec3t

𝝁2 =∫(cos3t × 9sec2(3t))/3 dt
=∫3 cost3t × (1/cos2(3t)dt
= ln⎮sec3t+tant3t⎮
Therefore 𝝁2 = ln⎮sec3t+tan3t⎮

Yp(t)= 𝝁1y1+ 𝝁2y2
=cos3t(-sec3t)+sin3t × ln⎮sec3t+tan3t⎮

Y(t)=Yc(t)+Yp(t)
=C1cos3t+C2sin3t+cos3t(-sec3t)+sin3t ln⎮sec3t+tan3t⎮

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##### Quiz-4 / TUT0801 QUIZ4
« on: October 18, 2019, 10:11:18 PM »
Solve the initial value problem
y’’ - 6y’ + 9y = 0
y(0) = 0
y’(0) =2

Convert the problem into the form r2 - 6r + 9 = 0
(r-3)(r-3)=0
r1 = r2 = 3
Since r1 = r2 , we use the formula
y(t) = C1e3t + C2te3t

Then we plug in the value y(0) = 0 and y’(0) =2
y(0) = C1e3t + C2te3t = 0
= C1e0 + C20e0
C1= 0
y’(t) = 3C1e3t + 3C2te3t + C2e3t
y’(0) = 3C1e0 + 3C20e0 + C2e0 = 2

3C1 + C2 = 2
Since  C1 = 0
Then C2 = 2

Therefore, the solution for this initial value problem is y(t) = 2te3t

3
##### Quiz-3 / TUT0801 QUIZ3
« on: October 11, 2019, 03:38:48 PM »
Question: Find the solution of 2y''- 3y' + y = 0 with y(0) = 2 and y'(0) = 1/2

Convert the original equation into
2r2 - 3r + 1 = 0

Then solve the equation as
(r-0.5) (r-1) = 0

Therefore,  r1 = 0.5 , r2 = 1

Since r1 ≠ r2

We use the formula y(t) = C1 er1t + C2 er2t
= C1 e0.5t + C2 et

y'(t) = (C1 /2) e0.5t + C2 et

Then we put the initial values: y(0) = 2 = C1 + C2

y'(0) = 0.5 = (C1 /2) + C2

C1 = 3 , C2 = -1
Thus, the final solution is: y(t) = 3e0.5t - et

4
##### Quiz-2 / TUT 0801 QUIZ2
« on: October 04, 2019, 04:23:42 PM »
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

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##### Quiz-1 / TUT0801 QUIZ1
« on: September 27, 2019, 04:39:40 PM »
Solution

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