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### Topics - XiaolongZhao

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##### Quiz-5 / TUT0801 Quiz5
« on: November 01, 2019, 02:12:20 PM »
Q：Find the general solution of y^''+4y^'+4y=t^(-2) e^(-2t)
A:

Firstly, find its homo solution

r^2 + 4r + 4 = 0

r = -2 (repeated solution)

y_c (t) = c_1 e^(-2t) + c_2 te^(-2t)

y_1 (t) = e^(-2t), y_2 (t) = te^(-2t)

Secondly, find the solution Y(t)

y_1' (t) = -2e^(-2t),  y_2' (t)=e^(-2t) - 2te^(-2t)

W(t) = e^(-4t)

W_1 (t) = -te^(-2t)

W_2 (t) = e^(-2t)

g(t) = t^(-2) e^(-2t)

Y(t) = e^(-2t) ∫(-te^(-2t) t^(-2) e^(-2t))/e^(-4t)  dt + te^(-2t) ∫(e^(-2t) t^(-2) e^(-2t))/e^(-4t)  dt

= -e^(-2t) ln(t) - e^(-2t)

Thus, the general solution is:

y(t) = y_c (t) + Y(t) = c_1 e^(-2t) + c_2 te^(-2t) -e^(-2t) ln(t) - e^(-2t)

The clearer answer is shown in the picture below

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##### Quiz-4 / TUT0801 Quiz4
« on: October 18, 2019, 05:03:08 PM »
Q: Solve t^2 y''+ty'+y=0 for t>0

A:

Let x=ln(t)

Thus, (d^2 y)/(dt^2 )=1/t^2 ((d^2 y)/(dx^2 ) - dy/dx)

dy/dt=1/t ∙ dy/dx

Plug them into the equation:

t^2 ∙ 1/t^2 ∙ ((d^2 y)/(dx^2 ) - dy/dx) + t ∙ 1/t ∙ dy/dx + y = 0

(d^2 y)/(dx^2 ) - dy/dx + dy/dx + y = 0

(d^2 y)/(dx^2 ) + y = 0

y'' (x) + y = 0

Set y''=r^2 , y=1

r^2 + 1 = 0

r = ±i

Thus, y(x) = c_1 e^0  cos(x) + c_2 e^0  sin(x)

y(x)= c_1  cos(x) + c_2  sin(x)

Substitute x = ln(t) in the solution above:

y(t) = c_1  cos(lnt) + c_2  sin(lnt)

Therefore, the general solution is y(t) = c_1  cos(lnt) + c_2  sin(lnt)  for t>0

Clearer answer is shown in the picture below:

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##### Quiz-3 / TUT0801 Quiz3
« on: October 11, 2019, 02:00:02 PM »
Q：Find the solution of 2y''- 3y' + y = 0 with y(0) = 2 and y'(0) = 1/2

A:
Set its characteristic equation as 2r^2 - 3r + 1 = 0

Then, r1 = 1/2 , r2 = 1 , r1 ≠ r2

Thus, the solution is y(t) = C1 e^(t/2) + C2 e^t

y'(t) = (C1 /2) e^(t/2) + C2 e^t

Plug in the initial values: y(0) = 2 = C1 + C2

y'(0) = 1/2 = (C1 /2) + C2

C1 = 3 , C2 = -1
Therefore, the final solution is: y(t) = 3e^(t/2) - e^t

Clearer answer shows in the picture below

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##### Quiz-2 / TUT0801 Quiz2
« on: October 04, 2019, 02:00:05 PM »
Q：Prove the equation is not exact, and becomes exact after multiplying the integrating factor, and then solve the equation:
x^2 y^3+x(1+y^2 ) y'=0   μ(x,y)=1/xy^3
A:
Step1: set M(x,y) and N(x,y), then get M_y and N_x to prove it is not exact
x^2 y^3 dx+x(1+y^2 )dy=0
Set M=x^2 y^3,N=x+xy^2
M_y=3x^2 y^2,N_x=1+y^2
Since M_y≠N_x, I’ve proven it is not exact

Step2: Multiply the integrating factor to both sides of the equation, and then get new M, N, M_y, and N_x, to prove it becomes exact
xdx+(1+y^2)/y^3  dy=0
Set M'=x, N'=(1+y^2)/y^3
M'_y=0, N'_x=0
Since M'_y=N'_x , I’ve proven it becomes exact

Step3: Solve the equation
Since the equation is exact, ∃φ(x,y)  s.t.φ_x=M',φ_y=N'
φ=∫M'dx=∫xdx=x^2/2+h(y)
φ_y=0+h' (y)=N'=(1+y^2)/y^3
h' (y)=(1+y^2)/y^3
h(y)=∫〖[(1+y^2)/y^3]dy〗=∫〖(1/y^3 )dy〗+∫〖(1/y)dy〗= -y^(-2)/2+ln|y|+C
φ=x^2/2 - y^(-2)/2 + ln|y|+C
Thus, x^2/2 - y^(-2)/2 + ln|y|=C is the solution of the differential equation

The clearer answer is shown in the picture below.

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##### Quiz-1 / TUT0801 Quiz1
« on: September 27, 2019, 02:59:40 PM »
Q: t^3 y'+4t^2 y=e^(-t),t<0
y(-1)=0
Find the general solution
A:see the picture below

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##### Quiz-1 / TUT0801 Quiz1
« on: September 27, 2019, 02:40:21 PM »
Q: t^3 y'+4t^2 y=e^(-t),t<0
y(-1)=0
Find the general solution

A:
Step1: let the coefficient of y' be '1'
y'+4/t y=e^(-t)/t^3

Step2: Set u = e^∫〖p(t) dt〗, where p(t) is the coefficient of y
Set u=e^∫〖4/t dt〗=e^(4ln|t|))=e^(4ln(-t))=e^(ln((-t)^4))=e^(ln(t^4))=t^4 (since t<0)

Step3: Multiply u to both sides
t^4 y'+4t^3 y=te^(-t)

(t^4 y)'=te^(-t)

Step4: Set integral to both sides
∫〖(t^4 y)' dt〗=∫〖te^(-t) dt〗

For ∫〖te^(-t) dt〗, use ‘by parts’ to solve it:
u=t,  du=1*dt,
dv=e^(-t)dt,   v= -e^(-t)
∫〖te^(-t) dt〗=-te^(-t)-∫〖-e^(-t) dt〗=-te^(-t)-e^(-t)+C

t^4 y= -te^(-t)-e^(-t)+C

y= -e^(-t)/t^3 -e^(-t)/t^4 +C/t^4

Step5: Plug in y(-1)=0

0=e-e+C, thus, C=0

Thus, y= -e^(-t)/t^3 -e^(-t)/t^4

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