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Messages - XiaolongZhao

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1
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

2
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:

3
Quiz-3 / Re: TUT0801 Quiz3
« on: October 11, 2019, 02:00:56 PM »
Clearer Answer

4
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

5
Quiz-2 / Re: TUT0801 Quiz2
« on: October 04, 2019, 02:12:38 PM »
I think you should prove the original equation is not exact before multiplying the integrating factor

6
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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