Author Topic: TUT0801 Quiz4  (Read 514 times)

XiaolongZhao

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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: