Find a second solution?

Question

Use the method of reduction of order to find a second solution of t²y"-4ty'+6y=0, t>0; y₁(t)=t².

Answer: y₂(t)=t³

 

Answer 1

You can notice that a partial solution of this equation can be written in the form

                                    y =  C t^k (C is  an arbitrary constant )                  (1)

If you substitute this solution into your equation you will come up with the follwoing eqation for "k":

                              k(k-1) - 4k +6 = 0           or

                                 k² - 6k + 6 = 0     or  (k-2)(k-3) = 0            (2)

This eqation gives you two possible solutions: k =2 and k = 3. Thus, generally speaking,

                         y₁(t) = C₁t²     and               y₂(t) = C₂t³                (3)             

To reduce the order of the equation you can put y₂(t) = y₁(t) v(t) = t²v(t)

If you plug this expression into the equation and simplify (I am skipping these steps because you can do it yourself) you will come up with the equation

               t v''(t) = 0    or   v"(t) = 0

Hence it follows v'(t) = const (say A₁) and v(t) = A₁ t + A₂       (in general)

Since y₁(t) = t²  you have to put    A₁ = 1 and A₂ = 0, Thus, your second solution is

                                                  y₂(t) = t³

Answer 2

Solve for y'',

y'' = (4/t)y' - (6/t^2)y

W = y1y2' - y1'y2

W' = y1y2'' - y1''y2

= y1{(4/t)y2' - (6/t^2)y2} - {(4/t)y1' - (6/t^2)y1}y2

= (4/t)(y1y2' - y1'y2)

= (4/t)W

Separate variables,

dW/W = (4/t)dt

Integrate,

lnW = 4lnt + c

W = Ct^4 = t^2 y2' - 2t y2

y2' - 2y2/t = Ct^2

y2 = (t^2)[∫(1/t^2) Ct^2 dt + D] = Ct^3 + Dt^2

Since y1 = t^2 is a solution, y2 = t^3 is another solution.