Question

How do you solve the initial value problem of `t^2y'' - 4ty' + 4y = -2t^2` given `y(1) = 2` and `y'(1) =0`?

Answer 1

Answer : `y : t \mapsto 2t - t^4 + t^2`

1) First, solve `t^2y''-4ty' + 4y = 0`.

• Consider `y(t) = t^k` where `k` is real. It's a solution if and only if
  `t^2k(k-1)t^(k-2) - 4tkt^(k-1) + 4t^k = 0`
  or if and only if `k^2 - 5k + 4 = 0`. The solutions are 1 et 4.

• Therefore, all the solutions of `t^2y''-4ty' + 4y = 0` are

`t \mapsto \lambda t + \mu t^4`

where `lambda` and `mu` are arbitrary real constants.

2) Second, find a particular solution of `t^2y''-4ty' + 4y = -2t^2`.

• Try a function like `y(t) = mt^k`. You find easily that `m=1` and `k=2` are ok.

• Therefore all the solutions of `t^2y''-4ty' + 4y = -2t^2` are

`y : t \mapsto \lambda t + \mu t^4 + t^2`

where `lambda` and `mu` are arbitrary real constants.

3) Finally, you want to have `y(1)=2` and `y'(1)=0`.

• Because `y(1) = \lambda + mu + 1`, you have #lambda + mu = & 1#

• Because `y'(t) = lambda + 4mu t^3 + 2t`, you have `lambda + 4 mu + 2 = 0`.

• You solve the system above. You get `mu = -1` and `lambda = 2`.

Conclusion The unique solution is `t \mapsto 2t^2 - t^4 + t^2`.