Consider the initial value problem x' = tx + x^3 , x(0) = (alpha), where (alpha) is a parameter.

(a) Prove that for = 1, the solution x(t) grows to infinity (explodes) somewhere on the interval (0, 1) completing the following steps

i. Find the solution of initial value problem y' = y^2 , y(0) = 1. We denote it y(t) and use in the next steps.

ii. Compute x'(0), x''(0), x'''(0), y'(0), y''(0), y'''(0) using differential equations.

iii. Conclude that for small t > 0 we have x(t) > y(t) using Taylor formula.

iv. Show that y(t) > t on the interval 0 < t < 1.

v. Show that on the interval 0 < t < 1 if x(t) > y(t), then x(t)(x^2 (t) t) > y(t)(y^2 (t) t).

vi. Show that y(t)(y^2 (t) t) > y^2 (t) on the interval 0 < t < 1.

vii. Conclude that if x(t) > y(t), then x'(t) > y'(t).

JUST DO from iii and on

viii. Conclude that while x(t) is defined, it is greater than y(t). Therefore it explodes to infinity at t = 1 or earlier.