Consider the initial value problem x′ = −tx + x3, x(0) = α,
where α is a parameter.
A) Prove that for α = 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′ = y2, 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 thaty(t)>√t on the interval 0< t <1.
v. Show that on the interval 0< t <1 if x(t)> y(t), then (t)((t) - t) > 0)(y(t) - t).
vi. Show that y(t)vº(t) - t) > y(t) on the interval 0 vii. Conclude that if (t) > y(t), then I'(t) > vt).
viii. Conclude that while x(t) is defined, it is greater than y(t). Therefore it explodes to infinity at t = 1 or earlier.
B) Use Matlab to draw a direction field for the differential equation (and attach your graph to your solutions). Observe that there is a critical value of a in the interval 0 Sasi such that if a < 0, then r(t) converges to zero and if a co, then r(t) explodes to infinity.