a) Verify that both y1(t)= 1-t and y2(t)= (-t^2)/4 are solutions of the initial value problem
y-prime = (-t + (t^2 + 4y)^(1/2)) / 2 , for y(2) = -1
Where are these solutions valid?
b) Explain why the existence of two solutions of the given problem does not contradict the uniqueness part of Theorem 2.4.2 (typed in section 2. below) .
C) Show that y=ct+c^2 where c is an arbitrary constant satisfies the differential equation in part (a) for t> or equal to -2c. If c=-1 the initial condition is also satisfied and the solution y=y1(t) is obtained. Show that there is no choice of c that gi es the second solution y=y2(t)
Let the functions f and df/dy be continiuous in some rectangle alpha < t < beta, gamma < y < theta containing the point (t0, y0). Then in some interval t0 - h < t < t0 + h contained in alpha < t < beta, there is a unique solution y = phi (t) of the initial value problem y-prime = f ( t,y ) , y(t0) = y0
i think this is the relevant equation, at least
The Attempt at a Solution
I am very unsure of my method in attempting to solve this problem. My problem is that I don't understand the concepts I'm employing or why I'm employing them. What I did to solve it was seperate the "-t/2" and "[(t^2+4y)^(1/2)]/2]" terms and add "-t+2" to each side. Then, I didn't know how to get the y on its own, so I squared both sides. I don't know if this is right to do. Now I have (d^2)y/(dt^2) + (t/2)^2 = [(t^2+4y)^(1/2)]/2]^2.
I got to (d^2)y/(dt^2) = y(1-t) . I don't know if this is right, what this means, or what to do here. I am very confused. My homework is due Monday, so I would really appreciate it if someone who understands this problem could explain any part of it to me, but especially a) since that is the part I am stuck on.