Please answer a, b, and c! Information about theorem one is attached. 3. Consider the initial-value problemdy = y²/³, y(0) = 0.dxUse separation of variables to obtain the solution y(x) = x³ /27.(b) Observe that another solution is y(x) 0.(c) Since this initial-value problem has two distinct solutions, uniqueness doesnot apply. What hypothesis in Theorem 1 in Section 1.2 does not hold? Theorem 1. If f(x, y) is continuous in an open rectangle R = (a, b) (c,d) in the xy-plane that contains the point (xo, yo), then there exists a solution y(x) to the initial-valueproblemdydx(1.11)that is defined in an open interval I = (a, b) containing xo. In addition, if the partialderivative df/dy is continuous in R, then the solution y(x) of (1.11) is unique.=f(x, y),Qualitative Analysisy(xo):= yo,This theorem is proved using successive approximations, but we shall not give thedetails; see, for example, [1]. However, let us make a couple of remarks that will helpclarify the need for the various hypotheses.Remark 1. We might hope that existence holds with a = a and ß = b, and indeed thisis the case for linear equations f(x, y) = a(x)y+b(x). But for nonlinear equations, thisneed not be the case. For example, if we plot the slope field and some solution curves forY= y², we see that the solutions seem to approach vertical asymptotes; this suggeststhat they are not defined on I = ( ) even though f y² continuous on thewhole xy-plane. (When we discuss separation of variables in Section 1.3, we will be ableto confirm the existence of these vertical asymptotes for many nonlinear equations.)==Remark 2. Regarding the condition for uniqueness, Exercise 3 in Section 1.3 shows thatthe initial-value problem y = y²/³, y(0) = 0 has two different solutions; the uniquenessclaim in the theorem does not apply to this example since df/dy is not continuous inany rectangle containing (0,0).

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