Question

Problem 1. In (a) and (b) below, consider the following intial-value problem. (x + 1)(x - 4)y" + (sin x) - (In xy = 0, y(1) = 3, (1) = -2 a. What is the largest interval on which the above initial-value problem has a unique solution? Explain why.

Answer 1

Existence and Uniqueness: Given, (x+1)(x-4)y"+(sin x) y'-(In x) y = 0 y (1)=3, y'(1) = -2 We have to find the largest interval in which given initial value problem has uniqueness solution. According to the existence and uniqueness theorem any second order differential equation of the form y"+ P(x)y'+q(x) y = g(x) with initial conditions y(x)= y, and y'(x) = y', has a unique solution in [a, b] if p(x), 9(x) and g(x) are continuous on[a,b]. Now rewrite the given differential equation in standard form as follows:

sinx In x y- (x+1)(x x-4) (x+1)(x-4) y=0 9(x)= sin x Therefore, p(x) = (x+1)(x-4) accept x=-1 and x=4. In x and g(x) = 0 are continuous at all x > 0 (x+1)(x-4) Therefore, interval of unique solution is (0,4),(4,0). But x =1 lies in (0,4), therefore, the largest interval is(0,4). Hence, the required interval is (0,4)