Theory and Applications
(Continuation of Exercise 3.) Assume the hypotheses of Exercise 3, and assume that y1(x) and y2(x) are both solutions to the first order linear equation satisfying the initial condition y(x0) = y0 .
a. Verify that y(x) = y1(x) - y2(x) satisfies the initial value problem
b. For the integrating factor show that
Conclude that v(x)[ y1(x) - y2(x)] ≡ constant
c. From part (a), we have y1(x0) - y2(x0) = 0. Since v(x) > 0 for a < x < b use part (b) to establish that y1(x) - y2(x) ≡ 0 on the interval (a, b). Thus y1(x) = y2(x) for all a < x < b
Reference: Exercise 3
a. Assume that P(x) and Q(x) are continuous over the interval [a, b]. Use the Fundamental Theorem of Calculus, Part 1 to show that any function y satisfying the equation.
for is a solution to the first-order linear equation
b. If then show that any solution y in part (a) satisfies the initial condition y(x0) = y0