Discontinuous Coefficients. Linear differential equations sometimes occur in which one or both of the functions and have jump discontinuities. If is such a point of discontinuity, then it is necessary to solve the equation separately for and . Afterward, the two solutions are matched so that is continuous at ; this is accomplished by a proper choice of the arbitrary constants.
Solve the initial value problem
I got the correct answer in the back of the book using g(t) = 1 and solving the initial value problem. For the first part I got for .
Perhaps I'm going about solving the initial value problem wrong when . But when It is set up as , I get a explicit solution of . When I go to solve for then, and will be , which will give me a value of for , so I must be doing something wrong on the second part, probably not fully understanding the meaning of the question. But the answer in the back of the book for the second part is the following: for . I just don't see how they got that unless I'm supposed to do something with my explicit solution I got from the first part of the problem.