Question

Linear differential equations sometimes occur in which one or both of the functions p(t) and g(t) for y' + p(ty-g(t) have jump discontinuities. If to is such a point of discontinuity, then it is necessary to solve the equation separately for t < to and t>to. Afterward, the two solutions are matched so that y is continuous at to: this is accomplished by a proper choice of the arbitrary constants. The following problem illustrates this situation. Note that it is impossible also to make y' continuous at to Solve the initial value problem. y' + 6y - 9(). WO) -0, where $(t)- 1, Osts 1, 10, - { Osts 1 t)

Answer 1

g'+69=ght) 910)=0 oz tel 941) . t21 for ostal y! +69= This is linean differenticel squation Integrating factor = soda =864 90 y est = fuest dit toi for ostsi y= t + ght for t> 4' +64 = 0 y (+)= Cz -6t T

08 79 s ttcelt i ostsi y (4= -6t 157 Ca since yit) is continuous at = ) lim YH) = Я(1) xtit G Czes = ++G8 = ) 9 3 (2= go + C 9 taest so 6 YH)= eb +9) 36t; t> 6 9500h thao tt a=-16

08 y(t)= lebt ; ostsi (loh i ett obt ; t'i G