Question

The graphs of a member of a family of solutions of a second order differential equation given. Match the solutions curve with a pair of initial conditions. 2-1(2, 4 v) 1. D 2. B y(0) = 2, y'(0) = -2 4 -2 y(0) = 1, y'(0) = 1/2 y(0) = 2, y'(0) = 0 3. с

Answer 1

day = f(niy, y). Cossicher ana Let's consides each solution and find the corresponding pair of initial conditions. Consider one graph As of the son you) Node that one functions gens yens tales the value a when n=o. y coo=a. Also note that at nzo, you changes it's nature bons sovrasing ato from increasing to decreasing. That is no beste a coibjcal point for ie, g'coo=0 : The pair of initial conditions is is y) y (o)=2. y col=0 CS Scanned vibreamScanner

low consider the the second options. obsence thert the function yen) at n=o. takes the value = 1 node dnet ie, yeos el Also, at nzo, y changes it's nature of concavity. ie, n=o is a point of inflections. ie, y"(0)=0 for some constant atos sinue y'co) is one tongent to y and goo=1 (Angent should meet c =) y' cos=c nao y'60) = 1. the crore at (0,1)) . The required pair of initial conditions is y con=1 y16) = 1. CS Scanned it Scanner