It was stated (on page 7) that "Norder sets of initial values are especially appropriate for N-order differential equations." The following problems illustrate one reason this is true. In particular, they demonstrate that, if y satisfies some N-order initial-value problem, then it automatically satisfies particular higher-order sets of initial values. Because of this, specifying the initial values for ym) with m 2 N is unnecessary and may even lead to problems with no solutions. a. Assume y satisfies the first-order initial-value problem i. Using the differential equation along with the given value for y(1), determine what value y'(1) must be ii. Is it possible to have a solution to i. Find the value of y(0) and of y(0) ii. Is it possible to have a solution to 3ry+2 with y(1)=2. that also satisfies both y(1)=2 and y'(1)-4? (Give a reason.) iii. Differentiate the given differential equation to obtain a second-order differential equation. Using the equation obtained along with the now known values for y(1) and y'(1), find the value of y(1). iv. Can we continue and find y(1), y) (1)....? b. Assume y satisfies the second-order initial-value problem that also satisfies all of the following: =3zy+z¹ +4-8y=0 with y(0) 3 and y(0) = 5. +44-8y=0 y (0) 3, y(0)-5 and "(0)=0 ?