a. Solve the initival value problem: y = -8 cos(2x)/(6+ 5y), y(0) = –1. y = (V1-40 sin(2x) -6) b. When solving an ODE, the solution is only valid in some interval. Furthermore, if an initial condition...

Extracted text: a. Solve the initival value problem: y = -8 cos(2x)/(6+ 5y), y(0) = –1. y = (V1-40 sin(2x) -6) b. When solving an ODE, the solution is only valid in some interval. Furthermore, if an initial condition is given, the solution will only be valid in the largest interval in the domain of the solution that is around the x-value given in the initial condition. In this case, since y(0) = –1, then the solution is only valid in the largest interval in the domain of y around x = = 0. For the following question, you do not need to find the interval exactly, but you should use a graphing utility, such as WolframAlpha, to plot the solution. Determine where the solution attains its maximum value. That is, you must compute the critical value of y using calculus. Note that there is only one x-value in the largest interval about x = 0 that works, which you can determine based on the graph. TC Xcrit %3D 4.