Consider the initial value problem y 2y y 2ety0 a y0 b

Consider the initial value problem y\" + 2y\' + y = 2e^-t,y(0) = a, y\'(0) = b| where a and b are constants. The DE describes a system with damped motion. But the specific behavior will depend on a and b. Some solutions for different values of a and b are shown below: Determine the solution y(t) of this IVP. Show that irrespective of the values of a and b. Now assume that b = - 1. What are the conditions on a so that the solution y(t) has no local extreme on the interval [0, infinity) has exactly one local minimum and one local maximum on the interval [0, infinity) has one local minimum or maximum on the interval [0, infinity) Is it possible to find a value for a so that y(t) has local extreme at both t = 1 and t = 2?

Solution

The characteristic equation has repeated roots -1,-1.

So the complimentary solution is

A e^-t+Bte^-t

A particular solutin is given by

t²e^-t

So a general solution is

y(t) =A e^-t+Bte^-t+ t²e^-t

The initial conditions are y(0)=a and y\'(0)=b.

(a) Plugging in these conditions , the solution of the initial value problem is

y(t) =a e^-t+(a+b)te^-t+ t²e^-t ......................................(1)

(b) as e^-t tends to 0 as t tends to infinity, y(t) tends to 0 irrespective of the values of a and b.

(c) (i) if b =-1, then

y(t) = a e^-t+(a-1)te^-t+ t²e^-t .

y\'(t) = -ae^-t + (a-1) [-te^-t +e^-t ] +2te^-t -t² e^-t

         =-e^-t [ t² +(a-3)+1]

(a) no local extrema if discriminant= (a-3)² -4 >0 (no real roots)

(b) if the discriminant is equal to 0

(c) if the discriminant is positive

(d) As the product of the roots is 1 , it is not possible to find such a value for a.