For each of the two equations below i Solve as in Chapter 4

For each of the two equations below, (i) Solve as in Chapter 4 (assuming a solution of the exponential form); (ii) Solve by rewriting the equation as a first-order system of two equations and 2 unknowns; (iii) Are the solutions the same? Explain.

1. y\'\'-2y\' + 17y = 0
2. y\'\'+8y\'+16y=0

Solution

1 .y\" - 2y \' + 17y = 0,

the exponential equation:

r 2 - 2r + 17 = 0 (r - 1)2 +16=0 s

r = 1 ± 4i.

Hence, y1(t) = e t cos(4t) and y2(t) = e t sin(4t) are linearly independent solutions to y\" 2y\' + 17y = 0,

and a general solution is y(t) = c1 e t cos(4t) + c2 e t sin(4t)

= (c1 cos(4t) + c2 sin(4t))e t so that y 0 (t)

= (c1 cos(4t) + c2 sin(4t))e t + (4c1 sin(4t) + 4c2 cos(4t))e t

= ((c1 + 4c2) cos(4t) (4c1 + c2) sin(4t))e t .

It follows that 1 = y(0) = c1 and 1 = y 0 (0) = 1 + 4c2 and thus c2. Hence, the solution is y1(t).

2. y\"+8y\'+16y=0

the general solution is y = (C + Cx) e^(2x) + (C + Cx) e^(-2x).