Consider the equation y′′ + 2ay′ + a2 y = 0. Show that the roots of the characteristic equation are r1 = r2 = - a so that one solution of the equation is e-at. b. Use Abel's formula [equation (23) of Section 3.2] to show that the Wronskian of any two solutions of the given equation is W (t) = y1 (t) y′2 (t) - y′1 (t) y2 (t) = c1e-2at, where c1 is a constant. c. Let y1 (t) = e-at and use the result of part b to obtain a differential equation satisfied by a second solution y2 (t). By solving this equation, show that y2 (t) = te-at.

Step-by-step explanation:

Given the differential equation

y'' + 2ay' + a²y = 0 ... (1)

The auxiliary equation to (1) is

m² + 2am + a² = 0

Solving the auxiliary equation, we have

(m + a) ² = 0

m + a = 0 twice

m = - a twice.

The complimentary solution for the repeated root is

y = C1e^ (-at) + C2te^ (-at)

So, obviously, one of the solutions is

y1 = C1e^ (-at).

For C1 = 1, we have

y1 = e^ (-at)

Let W be the Wronskian of y1 and y2.

The Wronskian satisfies the first order differential equation

W' + 2aW = 0

Solving this, we have

W'/W = - 2a

Integrating this

lnW = - 2at + C3

W = Ce^ (-2at)

(Where C = e^C3)

The Wronskian of y1 and y2 is the determinant

|y1 ... y2|

|y1' ... y2'|

= y1y2' - y1'y2

Since y1 = C1e^ (-at)

y1' = - aC1e^ (-at)

W = C1e^ (-at) y2' - (-aC1e^ (-at)) y2

But

W = Ce^ (-2at)

So, we have

C1e^ (-at) y2' + aC1e^ (-at) y2 = Ce^ (-2at)

Divide through by C1e^ (-at)

y2' + ay2 = C3e^ (-at) ... (2)

(where C3 = C/C1)

Now, solving (2), we have multiply by the integrating factor e^ (at) to have

d (y2e^ (at)) = C3

Integrating this, we have

y2e^ (at) = C3t + C4

y2 = C3te^ (-at) + C4e^ (-at)

Here, for C3 = 1, and C4 = 0, we have

y2 = te^ (-at).