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

|y1y2|

|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).

the mistake in the pattern is that :

b. the decimal point was moved one too many place values each time

the correct pattern could be made by :

b. moving the decimal point to the right the same numbers of place values as the exponents