Suppose that r1 and r2 are roots of ar2 + br + c = 0 and that r1 = r2; then exp (r1t) and exp (r2t) are solutions of the differential equation ay + by + cy = 0. Show that φ (t; r1, r2) = er2t - er1t r2 -

The Correct Question is:

Suppose that r1 and r2 are roots of ar² + br + c = 0 and that r1 = r2; then e^ (r1t) and e^ (r2t) are solutions of the differential equation

ay'' + by' + cy = 0.

Show that

φ (t; r1, r2) = [e^ (r2t) - e^ (r1t) ] / (r2 - r1)

is a solution of the differential equation.

Answer:

φ (t; r1, r2) is a solution of the differential equation, and it shown.

Step-by-step explanation:

Given the differential equation

ay'' + by' + cy = 0

and r1 and r2 are the roots of its auxiliary equation.

We want to show that

φ (t; r1, r2) = [e^ (r2t) - e^ (r1t) ] / (r2 - r1)

satisfies the given differential equation, that is

aφ'' + bφ' + cφ = 0 ... (*)

Where φ = φ (t; r1, r2)

We now differentiate φ twice in succession, with respect to t.

φ' = [r2e^ (r2t) - r1e^ (r1t) ] / (r2 - r1)

φ'' = [r2²e^ (r2t) - r1²e^ (r1t) ] / (r2 - r1)

Using these in (*)

We have

a[r2e^ (r2t) - r1e^ (r1t) ] / (r2 - r1) + [r2²e^ (r2t) - r1²e^ (r1t) ] / (r2 - r1) + c[e^ (r2t) - e^ (r1t) ] / (r2 - r1)

= [ (ar2² + br2 + c) e^ (r2t) - (ar1² + br1 + c) e^ (r1t) ] / (r1 - r2)

We know that r1 and r2 are the roots of the auxiliary equation

ar² + br + c = 0

and r1 = r2

This implies that

ar1² + br1 + c = ar2² + br2 + c = 0

And hence,

[ (ar2² + br2 + c) e^ (r2t) - (ar1² + br1 + c) e^ (r1t) ] / (r1 - r2) = 0

Therefore,

aφ'' + bφ' + cφ = 0