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
