Differential Equations Question please help

a. If ar^2 + br + c = 0 has equal roots r sub1, show that

L[e^rt] = a(e^rt)’’ + b(e^rt)’ + c = a(r - r sub1)^2 e^rt (i)


b. Differentiate Eq. (i) with respect to r and interchange differentiation with respect to r and with respect to t, thus showing that

(d/dr)L[e^rt ] = L[(d/dr) e^rt ] = L[te^rt ] = ate^rt (r - r sub1)^2 + 2ae^rt (r - r sub1) (ii)

Since the right side of the Eq. (ii) is zero when r = r sub1 conclude that t*exp(r1t) is also a solution of L[y]=0

Thank you

Assuming that L[y] = ay'' + by' + cy, and ar^2 + br + c = 0 has a double root r = r₁:

a) Letting y = e^(rt), we have
L[e^(rt)] = a(r^2 e^(rt)) + b(r e^(rt)) + ce^(rt)
............= (ar^2 + br + c) e^(rt).

However, we know that r = r₁ is a double root of ar^2 + br + c.
So, ar^2 + br + c = a(r - r₁)^2.

Hence, L[e^(rt)] = a(r - r₁)^2 e^(rt).
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b) Differentiate both sides of the last equation with respect to r:
(∂/∂r) L[e^(rt)] = a(r - r₁)^2 * re^(rt) + 2a(r - r₁) * e^(rt), by the product rule

However,
(∂/∂r) L[e^(rt)] = L[(∂/∂r) e^(rt)], due to equality of mixed partial derivatives
....................= L[te^(rt)].

So, L[te^(rt)] = a(r - r₁)^2 * re^(rt) + 2a(r - r₁) * e^(rt).

Finally, letting r = r₁ yields L[te^(r₁ t)] = 0 + 0 = 0.
==> y = te^(r₁ t) is also a solution of L[y] = ay'' + by' + cy = 0.
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I hope this helps!