Let (*) be a homogeneous equation of the form (*) y" + a1y' + aoy = 0 for constants a1, ao € R. To such an equation, we can also associate a polynomial x(t) = t² + a;t + ao- Suppose x(t) splits into linear factors, say x(t) = (t – a)(t - B) for a, 3 € R. Show that y = y(r) is a solution to (*) if and only if y is a solution to y" – (a + 3)y/ + a3y = 0. Next, show that (i) if a # 3, then y1 = eaz and y2 = e3z form a basis of solutions to (*); and

Algebra and Trigonometry (6th Edition)
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ISBN:9780134463216
Author:Robert F. Blitzer
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ChapterP: Prerequisites: Fundamental Concepts Of Algebra
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Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Let (*) be a homogeneous equation of the form
(*)
y" + a1y' + aoy = 0
for constants a1, ao E R. To such an equation, we can also associate a polynomial
x(t) = t² + a;t + ao-
Suppose x(t) splits into linear factors, say x(t) = (t – a)(t – 3) for a, 3 E R. Show that
y = y(x) is a solution to (*) if and only if y is a solution to
y" – (a + B)y/ + aßy = 0.
Next, show that
(i) if a + 3, then y1 = e# and y2 = e3% form a basis of solutions to (*); and
(ii) if a = B, then y1 = e0z and y2 =
rear form a basis of solutions to (*).
Transcribed Image Text:Let (*) be a homogeneous equation of the form (*) y" + a1y' + aoy = 0 for constants a1, ao E R. To such an equation, we can also associate a polynomial x(t) = t² + a;t + ao- Suppose x(t) splits into linear factors, say x(t) = (t – a)(t – 3) for a, 3 E R. Show that y = y(x) is a solution to (*) if and only if y is a solution to y" – (a + B)y/ + aßy = 0. Next, show that (i) if a + 3, then y1 = e# and y2 = e3% form a basis of solutions to (*); and (ii) if a = B, then y1 = e0z and y2 = rear form a basis of solutions to (*).
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