Let (*) be a homogeneous equation of the form(*)y" + a1y' + aoy = 0for constants a1, ao E R. To such an equation, we can also associate a polynomialx(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 thaty = y(x) is a solution to (*) if and only if y is a solution toy" – (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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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

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)

6th Edition

ISBN:9780134463216

Author:Robert F. Blitzer

Publisher:Robert F. Blitzer

ChapterP: Prerequisites: Fundamental Concepts Of Algebra

Section: Chapter Questions

Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...

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Differential Equations

Have you ever observed the movement of the satellites and rockets or how the planets go around the sun in their orbits? How will you determine their motion? Definitely they follow some scientific laws and these kinds of problems are framed as differential equations.

Question

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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