7. y" – 3y + 2y = 0;
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter3: Functions And Graphs
Section3.4: Definition Of Function
Problem 33E
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Question
please answer #7
section 3.2
![Exercises 1-15:
In these exercises, the t-interval of interest is -o <t < oo unless indicated otherwise.
(a) Verify that the given functions are solutions of the differential equation.
(b) Calculate the Wronskian. Do the two functions form a fundamental set of solutions?
(c) If the two functions form a fundamental set, determine the unique solution of the
initial value problem.
1. y" – 4y = 0; y1 (1) = e2".
2. y" – y = 0; y1(1) = 2e', y2(t) =e-+3;
3. y" + y = 0; y(t) = 0, y2(t) = sint; y(7/2) = 1, y (7/2) = 1
y2 (t) = 2e-2"; y(0) = 1, y'(0) = -2
y(-1) = 1, y(-1) = 0
4. y" + y = 0; y,(t) = cost, y2(t) = sint; y(7/2) = 1, y'(7/2) = 1
y, (t) = e", y2(t) = te: y(0) = 2, y(0) = 0
6. 2y" – y' = 0; y; (t) = 1, y2(t) = e'2; y(2) = 0, y'(2) = 2
7. y" – 3y + 2y = 0; y1(1) = 2e', y2(t) = e; y(-1) = 1, y'(-1) = 0
5. y" – 4y + 4y = 0;
8. 4y" + y = 0; y1(t) = sin[(t/2) + (7/3)], y2(t) = sin[(t/2) – (/3)];
y(0) = 0, y'(0) = 1
9. ty" + y' = 0, 0<t < o0; y; (f) = In t,
y2(t) = In 3t; y(3) = 0, y'(3) = 3](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3afa98d3-5d5f-4802-bbdd-7e6af6547327%2F1e03d0c6-1074-4977-ad6a-ea10f134650f%2Fj9n5vh6_processed.png&w=3840&q=75)
Transcribed Image Text:Exercises 1-15:
In these exercises, the t-interval of interest is -o <t < oo unless indicated otherwise.
(a) Verify that the given functions are solutions of the differential equation.
(b) Calculate the Wronskian. Do the two functions form a fundamental set of solutions?
(c) If the two functions form a fundamental set, determine the unique solution of the
initial value problem.
1. y" – 4y = 0; y1 (1) = e2".
2. y" – y = 0; y1(1) = 2e', y2(t) =e-+3;
3. y" + y = 0; y(t) = 0, y2(t) = sint; y(7/2) = 1, y (7/2) = 1
y2 (t) = 2e-2"; y(0) = 1, y'(0) = -2
y(-1) = 1, y(-1) = 0
4. y" + y = 0; y,(t) = cost, y2(t) = sint; y(7/2) = 1, y'(7/2) = 1
y, (t) = e", y2(t) = te: y(0) = 2, y(0) = 0
6. 2y" – y' = 0; y; (t) = 1, y2(t) = e'2; y(2) = 0, y'(2) = 2
7. y" – 3y + 2y = 0; y1(1) = 2e', y2(t) = e; y(-1) = 1, y'(-1) = 0
5. y" – 4y + 4y = 0;
8. 4y" + y = 0; y1(t) = sin[(t/2) + (7/3)], y2(t) = sin[(t/2) – (/3)];
y(0) = 0, y'(0) = 1
9. ty" + y' = 0, 0<t < o0; y; (f) = In t,
y2(t) = In 3t; y(3) = 0, y'(3) = 3
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