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10. This problem indicates why we can impose only n initial conditions on a solution of an nth-order linear differential equation. (a) Given the equation y" + py' + qy = 0, explain why the value of y"(a) is determined by the values of y(a) and y'(a). (b) Prove that the equation y"- 2y'-5y 0 has a solution satisfying the conditions y(0) = 1, y'(0) = 0, and y"(0) = C if and only if C= 5. %3D %3D (a) Let a be any number. Substitute x = a in y"+ py'+ qy = 0, thus (1) Rewrite y"(a) in terms of y(a) and y'(a). Thus, y"(a) = (2) %3D Thus the value of y"(a) is determined if the values of (3) are known. (b) Given the equation y"- 2y'- 5y = 0 and initial conditions y(0) = 1, y'(0) = 0, and y"(0) = C, what should be done to show that the differential equation has one solution if and only if C= 5? %3D %3D O A. Substitute y = 5 into the equation y" - 2y'- 5y = 0. O B. Substitute y = C into the equation y"- 2y' - 5y = 0. %3D %3D C. Substitute the initial conditions into the equation y"- 2y'- 5y 0. | D. Substitute C= 5 into the equation y"- 2y'- 5y = C. Applying the substitution above results in the equation (4) Why is the proof complete? O A. The proof is complete because solving this equation yields y = %3D B. The proof is complete because solving this equation yields C= C. The proof is complete because simplifying both sides of this equation yields a contradiction. (1) O -y"(a)+ p(a)y'(a) – q(a)y(a) = 0. O y" (a) + p(a)y'(a) + q(a)y(a) = 0. O y"(a) - p(a)y'(a) – q(a)y(a) = 0. O y"(a)- p(a)y'(a) + q(a)y(a) = 0. (2) O -p(a)y'(a) – q(a)y(a). O- p(a)y'(a) + q(a)y(a). p(a)y' (a) – q(a)y(a). O p(a)y' (a) + q(a)y (a). O y'(a), q(a) and y (a) O p(a), y'(a), q(a) and y(a) O y'(a) and y (a) (3) %3D %3D p(a), and q(a) (4) O y"-2y'- 5y = 5 O C- 2(0) - 5(1) = 0. O C- 2(C)- 5(C) = 0. O 5-2(5) - 5(5) = 0

10. This problem indicates why we can impose only n initial conditions on a solution of an nth-order linear differential equation. (a) Given the equation y" + py' + qy = 0, explain why the value of y"(a) is determined by the values of y(a) and y'(a). (b) Prove that the equation y"- 2y'-5y 0 has a solution satisfying the conditions y(0) = 1, y'(0) = 0, and y"(0) = C if and only if C= 5. %3D %3D (a) Let a be any number. Substitute x = a in y"+ py'+ qy = 0, thus (1) Rewrite y"(a) in terms of y(a) and y'(a). Thus, y"(a) = (2) %3D Thus the value of y"(a) is determined if the values of (3) are known. (b) Given the equation y"- 2y'- 5y = 0 and initial conditions y(0) = 1, y'(0) = 0, and y"(0) = C, what should be done to show that the differential equation has one solution if and only if C= 5? %3D %3D O A. Substitute y = 5 into the equation y" - 2y'- 5y = 0. O B. Substitute y = C into the equation y"- 2y' - 5y = 0. %3D %3D C. Substitute the initial conditions into the equation y"- 2y'- 5y 0. | D. Substitute C= 5 into the equation y"- 2y'- 5y = C. Applying the substitution above results in the equation (4) Why is the proof complete? O A. The proof is complete because solving this equation yields y = %3D B. The proof is complete because solving this equation yields C= C. The proof is complete because simplifying both sides of this equation yields a contradiction. (1) O -y"(a)+ p(a)y'(a) – q(a)y(a) = 0. O y" (a) + p(a)y'(a) + q(a)y(a) = 0. O y"(a) - p(a)y'(a) – q(a)y(a) = 0. O y"(a)- p(a)y'(a) + q(a)y(a) = 0. (2) O -p(a)y'(a) – q(a)y(a). O- p(a)y'(a) + q(a)y(a). p(a)y' (a) – q(a)y(a). O p(a)y' (a) + q(a)y (a). O y'(a), q(a) and y (a) O p(a), y'(a), q(a) and y(a) O y'(a) and y (a) (3) %3D %3D p(a), and q(a) (4) O y"-2y'- 5y = 5 O C- 2(0) - 5(1) = 0. O C- 2(C)- 5(C) = 0. O 5-2(5) - 5(5) = 0

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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10. This problem indicates why we can impose only n initial conditions on a solution of an nth-order linear differential equation. (a) Given the equation y" + py' + qy = 0, explain why the value of y"(a) is determined by the values of y(a) and y'(a). (b) Prove that the equation y"- 2y'-5y 0 has a solution satisfying the conditions y(0) = 1, y'(0) = 0, and y"(0) = C if and only if C= 5. %3D %3D (a) Let a be any number. Substitute x = a in y"+ py'+ qy = 0, thus (1) Rewrite y"(a) in terms of y(a) and y'(a). Thus, y"(a) = (2) %3D Thus the value of y"(a) is determined if the values of (3) are known. (b) Given the equation y"- 2y'- 5y = 0 and initial conditions y(0) = 1, y'(0) = 0, and y"(0) = C, what should be done to show that the differential equation has one solution if and only if C= 5? %3D %3D O A. Substitute y = 5 into the equation y" - 2y'- 5y = 0. O B. Substitute y = C into the equation y"- 2y' - 5y = 0. %3D %3D C. Substitute the initial conditions into the equation y"- 2y'- 5y 0. | D. Substitute C= 5 into the equation y"- 2y'- 5y = C. Applying the substitution above results in the equation (4) Why is the proof complete? O A. The proof is complete because solving this equation yields y = %3D B. The proof is complete because solving this equation yields C= C. The proof is complete because simplifying both sides of this equation yields a contradiction. (1) O -y"(a)+ p(a)y'(a) – q(a)y(a) = 0. O y" (a) + p(a)y'(a) + q(a)y(a) = 0. O y"(a) - p(a)y'(a) – q(a)y(a) = 0. O y"(a)- p(a)y'(a) + q(a)y(a) = 0. (2) O -p(a)y'(a) – q(a)y(a). O- p(a)y'(a) + q(a)y(a). p(a)y' (a) – q(a)y(a). O p(a)y' (a) + q(a)y (a). O y'(a), q(a) and y (a) O p(a), y'(a), q(a) and y(a) O y'(a) and y (a) (3) %3D %3D p(a), and q(a) (4) O y"-2y'- 5y = 5 O C- 2(0) - 5(1) = 0. O C- 2(C)- 5(C) = 0. O 5-2(5) - 5(5) = 0
Transcribed Image Text:10. This problem indicates why we can impose only n initial conditions on a solution of an nth-order linear differential equation. (a) Given the equation y" + py' + qy = 0, explain why the value of y"(a) is determined by the values of y(a) and y'(a). (b) Prove that the equation y"- 2y'-5y 0 has a solution satisfying the conditions y(0) = 1, y'(0) = 0, and y"(0) = C if and only if C= 5. %3D %3D (a) Let a be any number. Substitute x = a in y"+ py'+ qy = 0, thus (1) Rewrite y"(a) in terms of y(a) and y'(a). Thus, y"(a) = (2) %3D Thus the value of y"(a) is determined if the values of (3) are known. (b) Given the equation y"- 2y'- 5y = 0 and initial conditions y(0) = 1, y'(0) = 0, and y"(0) = C, what should be done to show that the differential equation has one solution if and only if C= 5? %3D %3D O A. Substitute y = 5 into the equation y" - 2y'- 5y = 0. O B. Substitute y = C into the equation y"- 2y' - 5y = 0. %3D %3D C. Substitute the initial conditions into the equation y"- 2y'- 5y 0. | D. Substitute C= 5 into the equation y"- 2y'- 5y = C. Applying the substitution above results in the equation (4) Why is the proof complete? O A. The proof is complete because solving this equation yields y = %3D B. The proof is complete because solving this equation yields C= C. The proof is complete because simplifying both sides of this equation yields a contradiction. (1) O -y"(a)+ p(a)y'(a) – q(a)y(a) = 0. O y" (a) + p(a)y'(a) + q(a)y(a) = 0. O y"(a) - p(a)y'(a) – q(a)y(a) = 0. O y"(a)- p(a)y'(a) + q(a)y(a) = 0. (2) O -p(a)y'(a) – q(a)y(a). O- p(a)y'(a) + q(a)y(a). p(a)y' (a) – q(a)y(a). O p(a)y' (a) + q(a)y (a). O y'(a), q(a) and y (a) O p(a), y'(a), q(a) and y(a) O y'(a) and y (a) (3) %3D %3D p(a), and q(a) (4) O y"-2y'- 5y = 5 O C- 2(0) - 5(1) = 0. O C- 2(C)- 5(C) = 0. O 5-2(5) - 5(5) = 0
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