1 e t c 2 e 4 t 1 2 e 2 t Joseph M Mahaffy h mahaffymathsdsuedu i Lecture Notes

# 1 e t c 2 e 4 t 1 2 e 2 t joseph m mahaffy h

• 32

This preview shows page 18 - 24 out of 32 pages.

1 e - t + c 2 e 4 t - 1 2 e 2 t Joseph M. Mahaffy, h [email protected] i Lecture Notes – Second Order Linear Equations — (18/32)

Subscribe to view the full document.

Introduction Theory for 2 nd Order DEs Linear Constant Coefficient DEs Homogeneous Equations Method of Undetermined Coefficients Forced Vibrations Method of Undetermined Coefficients Method of Undetermined Coefficients - Example 2: Consider y 00 - 3 y 0 - 4 y = 5 sin( t ) From before, the homogeneous solution is y c ( t ) = c 1 e - t + c 2 e 4 t Neither solution matches the forcing function , so try y p ( t ) = A sin( t ) + B cos( t ) so y 0 p ( t ) = A cos( t ) - B sin( t ) and y 00 p ( t ) = - A sin( t ) - B cos( t ) It follows that ( - A + 3 B - 4 A ) sin( t ) + ( - B - 3 A - 4 B ) cos( t ) = 5 sin( t ) or 3 A + 5 B = 0 and 3 B - 5 A = 5 or A = - 25 34 and B = 15 34 The solution combines these to obtain y ( t ) = c 1 e - t + c 2 e 4 t + 15 34 cos( t ) - 25 34 sin( t ) Joseph M. Mahaffy, h [email protected] i Lecture Notes – Second Order Linear Equations — (19/32)
Introduction Theory for 2 nd Order DEs Linear Constant Coefficient DEs Homogeneous Equations Method of Undetermined Coefficients Forced Vibrations Method of Undetermined Coefficients Method of Undetermined Coefficients - Example 3: Consider y 00 - 3 y 0 - 4 y = 2 t 2 - 7 From before, the homogeneous solution is y c ( t ) = c 1 e - t + c 2 e 4 t Neither solution matches the forcing function , so try y p ( t ) = At 2 + Bt + C It follows that 2 A - 3(2 At + B ) - 4( At 2 + Bt + C ) = 2 t 2 - 7 , so matching coefficients gives - 4 A = 2, - 6 A - 4 B = 0, and 2 A - 3 B - 4 C = - 7, which yields A = - 1 2 , B = 3 4 and C = 15 16 The solution combines these to obtain y ( t ) = c 1 e - t + c 2 e 4 t - t 2 2 + 3 t 4 + 15 16 Joseph M. Mahaffy, h [email protected] i Lecture Notes – Second Order Linear Equations — (20/32)

Subscribe to view the full document.

Introduction Theory for 2 nd Order DEs Linear Constant Coefficient DEs Homogeneous Equations Method of Undetermined Coefficients Forced Vibrations Method of Undetermined Coefficients Superposition Principle: Suppose that g ( t ) = g 1 ( t ) + g 2 ( t ). Also, assume that y 1 p ( t ) and y 2 p ( t ) are particular solutions of ay 00 + by 0 + cy = g 1 ( t ) ay 00 + by 0 + cy = g 2 ( t ) , respectively. Then y 1 p ( t ) + y 2 p ( t ) is a solution of ay 00 + by 0 + cy = g ( t ) From our previous examples, the solution of y 00 - 3 y 0 - 4 y = 3 e 2 t + 5 sin( t ) + 2 t 2 - 7 satisfies y ( t ) = c 1 e - t + c 2 e 4 t - 1 2 e 2 t + 15 34 cos( t ) - 25 34 sin( t ) - t 2 2 + 3 t 4 + 15 16 Joseph M. Mahaffy, h [email protected] i Lecture Notes – Second Order Linear Equations — (21/32)
Introduction Theory for 2 nd Order DEs Linear Constant Coefficient DEs Homogeneous Equations Method of Undetermined Coefficients Forced Vibrations Method of Undetermined Coefficients Method of Undetermined Coefficients - Example 4: Consider y 00 - 3 y 0 - 4 y = 5 e - t From before, the homogeneous solution is y c ( t ) = c 1 e - t + c 2 e 4 t Since the forcing function matches one of the solutions in y c ( t ), we attempt a particular solution of the form y p ( t ) = Ate - t , so y 0 p ( t ) = A (1 - t ) e - t and y 00 p ( t ) = A ( t - 2) e - t It follows that ( A ( t - 2) - 3 A (1 - t ) - 4 At ) e - t = - 5 Ae - t = 5 e - t , Thus, A = - 1 The solution combines these to obtain y ( t ) = c 1 e - t + c 2 e 4 t - te - t Joseph M. Mahaffy, h [email protected] i Lecture Notes – Second Order Linear Equations — (22/32)

Subscribe to view the full document.

Introduction Theory for 2 nd Order DEs Linear Constant Coefficient DEs Homogeneous Equations Method of Undetermined Coefficients
• Fall '08
• staff

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern