lecture_18

# lecture_18 - MA 36600 LECTURE NOTES FRIDAY FEBRUARY 27...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MA 36600 LECTURE NOTES: FRIDAY, FEBRUARY 27 Repeated Roots Review. Say that we wish to find the general solution to the differential equation ay 00 + by + cy = 0 where b 2- 4 ac = 0. First note that the roots of the characteristic polynomial ar 2 + br + c = 0 are r 1 = r 2 =- b 2 a ± √ b 2- 4 ac 2 a =- b 2 a . Then { y 1 ,y 2 } forms a fundamental set of solutions, where y 1 ( t ) = e rt and y 2 ( t ) = te rt . Example. Consider the homogeneous differential equation y 00 + 4 y + 4 y = 0 . We find the general solution using d’Alembert’s Method. First we find one solution in the form y 1 ( t ) = e rt . Indeed, the characteristic equation is r 2 + 4 r + 4 = 0 = ⇒ ( r + 2) 2 = 0 . Hence r 1 = r 2 =- 2 is the only root, so one solution is y 1 ( t ) = e- 2 t . Assume that the general solution is in the form y ( t ) = v ( t ) y 1 ( t ) for some function v = v ( t ) to be found. We have the derivatives y = v e- 2 t y = ( v- 2 v ) e- 2 t y 00 = ( v 00- 4 v + 4 v ) e- 2 t This gives the expression 0 = y 00 + 4 y + 4 y = ( v 00- 4 v + 4 v ) e- 2 t + 4( v- 2 v ) e- 2 t + 4 v e- 2 t = v 00 e- 2 t so that v 00 = 0. We know that the general solution to this differential equation is v ( t ) = c 1 + c 2 t for constants c 1 and c 2 . Hence the general solution to the differential equation y + 4 y + 4 y = 0 is the function y ( t ) = c 1 e- 2 t + c 2 te- 2 t . 1 2 MA 36600 LECTURE NOTES: FRIDAY, FEBRUARY 27 Summary. We summarize the general solution to the differential equation ay 00 + by + cy = 0 . Say that the characteristic equation ar 2 + br + c = 0 has roots r 1 and r 2 . A fundamental set of solutions for this constant coefficient differential equation is { y 1 ,y 2 } in terms of the functions y 1 ( t ) = e r 1 t if b 2- 4 ac > 0, e r 1 t if b 2- 4 ac = 0, e λt cos μt if b 2- 4 ac < 0....
View Full Document

## This note was uploaded on 05/24/2011 for the course MA 36600 taught by Professor Ma during the Spring '09 term at Purdue.

### Page1 / 4

lecture_18 - MA 36600 LECTURE NOTES FRIDAY FEBRUARY 27...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online