{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

20d_07s_fs

# 20d_07s_fs - Math 20D Final Exam Solutions 2 2 2 1(a An...

This preview shows pages 1–2. Sign up to view the full content.

Math 20D Final Exam Solutions June 12, 2007 1. (a) An integrating factor is e t/ 2 / 2 and so e t 2 / 2 y = i t 3 e t 2 / 2 dt . The integral can be done using integration by parts with u = t or it can be done by substitution with x = t 2 / 2 and then integration by parts. The result is ( t 2 - 2) e t 2 / 2 + C and so y = t 2 - 2 + Ce t 2 / 2 . Using the initial condition, 0 = - 2 + C and so C = 2. (b) Separate variables: e y dy = - t 2 dt . Thus - e y = t 1 + C . (c) Set y = e rt to obtain r 2 - 2 r +2 = 0 and so r = 1 ± - 1. Hence the general solution is y = C 1 e (1+ i ) t + C 2 e (1 i ) t , which can be written y = e t ( D 1 sin t + D 2 cos t ). (d) The homogeneous equation y ′′ - 4 y + y = 0 gives r 2 - 4 r + 4 = 0 and so r = - 2 is a double root. Thus the general solution to the homogeneous equation is y = ( C 1 + C 2 t ) e 2 t . One can use variation of parameters or undetermined coe±cients to ²nd a particular solution. Using the latter, we set y = Ce t and

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.

{[ snackBarMessage ]}

### Page1 / 2

20d_07s_fs - Math 20D Final Exam Solutions 2 2 2 1(a An...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online