{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

61_pdfsam_math 54 differential equation solutions odd

# 61_pdfsam_math 54 differential equation solutions odd - etc...

This preview shows page 1. Sign up to view the full content.

Exercises 2.3 The last integral can be found by integrating by parts twice which leads back to an integral similar to the original. Combining these two similar integrals and simplifying, we obtain e 2 t cos πt 12 dt = e 2 t 2 cos ( πt 12 ) + π 12 sin ( πt 12 ) 4 + ( π 12 ) 2 + C. Thus we see that x ( t ) = 1 2 2 cos ( πt 12 ) + π 12 sin ( πt 12 ) 4 + ( π 12 ) 2 + Ce 2 t . Using the initial condition, t = 0 and x = 10, to solve for C , we obtain C = 19 2 + 2 4 + ( π 12 ) 2 . Therefore, the desired solution is x ( t ) = 1 2 2 cos ( πt 12 ) + π 12 sin ( πt 12 ) 4 + ( π 12 ) 2 + 19 2 + 2 4 + ( π 12 ) 2 e 2 t . 39. Let T j ( t ), j = 0 , 1 , 2 , . . . , denote the temperature in the classroom for 9 + j t < 10 + j , where t = 13 denotes 1 : 00 p.m. , t = 14 denotes 2 : 00 p.m.
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: , etc. Then T (9) = 0 , (2.12) and the continuity of the temperature implies that lim t → 10+ j = T j +1 (10 + j ) , j = 0 , 1 , 2 , . . . . (2.13) According to the work of the heating unit, the temperature satisFes the equation dT j dt = · 1 − T j , if j = 2 k − T j , if j = 2 k + 1 , 9 + j < t < 10 + j k = 0 , 1 , . . . . The general solutions of these equations are: for j even dT j dt = 1 − T j ⇒ dT j 1 − T j = dt ⇒ ln | 1 − T j | = − t + c j ⇒ T j ( t ) = 1 − C j e − t ; 57...
View Full Document

{[ snackBarMessage ]}