298_pdfsam_math 54 differential equation solutions odd

# 298_pdfsam_math 54 differential equation solutions odd -...

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

Chapter 5 dy dt = d dt ( e t ) = e t = y. Thus, given pair of functions is a solution. To sketch the trajectory of this solution, we express x as a function of y . x = e 3 t = ( e t ) 3 = y 3 for y = e t > 0 . Since y = e t is an increasing function, the ﬂow arrows are directed away from the origin. See Figure B.29 in the answers of the text. 3. In this problem, f ( x, y )= x y , g ( x, y x 2 + y 2 1. To ±nd the critical point set, we solve the system x y =0 , x 2 + y 2 1=0 x = y, x 2 + y 2 =1 . Eliminating y yields 2 x 2 x = ± 1 2 . Substituting x into the ±rst equation, we ±nd the corresponding value for y . Thus the critical points of the given system are (1 / 2 , 1 / 2) and ( 1 / 2 , 1 / 2). 5.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Ask a homework question - tutors are online