{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

272_pdfsam_math 54 differential equation solutions odd

# 272_pdfsam_math 54 differential equation solutions odd -...

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

Chapter 5 We use initial conditions to determine constants C 1 , C 2 , C 3 , and C 4 . Differentiating (5.7) and (5.8), we get 3= x (0) = C 1 e 0 + C 2 e 0 + C 3 cos 0 + C 4 sin 0 = C 1 + C 2 + C 3 , 1= x (0) = C 1 e 0 C 2 e 0 C 3 sin 0 + C 4 cos 0 = C 1 C 2 + C 4 , 1= y (0) = C 1 e 0 + C 2 e 0 C 3 cos 0 C 4 sin 0 = C 1 + C 2 C 3 , 1 = y (0) = C 1 e 0 C 2 e 0 + C 3 sin 0 C 4 cos 0 = C 1 C 2 C 4 C 1 + C 2 + C 3 = 3 , C 1 C 2 + C 4 = 1 , C 1 + C 2 C 3 = 1 , C 1 C 2 C 4 = 1 . Solving we obtain C 1 = C 2 = C 3 = C 4 = 1. So, the desired solution is x ( t ) = e t + e t + cos t + sin t, y ( t ) = e t + e t cos t sin t. 23. We will attempt to solve this system by first eliminating the function y ( t ). Thus, we multiply the first equation by ( D + 2) and the second by ( D 1). Therefore, we obtain ( D + 2)( D 1)[ x ] + ( D + 2)( D 1) D [ y ] = ( D + 2) 3 e 2 t = 6 e 2 t 6 e 2 t = 0 , ( D 1)( D + 2)[ x ] ( D 1)( D + 2)[ y ] = ( D 1) 3 e t = 3 e t + 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Ask a homework question - tutors are online