{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

270_pdfsam_math 54 differential equation solutions odd

# 270_pdfsam_math 54 differential equation solutions odd -...

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

Chapter 5 17. Expressed in operator notation, this system becomes ( D 2 + 5) [ x ] 4[ y ] = 0 , [ x ] + ( D 2 + 2) [ y ] = 0 . In order to eliminate the function x ( t ), we apply the operator ( D 2 +5) to the second equation. Thus, we have ( D 2 + 5 ) [ x ] 4[ y ] = 0 , ( D 2 + 5 ) [ x ] + ( D 2 + 5 ) ( D 2 + 2 ) [ y ] = 0 . Adding these two equations together yields the differential equation involving the single func- tion y ( t ) given by ( D 2 + 5)( D 2 + 2) 4 [ y ] = 0 ( D 4 + 7 D 2 + 6 ) [ y ] = 0 . The auxiliary equation for this homogeneous equation, r 4 + 7 r 2 + 6 = ( r 2 + 1)( r 2 + 6) = 0, has roots r = ± i , ± i 6. Thus, a general solution is y ( t ) = C 1 sin t + C 2 cos t + C 3 sin 6 t + C 4 cos 6 t. We must now find a function x ( t ) that satisfies the system of differential equations given in the problem. To do this we solve the second equation of the system of differential equations
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Ask a homework question - tutors are online