{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

285lect18 - 1 1.1 Lecture 18 Non-homogeneous linear ODE of...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
1 Lecture 18 1.1 Non-homogeneous linear ODE of order n Let L = D n + p 1 ( x ) D n ° 1 + p 2 ( x ) D n ° 2 + ::: + p n ° 1 ( x ) D + p n ( x ) D 0 : Consider non-homogeneous ODE y ( n ) + p 1 ( x ) y ( n ° 1) + p 2 ( x ) y ( n ° 2) + ::: + p n ( x ) y ( x ) = f ( x ) , or L [ y ] = f: The complementary homogeneous equation is y ( n ) + p 1 ( x ) y ( n ° 1) + p 2 ( x ) y ( n ° 2) + ::: + p n ( x ) y ( x ) = 0 , or L [ y ] = 0 : Solutions of complementary equation are called complementary solutions y c ( x ) . So, L [ y c ] = 0 . Let y p ( x ) be a particular solution of the non° homogeneous equation L [ y ] = f . Hence, L [ y p ] = f . Then y ( x ) = y c ( x ) + y p ( x ) is a solution of the non-homogeneous equation. Indeed, L [ y c + y p ] = L [ y c ] + L [ y p ] = 0 + f ( x ) = f ( x ) : Notice, that if y c is a general solution of the complimentary equation, then, there are n constants in the formula for y c . Thus, y ( x ) = y c ( x ) + y p ( x ) contains n constants C 1 ; C 2 ; :::; C n . It can be showed that by selecting the constants C 1 ; C 2 ; :::; C n , we can solve a given initial value problem. Example 1 Consider the following initial value problem: y 000 ° 6 y 00 + 11 y 0 ° 6 y = 1; y (0) = 0 ; y 0 (0) = 0 and y 00 (0) = 0 : We already found y c : y c ( x ) = C 1 e x + C 2 e 2 x + C 3 e 3 x : 1
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
It is not di¢ cult to see that y p ( x ) = ° 1 = 6
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}