section%204.2

section%204.2 - x And y 2 x = y 1 x ° e − ± P x dx y 2...

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

View Full Document Right Arrow Icon
Chapter 4. Higher-Order Diferential Equations § 4.2 Reduction oF Order Suppose that we have a second order linear homogeneous DE y °° + P ( x ) y ° + Q ( x ) y =0 where P ( x ) and Q ( x ) are continuous on I . Suppose further that y 1 ( x ) is a known solution on I such that y 1 ( x ) ± =0forevery x in I . Then we can Fnd a second solution by deFning y 2 ( x )= u ( x ) y 1
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ( x ) . And y 2 ( x ) = y 1 ( x ) ° e − ± P ( x ) dx y 2 1 ( x ) dx Then the general solution is y = c 1 y 1 + c 2 y 2 . Ex: The function y 1 = x 2 is a solution of x 2 y °° − 3 xy ° + 4 y = 0. ±ind the general solution of the di²erential Equation . 1...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online