section%204.2

section%204.2 - ( x ) . And y 2 ( x ) = y 1 ( x ) e P ( x )...

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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
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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 dierential Equation . 1...
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