# 285lect25 - 1 Lecture 26 Example 1 y 00 4y = cosh 2x Recall...

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1 Lecture 26 Example 1 y 00 4 y = cosh 2 x: Recall that cosh t = e t + e t 2 ; sinh t = e t e t 2 ; cosh 0 t = sinh t; sinh 0 t = cosh t: Complimentary equation: y 00 4 y = 0 : Characteristic equation: r 2 4 = 0 ) r 1 = 2 and r 2 = 2 . Because r = 2 is a root of multiplicity 1, according to the table, (b), we should consider L [ y 1 p ] = e 2 x 2 and L [ y 2 p ] = e 2 x 2 and write y p ( x ) = x ae 2 x + be 2 x ± . Because e 2 x = cosh 2 x + sinh 2 x 2 and e 2 x = cosh 2 x sinh 2 x 2 ; it is not di¢ cult to see that we can write solution in the following form: y p ( x ) = x ( A cosh 2 x + B sinh 2 x ) : y 0 p ( x ) = A cosh 2 x + B sinh 2 x + x (2 A sinh 2 x + 2 B cosh 2 x ) ; y 00 p ( x ) = 2 A cosh 2 x + 2 B sinh 2 x + 2 A sinh 2 x + 2 B cosh 2 x + x (4 A cosh 2 x + 4 B sinh 2 x ) = 4 ( A sinh 2 x + B cosh 2 x ) + 4 x ( A cosh 2 x + B sinh 2 x ) : 1

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y 00 4 y = 4 ( A sinh 2 x + B cosh 2 x ) + 4 x ( A cosh 2 x + B sinh 2 x ) 4 x ( A cosh 2 x + B sinh 2 x ) = 4 ( A sinh 2 x + B cosh 2 x ) ) 4 ( A sinh 2 x
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285lect25 - 1 Lecture 26 Example 1 y 00 4y = cosh 2x Recall...

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