9_pdfsam_math 54 differential equation solutions odd

9_pdfsam_math 54 differential equation solutions odd - and...

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Exercises 1.2 So, dx dt + tx = 2sin2 t + t cos 2 t 6≡ sin 2 t on any interval. Therefore, x ( t ) is not a solution to the given diFerential equation. 7. We diFerentiate y = e 2 x 3 e x twice: dy dx = d dx ( e 2 x 3 e x ) = e 2 x (2) 3 e x ( 1) = 2 e 2 x +3 e x ; d 2 y dx 2 = d dx ± dy dx ² = d dx ( 2 e 2 x +3 e x ) =2 e 2 x (2) + 3 e x ( 1) = 4 e 2 x 3 e x . Substituting y , y 0 ,and y 0 into the diFerential equation and collecting similar terms, we get d 2 y dx 2 dy dx 2 y = ( 4 e 2 x 3 e x ) ( 2 e 2 x +3 e x ) 2 ( e 2 x 3 e x ) =( 4 2 2) e 2 x +( 3 3 2( 3)) e x =0 . Hence y = e 2 x 3 e x is an explicit solution to the given diFerential equation. 9. DiFerentiating the equation x 2 + y 2 = 6 implicitly, we obtain 2 x +2 yy 0 =0 y 0 = x y . Since there can be no function y = f ( x ) that satis±es the diFerential equation y 0 = x/y
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Unformatted text preview: and the diFerential equation y = x/y on the same interval , we see that x 2 + y 2 = 6 does not dene an implicit solution to the diFerential equation. 11. DiFerentiating the equation e xy + y = x 1 implicitly with respect to x yields d dx ( e xy + y ) = d dx ( x 1) e xy d dx ( xy ) + dy dx = 1 e xy y + x dy dx + dy dx = 1 ye xy + dy dx ( xe xy + 1) = 1 dy dx = 1 ye xy 1 + xe xy = e xy ( e xy y ) e xy ( e xy + x ) = e xy y e xy + x . 5...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.

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