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# 285hw1 - HW1 1 Sec 1.1 10 Let y(x = x ln x Then 1 x dy(x =1...

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HW1 1. Sec. 1.1: 10. Let y ( x ) = x ° ln x . Then dy ( x ) dx = 1 ° 1 x and d 2 y ( x ) dx 2 = 1 x 2 : We have x 2 y 00 + xy 0 ° y = x 2 ° 1 x 2 ± + x ° 1 ° 1 x ± ° ( x ° ln x ) = ln x: Now let y ( x ) = 1 x ° ln x: Then dy ( x ) dx = ° 1 x 2 ° 1 x and d 2 y ( x ) dx 2 = 2 x 3 + 1 x 2 : We have x 2 y 00 + xy 0 ° y = x 2 ° 2 x 3 + 1 x 2 ± + x ° ° 1 x 2 ° 1 x ± ° ° 1 x ° ln x ± = ln x: 2. Sec. 1.1: 15. y ( x ) = e rx . Then y 0 ( x ) = re rx and y 00 ( x ) = r 2 e rx . We have y 00 + y 0 ° 2 y = r 2 e rx + re rx ° 2 e rx = ² r 2 + r ° 2 ³ e rx = 0 ; whence r 2 + r ° 2 = 0 , Solution is: ° 2 ; 1 r 2 + r ° 2 = 0 : Hence r 1 = ° 2 and r 2 = 1 . Solutions: y 1 ( x ) = e ° 2 x and y 2 ( x ) = e x : 1

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3. Sec. 1.1 :21. Substitute y ( x ) = Ce ° x 3 into the equation y 0 + 3 x 2 y = 0 : We °nd that dy ( x ) dx = ° 3 Cx 2 e x 3 ; then y 0 + 3 x 2 y = ° 3 Cx 2 e x 3 + 3 x 2 ´ Ce ° x 3 µ = 0 : 4. Sec. 1.1: 31. The equation of the tangent line to the graph of g at the point ( x; y ) is Y ° y = g 0 ( x ) ( X ° x ) . Now because ( ° y; x ) is on the line, we have x ° y = g 0 ( x ) ( ° y ° x ) , whence g 0 ( x ) ( x + y ) = x ° y ) g 0 ( x ) = y ° x x + y : Because y = g ( x ) ; y 0 ( x ) = y ° x x + y : 5. Sec. 1.1: 33.
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285hw1 - HW1 1 Sec 1.1 10 Let y(x = x ln x Then 1 x dy(x =1...

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