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30. } d d y x } 5 y ˇ x w E } d y y } 5 E x 1/2 dx ln ) y ) 5 } 2 3 } x 3/2 1 C ) y ) 5 e (2/3) x 3/2 1 C y 5 6 e C e (2/3) x 3/2 y 5 Ae (2/3) x 3/2 Initial condition: y (0) 5 1 1 5 Ae 0 1 5 A Solution: y 5 e (2/3) x 3/2 31. (a) Note that k . 0 and M . 0, so the sign of } d d P t } is the same as the sign of ( M 2 P )( P 2 m ). For m , P , M , both M 2 P and P 2 m are positive, so the product is positive. For P , m or P . M , the expressions M 2 P and P 2 m have opposite signs, so the product is negative. (b) } d d P t } 5 } M k } ( M 2 P )( P 2 m ) } d d P t } 5 } 12 k 00 } (1200 2 P )( P 2 100) } d d P t } 5 k } d d P t } 5 } 1 1 1 2 } k } d d P t } 5 } 1 1 1 2 } k 1 } 1200 1 2 P } 1 } P 2 1 100 } 2 } d d P t } 5 } 1 1 1 2 } k E 1 } 1200 1 2 P } 1 } P 2 1 100 } 2 dP 5 } 1 1 1 2 } k dt 2 ln ) 1200 2 P ) 1 ln ) P 2 100 ) 5 } 1 1 1 2 } kt 1 C ln ) } 1 P 20 2 0 1 2 00 P } ) 5 } 1 1 1 2 } kt 1 C } 1 P 20 2 0 1 2 00 P } 5 6 e C e 11 kt /12 } 1 P 20 2 0 1 2 00 P } 5 Ae 11 kt /12 P 2 100 5 1200 Ae 11 kt /12 2 APe 11 kt /12 P (1 1 Ae 11 kt /12 ) 5 1200 Ae 11 kt /12 1 100 P 5 (c) 300 5 300(1 1 A ) 5 1200 A 1 100 300 2 100 5 1200 A 2 300 A 200 5 900 A A 5 } 2 9 } P ( t ) 5 P ( t ) 5 P ( t ) 5 (d) [0, 75] by [0, 1500] Note that the slope field is given by } d d P t } 5 } 1 0 2 . 0 1 0 } (1200 2 P )( P 2 100). (e) } d d P t } 5 } M k } ( M 2 P )( P 2 m ) } ( M 2 P M )( P 2 m ) } } d d P t } 5 k } M M 2 m } } ( M 2 M P ) 2 ( P m 2 m ) } } d d P t } 5 k } d d P t } 5 } M M 2 m } k 1 } M 2 1 P } 1 } P 2 1 m } 2 } d d P t } 5 } M M 2 m } k E 1 } M 2 1 P } 1 } P 2 1 m } 2 dP 5 E } M M 2 m } k dt 2 ln ) M 2 P ) 1 ln ) P 2 m ) 5 } M M 2 m } kt 1 C ln ) } M P 2 2 m P } ) 5 } M M 2 m } kt 1 C } M P 2 2 m P } 5 6 e C e ( M 2 m ) kt / M } M P 2 2 m P } 5 Ae ( M 2 m ) kt / M P 2 m 5 ( M 2 P ) Ae ( M 2 m ) kt / M P (1 1 Ae ( M 2 m ) kt / M ) 5 AMe ( M 2 m ) kt / M 1 m P 5 P (0) 5 } A 1 M 1 e 0 A 1 e 0 m } 5 } A 1 M 1 1 A m } P (0)(1 1 A ) 5 AM 1 m A ( P (0) 2 M ) 5 m 2 P (0) A 5 } P m (0 2 ) 2 P (0 M ) } 5 } M P (0 2 ) 2 P (0 m ) } Therefore, the solution to the differential equation is P 5 where A 5 } M P (0 2 ) 2 P (0 m ) } . AMe ( M 2 m ) kt / M 1 m }} 1 1 Ae ( M 2 m ) kt / M AMe ( M 2 m ) kt / M 1 m }} 1 1 Ae ( M 2 m ) kt / M ( P 2 m ) 1 ( M 2 P ) }}} ( M 2 P )( P 2 m ) 300(8 e 11 kt /12 1 3) }} 9 1 2 e 11 kt /12 1200(2) e 11 kt /12 1 100(9) }}} 9 1 2 e 11 k /12 1200(2/9) e 11 kt

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