lecture_7 (dragged) 1 - y = 1. The square root is only...

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2 MA 36600 LECTURE NOTES: WEDNESDAY, JANUARY 28 First we fnd the general solution to the di±erential equation. (Again, we solved this equation in Lecture 5.) This di±erential equation is separable; we can bring all o² the y terms to one side, and all o² the t terms to the other. dy dt = ° 3 t 2 +4 t +2 ± / (2 y 2) (2 y 2) dy dt = ° 3 t 2 +4 t +2 ± d dt ² y 2 2 y ³ =3 t 2 +4 t +2 = y 2 2 y = t 3 +2 t 2 +2 t + C ²or some constant C . These are the level curves i.e., this is the implicit solution. Second, we fnd the particular solution to the initial value problem. This curve must go through the point ( t 0 ,y 0 )=(0 , 1): y (0) 2 2 y (0) = 0 + C = C =( 1) 2 2 · ( 1) = 3 . We may solve ²or y = y ( t ) to fnd the explicit solution. Upon completing the square with respect to y we fnd that y 2 2 y = t 3 +2 t 2 +2 t +3 y 2 2 y +1= t 3 +2 t 2 +2 t +4 ( y 1) 2 =( t + 2) ° t 2 +2 ± = y ( t )=1 ± ´ ( t + 2) ( t 2 + 2) We choose the negative square root because y (0) = 1. A plot o² this curve can be ²ound in Figure 1. Note the line o² symmetry
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Unformatted text preview: y = 1. The square root is only defned when ( t +2) ( t 2 +2) is nonnegative. The polynomial t 2 +2 is always positive, so we also need t 2. Since we choose the negative square root, the region R is R = ( t, y ) R 2 2 < t < , < y < 1 . Figure 1. Plot o y 2 2 y = t 3 + 2 t 2 + 2 t + 3-4.8-4-3.2-2.4-1.6-0.8 0.8 1.6 2.4 3.2 4 4.8-3.2-2.4-1.6-0.8 0.8 1.6 2.4 3.2 Population Dynamics Autonomous Equations. Recall that a frst order dierential equation is always in the orm dy dt = G ( t, y ) or some unction G ( t, y ). We say that such a dierential equation is an autonomous equation i this unction is a unction o y alone i.e., i G ( t, y ) = f ( y )...
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This note was uploaded on 11/30/2011 for the course MATH 366 taught by Professor Edraygoins during the Spring '09 term at Purdue University-West Lafayette.

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