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lecture_7 (dragged) 1

# 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 = µ ( t, y ) ∈ R 2 ¶ ¶ ¶ ¶ − 2 < t < ∞ , −∞ < y < 1 · . Figure 1. Plot o² 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 di±erential equation is always in the ²orm dy dt = G ( t, y ) ²or some ²unction G ( t, y ). We say that such a di±erential equation is an autonomous equation i² this ²unction is a ²unction o² y alone i.e., i² G ( t, y ) = f ( y )...
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