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Unformatted text preview: MA 36600 LECTURE NOTES: FRIDAY, JANUARY 23 First Order Nonlinear Equations (cont’d) Example. Say we wish to solve the initial value problem dy dt = 3 t 2 + 4 t + 2 2( y 1) , y (0) = 1 . First we find the general solution to the differential equation, then we determine the constants of integration from the initial condition. This differential equation is separable because we have dy dt = Y ( y ) T ( t ) where Y ( y ) = 1 2( y 1) , T ( t ) = 3 t 2 + 4 t + 2 . We can bring all of the y terms to one side, and all of the t terms to the other: 2( y 1) dy dt = ( 3 t 2 + 4 t + 2 ) . The lefthand side is 2( y 1) dy dt = d dt y 2 2 y = ⇒ d dt y 2 2 y = 3 t 2 + 4 t + 2 . Upon integrating both sides, we see that y 2 2 y = t 3 + 2 t 2 + 2 t + C for some constant C . These are the level curves. For the initial condition, set t = 0: C = y (0) 2 2 y (0) = ( 1) 2 2 · ( 1) = 3 . Hence the solution to the initial value problem is the implicit solution y 2 2 y = t 3 + 2 t 2 + 2 t + 3 . The graph of this solution is called an elliptic curve . Modeling with First Order Equations We give several examples of first order equations by discussing several types of mathematical models....
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This note was uploaded on 05/24/2011 for the course MA 36600 taught by Professor Ma during the Spring '09 term at Purdue.
 Spring '09
 MA
 Linear Equations, Equations

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