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# lecture_5 (dragged) - MA 36600 LECTURE NOTES FRIDAY JANUARY...

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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 di ff erential equation, then we determine the constants of integration from the initial condition. This di ff erential 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 left-hand 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 .
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