MA 36600 LECTURE NOTES: FRIDAY, JANUARY 23First Order Nonlinear Equations (cont’d)Example.Say we wish to solve the initial value problemdydt=3t2+ 4t+ 22 (y−1),y(0) =−1.First we find the general solution to the differential equation, then we determine the constants of integrationfrom the initial condition.This differential equation is separable because we havedydt=Y(y)T(t)whereY(y) =12 (y−1),T(t) = 3t2+ 4t+ 2.We can bring all of theyterms to one side, and all of thetterms to the other:2 (y−1)dydt=3t2+ 4t+ 2.The left-hand side is2 (y−1)dydt=ddty2−2y=⇒ddty2−2y= 3t2+ 4t+ 2.Upon integrating both sides, we see thaty2−2y=t3+ 2t2+ 2t+Cfor some constantC. These are the level curves. For the initial condition, sett= 0:C=y(0)2−2y(0) = (−1)2−2·(−1) = 3.Hence the solution to the initial value problem is the implicit solutiony2−2y=t3+ 2t2+ 2t+ 3.
This is the end of the preview.
access the rest of the document.