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# lecture_7 - MA 36600 LECTURE NOTES WEDNESDAY JANUARY 28...

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MA 36600 LECTURE NOTES: WEDNESDAY, JANUARY 28 Linear vs. Nonlinear Solutions Example 1. Consider the initial value problem t y 0 + 2 y = 4 t 2 , y (1) = 2 . What is the largest region R such that there exists a unique solution? We solve this problem in three steps: first we find the general solution to the differential equation, then we find the particular solution using the initial conditions, and finally we compute the region R by considering properties of this solution. We begin by finding the general solution to the differential equation. (Recall that we solved this equation in Lecture 5.) Upon dividing this equation by t , we see that dy dt + p ( t ) y = g ( t ) where p ( t ) = 2 t , g ( t ) = 4 t. The integrating factor is μ ( t ) = exp Z t p ( τ ) = exp Z t 2 τ = t 2 . Now multiply both sides of the differential equation by this function: μ ( t ) dy dt + μ ( t ) p ( t ) y = μ ( t ) g ( t ) t 2 dy dt + 2 t y = 4 t 3 d dt t 2 y = 4 t 3 = t 2 · y ( t ) = t 4 + C for some constant C . Now we can find the particular solution to the initial value problem. If we set t = 1 we find the equation 2 = 1 + C . This gives C

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