University:
Whitman College
Course:
MATH-125 | Calculus I
Academic year:

2022
Views:

137

Pages:

1
Author:

Evelyn Harrell

Section 2.6, Problem 26 Write Up Problem: Given that 2 1 y 0 + y = 1 − t, y(0) = y0 3 2 Find the value of y0 so that the solution just touches, but does not cross, the t−axis. SOLUTION: First, if y(t) touches, but does not cross, the t−axis at some point, t∗, then: y(t∗) = 0, y 0 (t∗) = 0 y 00 (t∗) 6= 0 Secondly, the differential equation, y 0 + 32 y = 1 − 12 t, must hold for y at all t, and in particular, for t∗. Putting this together, 2 1 0 + 0 = 1 − t∗ ⇒ t∗ = 2 3 2 Now, our solution also satisfies: y(2) = 0 and y 0 (2) = 0. We solve the differential equation using an integrating factor to get: 2t ye 3 0 2t 1 2t = e 3 − te 3 2 Integrate the right hand side by Integration by Parts to get: 2t 3 2t 1 3 2t 9 2t 3 3 3 3 ye = e − te − e +C 2 2 2 4 Simplifying, we get that: y= and, using y(0) = y0 , we get that C = y0 − y= 2 21 3 − t + Ce− 3 t 8 4 21 8 . So, 2 21 21 3 − t + (y0 − )e− 3 t 8 4 8 Finally, using y(2) = 0, we solve for y0 : 0= 2 21 3 21 − 2 + (y0 − )e− 3 2 8 4 8 From which we get that: y0 = 3 4/3 21 e + (1 − e4/3 ) ≈ −1.64 2 8 1

Calculus Section 2.6, Problem 26 Write Up

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