MATH 2850 Quiz: MATH 2850 Iowa 2850 Sp 2017 Solution to Quiz 4
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Find the general solution to ty 2y = t3et 8. Also nd the solution that passes thru the point (1, 3). How does the solution passing thru (1, 3) behave as t ? ty 2y = t3et 8 implies 1y + ( 2. Integrating factor: u = e p(t)dt = e 2 t dt = e 2ln|t| = eln(|t| 2) = t 2 t )y = t2et 8t 1. Let u(t) = t 2 t 2y 2t 3y = et 8t 3 (t 2y) = et 8t 3. (t 2y) dt = (et 8t 3)dt t 2y = et + 4t 2 + c y = t2et + 4 + ct2. Check: (t 2y) = t 2y 2t 3y y(1) = 3: 3 = e + 4 + c. thus c = 1 e. Thus y = t2(et e 1) + 4 + . General solution: y = t2et + 4 + ct2.
