MATH 240 Final: MATH 240 KSU Final Exam Solutions14

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Problem 1. (25 pts) solve the initial value problem dy dx. This is a bernoulli equation; use the substitution v = 1/y to get which is linear. Multiplying both sides by the integrating factor dv dx. Since y(0) = 1, c = 1, and. E xv = x + c; v = ex(c x); y = e x. Problem 2. (25 pts) solve the initial value problem dy dx. Re-write the equation as (3y 8x)dx + (3x y)dy = 0, which is exact because. Hence the general solution is of the form f (x, y) = c, where. F (x, y) = 3xy 4x2 y2/2; 3xy 4x2 y2/2 = c or 8x2 6xy + y2 = c. Since y(0) = 1, c = 1 and. 8x2 6xy + y2 = 1 is the solution of the ivp. Alternatively, one may treat the equation as homogeneous and use the substitution y = xv which leads to x dv dx.

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