MATH 251 Midterm: MATH 251 PSU s251Exam 1(sp04)

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15 Feb 2019
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1: both f (t, y) = needed is that y0 3t2 y 3t2 and. 1 (y 3t2)2 have to be continuous. So the equation is an exact equation: exy2. + x4 + 2y = 5: a) the roots of the characteristic equation r2 4r 5 = 0 are r = 1, 5. Therefore, the functions y1 = e t and y2 = e5t form a pair of fundamental solutions for this second order linear di erential equation: general solution is y(t) = c1y1 + c2y2 = c1e t + c2e5t, y(t) = 7. 6e5t: a) (solved as a separable equation. ) y = ln(sin t + t + e3) (explicit solution). ey = sin t + t + e3 (implicit solution), or, (solved as a rst order linear equation. ) y(t) = By abel"s theorem, w (y1, y2) = ce r p(t) dt = cer 4.

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