MATH 1005 Study Guide - Midterm Guide: Frobenius Method

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turquoisegnu128

24 Oct 2018

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Math1005c solution-test 2 4:35 5:25, feb. 27 2013. [3] the general solution of x2y(cid:48)(cid:48) 7xy(cid:48) + 16y = 0, x > 0 is given by: c1e4x + c2xe4x, c1 ln(x4) + c2x ln(x4, c1x4 + c2x5, c1x4 + c2x4 ln x (**, none of these. [4] find the general solution of the equation y(cid:48)(cid:48) + y(cid:48) 2y = 9ex. Particular solution is in the form yp = axex, y(cid:48) = aex + axex, y(cid:48)(cid:48) = aex + aex + axex, after plug in a = 3. The general solution is y = c1ex + c2e 2x + 3xex. [5] find the general solution by variation of parameters of the equation x2y(cid:48)(cid:48) 4xy(cid:48) + 6y = x4ex. To nd a particular solution; the standard form is y(cid:48)(cid:48) 4 xy(cid:48) + 6. Then u1 =(cid:82) y2f (x) (cid:82) exdx = ex, so yp = (ex xex)x2 + x3ex = x2ex.

Related Questions

[1] 1. (a) Solve 2 – 31 < 7. Solution: We have – 7<-3 < 7 which gives -4 0 on (-0, 1] U [2,). Next, we need to see where Vr2 – 3r +2 -1 > 0. If r2 - 3.c + 2 – 1 > 0, then V.22 – 3.0 +2 > . Observe that this is definitely true if r < 0. If r > 0, then squar- ing both sides we get 22 - 3.+2 > 1 2 > 3. V A VICO Thus, the domain of f is (-00, ). [2] (c) Find the exact value of tan(sec-14). Solution: Let y = sec-14. Then, sec y = 4. From the diagram we get that tan(sec-? 4) = tan y = 15 [2] (d) Find all r such that sin(2.x) = COS I. Solution: We have 2 sin r cos r = COS I, SO 0 = 2 sin Cos I -Cos I = cos .r (2 sin c - 1). Hence, sin(2x) = cost when cos x = 0, or when sin r = Thus, r= +ik, kez, I= +2nk, ke Z, or r= +2nk, kez [2] (e) State the precise mathematical definition of lim f(x) = 0. Solution: For every e > 0 there exists a 8 >0 such that if 0 < x-al <, then f(c) >

please answer all questions,