MTH 311 Study Guide - Midterm Guide: System Of Linear Equations, Wronskian, Linear Independence

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3. 1: second order linear equations: theorem on existence and uniqueness for linear equations, theorem on wronskians of solutions, theorem on general solutions of homogeneous equations, linear independence, wronskian, homogeneous linear equation, theorem on principle of superposition. Two functions are said to be linearly independent if neither is a constant multiple of the other. (cid:1827)(cid:4666)(cid:1876)(cid:4667)(cid:1877) +(cid:1828)(cid:4666)(cid:1876)(cid:4667)(cid:1877) +(cid:1829)(cid:4666)(cid:1876)(cid:4667)(cid:1877)=(cid:882) Let (cid:1877)(cid:2869) and (cid:1877)(cid:2870) be two solutions of the homogeneous linear equation. Then, if (cid:1855)(cid:2869) and (cid:1855)(cid:2870) are constants, the linear combination (cid:1877)=(cid:1855)(cid:2869)(cid:1877)(cid:2869)+(cid:1855)(cid:2870)(cid:1877)(cid:2870) is also a solution. Suppose the functions (cid:1868),(cid:1869),(cid:1858) are continuous on the open interval containing the point (cid:1853). Then, given any two numbers (cid:1854)(cid:2868) and (cid:1854)(cid:2869), the equation (cid:1877) +(cid:1868)(cid:4666)(cid:1876)(cid:4667)(cid:1877) +(cid:1869)(cid:4666)(cid:1876)(cid:4667)(cid:1877)=(cid:1858) has a unique solution on that satisfies (cid:1877)(cid:4666)(cid:1853)(cid:4667)=(cid:1854)(cid:2868) and (cid:1877) (cid:4666)(cid:1853)(cid:4667)=(cid:1854)(cid:2869). i. e. , (cid:1863) ,(cid:1863)(cid:1858)(cid:2869) (cid:1858)(cid:2870). (cid:1859) Given two functions (cid:1858) and (cid:1859), the wronskian is the determinant, =|(cid:1858) (cid:1858) (cid:1859) |=(cid:1858)(cid:1859) (cid:1858) (cid:1859). Suppose that (cid:1877)(cid:2869) and (cid:1877)(cid:2870) are two solutions of (cid:1877) +(cid:1868)(cid:4666)(cid:1876)(cid:4667)(cid:1877) +(cid:1869)(cid:4666)(cid:1876)(cid:4667)(cid:1877)=(cid:882).

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Let r in and u =(1 1 0), v = (1 4 1), w = (r2 1 r2), b = (3+2r 5+12r 2r). For which values r in is the set {u. v, w} linearly independent? For which values r in is the vector b a linear combination of u. v and w? For which of these values of r can b be written as a linear combination of u. v and w in more than one way? The set 3 of all column vectors of length three, with real entries, is a vector space. Is the subset B = {(x y z) in 3 | xy + yz = 0} a subspace of 3? Justify your answer. Show that the set of all twice differentiable functions f : satisfying the differential equation sin(x) f"(x) + x2f(x) = 0 is a vector space with respect to the usual operations of addition of functions and multiplication by scalars. Here, f" denotes the second derivative of f. Let S be the following subset of the vector space 3 of all real polynomials p of degree at most 3: S = {p in 3 | p(l) = 0, p'(l) = 0}, where p' is the derivative of p. Determine whether S is a subspace of 3. Determine whether the polynomial q(x) = x - 2x2 + x3 is an element of S. u = (1 1 0), v = (1 4 1), w = (r2 1 r2) if {u, v, w} is linearly independent set 1(4r2 - 1)-1(r2 - r2) ne 0 4r2 ne 1 r ne plusmn1/2 this set is linearly independent set for values of e ne plusmn1/2 - - - - - - - (3+2r 5+12r 2r) = a1 (1 1 0) + a2 (1 4 1) + a3 (r2 1 r2) this forms system of equations in a1, a2, a3 if 4r2 - 1 ne 0 then system has unique solution if 4r2 - 1 = 0 and -2 - 4r = 0 rArr r = -1/2, the system has more than one solution if 4r2 - 1 ne 0 b can be written as linear combination of u,v,w as uniquely if r = -1/2 then b can be written as linear combination of u,v,w more than one way - - - - - - - - - (1 1 -1) in B and (1 0 1) in B (1 1 -1) + (1 0 1) = (2 1 0) not in B since 2times1+1times0 ne 0 so addition of vector is failed, this set is not forms subspace consider the following differential equation (sinx)f"(x) + x2f (x) = 0 each solution is continuous function the set of all solutions is subset of space of set of all continuous functions n from R to R let f,g are two solutions (sinx)f"(x) + x2f (x) = 0 (sinx)g"(x)+x2g(x) = 0 ccheck whether af+bg is solution or not (sinx)f"(x) + x2f(x) = 0 rArr (sinx) af"(x) + x2?f (x) = 0 (sinx)g"(x) + x2g (x) = 0 rArr (sinx)bg" (x) + x2bg (x) = 0 (sinx)(af"(x)+bg"(x)) +x2 (af (x) + bg(x)) = 0 af+bg is also solution to differential equation the set of all solutions to differential equation forms subspace - - - - - - - - - - consider the following subset of P3 {p in P3/.p(l)=0, p'(l) = 0} let p, q in S rArr p(l)= 0, p'(l) = 0 and q(l) = 0, q'(l) = 0 (ap + bq) (l) = ap (l) +b q(l) = a (0) + b (0) = 0 (ap + bq)'(1) = ap'(l) +bq'(1) = a (0) + b(0) = 0 p, q in S rArr ap +bq in S then S is subspace q(x) = x - 2x2 + x3 q(1) = 1 - 2 + 1 = 0 q'(x)=1 - 4x + 3x2 q'(l) = 1 - 4 + 3 = 0 q is element of S

5. (18 pts) Circle True if the statement is ALWAYS true; circle False otherwise. No explanations are required. Let f be a one-to-one function and f its inverse. (a) If the point (2, 5) lies on the graph off, then the point (5,2) lies on the graph of f. True False f (2) = 5 f-1 (5) = 2 (b) sin-1 (T) = 0 True False Since >1, ris not in the domain of sin-?; | or sin-1 (7) #0, since sin (0) = 0 # Iff and g are two functions defined on (-1, 1), and if (c) limg (x) = 0, then it must be true that lim[ f(x) g (x)] = 0. True False Let f (x) = 1 / x, if x #0, and f (0) = 4 and g(x) = x. Then lim g (x) = lim (x) = 0, but Lim ( f (x) g(x)] = lim [() x] = 18 0. Iffis continuous on (-1, 1), and if f (0) = 10 and lim g (x) = 2, (a) True False then lin (= 5. Because lim : (*) - Limx-of (*) .f(0) = 20 = 5 If f is continuous on [1, 3], and if f(1) = 0 and f(3) = 4, then the equation f(x) = (e) has a solution in (1, 3). True False Since f continuous on [1, 3), f(1) = 0 < < 4 = f (3), by the Intermidiate Value Theorem there exists a cin (1, 3) such that f (c) = . (f) Let f be a positive function with vertical asymptote x = 5. Then lim f (x) = +0. True False Let f (x) = 1, if x < 5, and f (x) ==, ifx>5. Then lim f (x) = +00, and therefore, X = 5 is a vertical asymptote. 5 - X So, f is a positive function with vertical asymptote x = 5 but lim f (x) + +co, since lim f (x) = + 0 + lim f (x) = 1, which means that lim f (x) does not exist.

MTH 311 Study Guide - Midterm Guide: System Of Linear Equations, Wronskian, Linear Independence

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