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MATH-M 344 Chapter 7: M344 7.4 Notes (Jan. 31)

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16 Feb 2017

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Benjamin Melinand

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M344 section 7. 4 notes- basic theory of systems of first order linear equations. Consider system {(cid:1876)(cid:2869) =(cid:2869)(cid:2869)(cid:4666)(cid:4667)(cid:1876)(cid:2869)+(cid:2869)(cid:2870)(cid:4666)(cid:4667)(cid:1876)(cid:2870)+(cid:1710)+(cid:2869)(cid:4666)(cid:4667)(cid:1876)+(cid:2869)(cid:4666)(cid:4667) (cid:1876) =(cid:2869)(cid:4666)(cid:4667)(cid:1876)(cid:2869)+(cid:2870)(cid:4666)(cid:4667)(cid:1876)(cid:2870)+(cid:1710)+(cid:4666)(cid:4667)(cid:1876)+(cid:4666)(cid:4667) ; closely parallels single linear differential equation of order theory of order odes) Consider (cid:1876)(cid:2869)=(cid:2869)(cid:4666)(cid:4667),(cid:1876)(cid:2870)=(cid:2870)(cid:4666)(cid:4667), ,(cid:1876)=(cid:4666)(cid:4667) to be components of vector =(cid:4666)(cid:4667) Let polynomials (cid:2869)(cid:2869)(cid:4666)(cid:4667), ,(cid:4666)(cid:4667) be elements of matrix (function) (cid:4666)(cid:4667: can rewrite system as =(cid:4666)(cid:4667)+(cid:4666)(cid:4667) Directly follows that each scalar function (cid:2869)(cid:2869), ,(cid:2869), , continuous on. By theorem 7. 1. 2, we have enough to guarantee existence of solutions to system on. Vector =(cid:4666)(cid:4667) is a solution of system if its components satisfy system of equations. Convenient to work with homogeneous equation =(cid:4666)(cid:4667) where (cid:4666)(cid:4667)=(cid:2777: for specific solutions of this system, write (cid:2778)(cid:4666)(cid:4667)=((cid:1876)(cid:2869)(cid:2869)(cid:4666)(cid:4667),(cid:1876)(cid:2870)(cid:2869)(cid:4666)(cid:4667), ,(cid:1876)(cid:2869)(cid:4666)(cid:4667)), ,(cid:2193)(cid:4666)(cid:4667)= ((cid:1876)(cid:2869)(cid:3038)(cid:4666)(cid:4667),(cid:1876)(cid:2870)(cid:3038)(cid:4666)(cid:4667), ,(cid:1876)(cid:3038)(cid:4666)(cid:4667)), , (cid:1876)(cid:3036)(cid:3037)(cid:4666)(cid:4667) is (cid:1861) component of (cid:1862) solution (cid:2192)(cid:4666)(cid:4667) (cid:2191) are linearly independent at given iff their wronskian is nonzero there (cid:1877) +(cid:1877)=(cid:882: let =(cid:4666)(cid:1877),(cid:1877) (cid:4667, equation becomes +[(cid:882) (cid:883) (cid:882)]=(cid:2777) (cid:883) (cid:2870)+(cid:883)=(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