S-Class. Differential Eqns MAP2302. Prof. JLF King Wedn. 30Jan2019

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1 Differential Eqns MAP2302 S-Class Prof JLF King Wedn 30Jan209 Welcome Write DNE in a blank if the described object does not exist or if the indicated operation cannot be performed Write expressions unambiguously eg, /a + b should be bracketed either [/a] + b or /[a + b] (Be careful with negative signs!) Use f(x) notation when writing fncs; in particular, for trig and log fncs Eg, write sin(x) rather than the horrible sin x or [sin x] Recall x K := x [x ] [x 2 ] [x [K ] ], is read as N falling-factorial K S: Show no work a Prof King wears bifocals, and cannot read small handwriting Circle one: True! Yes! Who??

2 Prof JLF King Page 2 of 7 b [ A soln to f 3f ] (x) = 4 6x is polynomial f(x)= Using parameters α and β, then, the general solution to [ h 3h ] (x) = 4 6x is h α,β (x)= And the h with h(0) = 0 and h (0) = 0 is h(x)= PolyTar soln: The DOp is V := D 2 3D; so in the notation from Polynomial target, L = and N = 2 Our target polynomial, 4 6x, has degree K=, so our candidate soln has form f(x) := Ux 2 + W x Computing, D(f) = 2Ux + W and D 2 (f) = 2U Hence 4 6x Goal === V(f) = [2U 3W ] [3 2U] x So 6 = 3 2U; whence U = Also, 4 = 2U 3W = 2 3W, so W = 4 We happily conclude that f(x) = x 2 4x [You can add a constant, if desired] The aux-poly for D 2 3D factors as [z 0][z 3] Consequently, e 0 xnote === and e 3x form a fund-pair of fncs annihilated by V() [ Putting it all together, the general solution to h 3h ] (x) = 4 6x is h α,β (x) = α + βe 3x + [x 2 4x] h (x) = 0 + 3βe 3x + [2x 4] Note, To compute the α, β for the IVP, observe that 0 = h (0) = 0 + 3β + [0 4] and 0 = h(0) = α + β + [0 + 0] 3 So β = 4 3 and thus α = 4 Ie, the function solving our IVP is h(x) = e3x + [x 2 4x] Filename: Classwork/DiffyQ/D209g/s-clDfyQ209glatex

3 Prof JLF King Page 3 of 7 c The simplest soln to y + 2y + y = [t 2 + ] / e t is y(t) = PolyExp: For L := D 2 + 2D + I, fnc f, and E := e t, note [fe] = [f f]e and [fe] = [f 2f + f]e So L(fE) = f E Target G := t 2 + has degree 2 [is 3-topped], so f has form wt 4 + vt 3 + ut 2 Hence f = 2wt 2 + 6vt + 2u Equality f = G says w = 2 and v = 0 and u = 2 [ t 4 ] Thus y = 2 + t2 e t As usual, the general 2 solution adds α e t + β te t Filename: Classwork/DiffyQ/D209g/s-clDfyQ209glatex

4 Prof JLF King Page 4 of 7 d Fnc yβ (t) := is the general soln to dy dt = 8t 3 [y 5] [SoV] The particular y() with y(0) = 8 is y(t) := function has y() = SoV Soln to (d) Rewrite the given DE as And this Adding a CoI β and dividing by M(t) yields y β (t) = Q(t) M(t) + β M(t) which is what we obtained in ( ) note === 5 + β M(t) = 5 + β e 2t4, : [y 5] dy = 8t3 dt, noting that this has lost the y() 5 soln Firstly, y LhS( ) = log ( y 5 ) And t RhS( ) equals 2t 4 For an arbitrary constant α, then, : log ( y 5 ) = α + 2t 4 With β := e α, applying exp() to ( ) gives y 5 = β exp(2t 4 ) IOWords, y 5 = ±β exp(2t 4 ) Letting β vary over all numbers, we can rewrite this simply as y 5 = β exp(2t 4 ) Thus : y β (t) = 5 + β e 2t note Solving for β in 8 = 5 + β e === 5 + β, yields β = 3 So y(t) = e 2t4 is the particular soln we sought Evaluating at t= gives y() = 5 + 3e 2 FOLDE Soln to (d) Rewrite the given DE as : dy dt 8t3 y = 5 8 t 3 Using the notation from FOLDE, our C(t) = 8t 3 ; so an anti-deriv is B(t) = 2t 4 Hence the multiplier is M(t) = exp( 2t 4 ) The product function is thus P (t) = 5 8 t 3 exp( 2t 4 ) An anti-deriv of P is Q(t) = 5 exp( 2t 4 ) note === 5 M(t) Filename: Classwork/DiffyQ/D209g/s-clDfyQ209glatex

5 Prof JLF King Page 5 of 7 S2: Show no work e : Degree-N polynomial y = y(t) satisfies 4y 2 t 9 y = 5t 9 + 4t 2 Thus N = 8 [Hint: Don t compute y; just the polynomial s degree] Soln: As polynomials, let E := Deg(4y 2 ) and B := Deg(t 9 y ) If E B, then Max(E, B) must == Deg(RhS( ))=9 But E is even, thus B = 9 So y = 5, whence y = 5t + K, for some number K Hence the t 2 -coefficient of LhS( ) is 4 [ 5] 2, which does not equal 4, the t 2 -coeff of RhS( ) Thus E = B, so 2N = 9 + [N ] [DE ( ) has no const-solns, hence Deg(y ) really is N ] So N = 8 [Aside: Polynomial y(t) := t + 2t 8 satisfies ( )] Aside: For u=u(t), define operator E(u) := 4u 2 t 9 u A zerotar version of ( ) is E(u) = 0 This DE can be written as du u 2 = 4 dt t 9 ; a separable DE Integrating gives u = α + 2 t 8 = α 2 t 8 = 2αt8 + 2 t 8 So, : u α (t) = 2 t 8 2αt 8 + Caveat: Must [ [t + 2t 8 ] ] + u α be a soln to ( )? No! For note that operator E() is not linear, due to the u 2 term BTWay, note DE 4u 2 t 9 u = 0, is a Bernoulli DE, when written u = 4 / t u [ ] Multiplying by 9 u [ ] yields u 2 u = 4/t 9 CoV z := u produces FOLDE z = 4/t 9 Antidiffing gives z α (t) = α t 8 Hence With β := 2α, then, u α (t) = /[α + 2 t 8 ] β u (t) = βt 8 + 2t 8 Filename: Classwork/DiffyQ/D209g/s-clDfyQ209glatex

6 Prof JLF King Page 6 of 7 f DiffOperators P, Q, R, S are defined as P(f) := f(3) f, Q(f) := cos(3) f (3), R(f) := [cos(3) f] + f, S(f) := cos(3) + [3f ] Then P is linear: T F Q is linear: T F R is linear: T F S is linear: T F Filename: Classwork/DiffyQ/D209g/s-clDfyQ209glatex

7 Prof JLF King Page 7 of 7 g Write cos( 2i), which is real, ITOf exp() and add/sub/mul/div: cos( 2i)= And cos( 2i) lies in circle the correct interval (, 5 ] ( 5, 5 ] ( 5, 2] (2, 5] (5, 5] (5, 45] (45, ) Soln: Recall cos(z) = [e iz + e iz ]/2, for all z C Hence cos( 2i) equals 2 e e2 The e 2 /2 term is negligible, here As 2 < e < 3, so 4 < e 2 < 9 Thus 2 < 2 e2 < 9 2 End of S-Class S: 25pts S2: 60pts Total: 85pts Filename: Classwork/DiffyQ/D209g/s-clDfyQ209glatex