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Review Key 1. A 8 lb weight is attached to a spring. It takes a force of 10 lb to stretch the spring 8 feet. The damping force is approximately half of the velocity. If the mass is pulled 18 inches to the right of the equilibrium position and given an initial velocity of 3 ft/sec to the left, find the position function of the mass as well as the amplitude and frequency. Write the position function in terms of a single cosine function. Initial conditions: x(0) = 4/3 feet & x'(0) = 4 ft/sec Find the spring constant: F = kx »» 10lb = k(8 ft) »» k = 1.25 lb/ft The differential equation representing the motion is: mx" + cx' + kx = 0 »» x" + x' + 1.25x = 0 »» 2 x" + x' + 1.25x = 0 2 »» r + 2r + 5 = 0 »» (r+1) = -4 »» r = -1 ± 2i »» x = e-t[Acos(2t) + Bsin(2t)] Since x(0) = 4/3 »» A = 4/3 »» x = e-t[Acos(2t) + Bsin(2t)] x'(t) = -e-t[Acos(2t) + Bsin(2t)] + e-t[-2Asin(2t) + 2Bcos(2t)] x'(0) = 4 = -A + 2B and A = 4/3 »» B = 8/3 »» x(t) = e-tcos(2t) + e-tsin(2t) = e-t[cos(2t) + sin(2t) ] Now to write as a single trigonometric function: C = √ + = () + () = = and tan(α) = 2 with cos(α) > 0 & sin(α) > 0 »» α is in Quadrant I »» α = tan-1(2) = 1.107 and Time varying amplitude = √ we have x(t) = √ -t e cos e-2t ft Period = 2π/2 = π sec Frequency = 1/π ≈ 0.318 hz 2. Solve the differential equation: y(3) + 125y = 0 Characteristic equation: r3 +125 = 0 »» r3 = -125 (2t - 1.107) √ r3 = 125cisπ »» r1 = (125cisπ)య = (125)య cis( ) = 5[cos + isin ] = 5( + భ భ √ i) = + √ i Since the roots will have the same modulus and will be equally spaced within 2π we have ω= r2 = 5cis( + ) = 5cis(π) = 5[cos(π) + isin(π ) ] = -5 r3 = 5cis(+ ) = 5cis( ) = 5[cos( ) + isin( ) ] = »» Our solution is: y = c1e-5x + మ [c2cos( ఱ √ √ x) + c3sin( − √ i x)] 3. Solve the differential equation: y(3) + y" - 2y = xex Since xex is a UC function we can use the Method of Undetermined Coefficients. First find yc: the characteristic equation is r3 + r2 - 2r = 0 Using the Rational Zero Theorem we find that the roots are r = 1, r = -1 ± i and yc = c1ex + [c2cos(x) + c3sin(x)] Now consider f(x), the nonhomogenous function: f(x) = xex The typical form of yp = ( A + Bx) ex but since this overlaps with ex in the complementary solution we must multiply by the appropriate power of x to obtain: yp = x(A+Bx)ex =(Ax+Bx2) ex Now, we need to find the derivatives: (yp)' = (A + 2Bx)ex + (Ax+Bx2) ex = [A + (2B + A)x + Bx2]ex combining the like terms to ease the differentiation process. (yp)'' = [(2B + A) + 2Bx]ex + [A + (2B + A)x + Bx2]ex = [(2B+2A) + (4B + A)x + Bx2]ex (yp)''' = [(4B + A) + 2Bx]ex + [(2B+2A) + (4B + A)x + Bx2]ex = [(6B+3A) +(6B + A)x + Bx2]ex Now, to substitute into the DE: [(6B+3A) +(6B + A)x + Bx2]ex + [(2B+2A) + (4B + A)x + Bx2]ex - 2(Ax+Bx2) ex = xex »» Equating coefficients: ex-term: (6B+3A) + (2B+2A) = 0 »» 5A + 8B = 0 »» A = − B xex-term: (6B + A) + (4B + A) - 2A = 1 »» 10B = 1 »» B = and A = − ()= − Finally, yp = ( + x) ex And, yg = c1ex + [c2cos(x) + c3sin(x)] + (− + x2) ex 4. Use Variation of Parameters to solve the differential equation: y" + 4y' + 4 = e-2xlnx The characteristic equation: r2 + 4r + 4 = 0 »» (r + 2)2 = 0 and yc = c1e-2x + c2xe-2x Here we have yp = Ae-2x + Bxe-2x with y1 = e-2x and y2 = xe-2x so that (y1)' = -2e-2x and (y2)' = e-2x-2xe-2x Our system of equations we have from the 2 conditions is: (*) A'e-2x + B'xe-2x = 0 »» Multiplying by 2 »» -2A'e-2x +B'e-2x-2B'xe-2x=e-2xlnx »» 2A'e-2x + 2B'xe-2x = 0 -2A'e-2x +B'e-2x-2B'xe-2x=e-2xlnx Adding these last 2 equations yields: B'e-2x = e-2xlnx »» »» B = = xlnx - 1 = xlnx - x B' = lnx (**) Tabular Integration by Parts u→ u & du Lnx dv & v dx x From (*): A'e -2x = -B'xe -2x du → »» A' = -B'x = uv - and using (**)»» A' = -B'x = -xlnx »» A = − Again Tabular By Parts మ A= + − = మ Tabular Integration By Parts + x2 u→ u & du Lnx du → మ »» yp = ( ←v dv & v -x మ + x2)e-2x + (xlnx - x)xe-2x ←v మ and yg = yc + yp = c1e-2x + c2x e-2x + ( + x2)e-2x + (xlnx - x)xe-2x 5. Use Variation of Parameters to solve the differential equation: y" + 4y = csc2(2x) The characteristic equation: r2 + 4 = 0 »» r = ±2i and yc = c1cos(2x) + c2sin(2x) Here we have yp = Acos(2x) + Bsin(2x) and y1 = cos(2x) and y2 = sin(2x) so that (y1)' = -2sin(2x) and (y2)' = 2cos(2x) Our system of equations we have from the 2 conditions is: (*) A'cos(2x) + B'sin(2x) = 0 »» Multiplying by 2sin(2x) » 2A'sin(2x)cos(2x) + 2B'sin2(2x) = 0 -2A'sin(2x)+2B'cos(2x)=csc2(2x)» Multiplying bycos(2x) » -2A'sin(2x)cos(2x)+2B'cos2(2x) = cos(2x) csc2(2x) Adding these last two equations »» 2B'= csc(2x)cot(2x) »» B' = csc(2x)cot(2x) »» B = - csc(2x) »» A = From (*) we see that A' = -B'tan(2x) = - csc(2x) ln|csc(2x) - cot(2x)| and finally: yp = () ( And, yg ln|csc(2x) - cot(2x)|) - = yc + yp = yc = c1cos(2x) + c2sin(2x) + () ( ln|csc(2x) - cot(2x)|) - 6. Use the Series Method to find the power series solution of the form ∑ . Express in terms of elementary functions. y" - 4y = 0, y(0) = 2, y'(0) = 0 From the initial conditions we see that c0 = 2 and c1 = 0. ஶ ିଶ Subtituting y = ∑ஶ into the DE: and y" = ∑ଶ ( − 1) ିଶ ∑ஶ - 4∑ஶ =0 ଶ ( − 1)ାଶ Re-index the first series Combining the series ∑ஶ - 4∑ஶ =0 ( + 2)( + 1)ାଶ »» ∑ஶ [( + 2)( + 1)ାଶ − 4 ] = 0 »» Which yields the coefficient recursion formula: = n = 0: , n≥0 ()() = ∗ with = 2 Waiting until the end to replace with 2 for ease in calculations n = 1: = ∗ with = 0 We see that all odd-powered terms drop out! n = 2: = ∗ = ∗ (∗ ) = ∗∗∗ ()మ n = 4: = ∗ = ∗ (∗∗∗ )= ()మ ()య ! The coefficient pattern becomes obvious: = »» y = [ 2 + (2)x2 + ! () ! ()మ ! = (2)x4 + () ! ()య ! ∗ 2 but with only even powered terms. (2)x6 + ...] = 2[ 1 + !x2 + ()మ 4 x ! + ()య 6 x ! + ...] This series looks similar to the cosh(x) series except for the powers of 4 in the numerator. These can be converted to powers of 2: y = 2[ 1 + ()మ 2 x ! + ()ర 4 x ! + ()ల 6 x ! + ...] = 2[ 1 + ()మ ! + ()ర 4 x ! + ()ల ! + ...] = 2∑ஶ (2)݊ ! Which is the elementary function: y = 2cosh(2x) 7. Use the Series Method to find the power series solution of the form ∑ . Express in terms of elementary functions. y" - 2y' + y = 0, y(0) = 0, y'(0) = 1 From the initial conditions we find: c0 = 0 and c1 = 1. ିଵ ିଶ Substituting y = ∑ஶ , y' = ∑ஶ and y" = ∑ஶ into the DE: ଵ ଶ ( − 1) ିଶ ିଵ ∑ஶ - 2∑ஶ + ∑ஶ =0 ଶ ( − 1) ଵ Adjusting the indices of the first two series »» ஶ ∑ஶ - ∑ஶ =0 ( + 2)( + 1)ାଶ 2( + 1)ାଵ + ∑ Combining the series »» ∑ஶ [( + 2)( + 1)ାଶ − 2 + 1ାଵ + ] =0 »» ାଶ = ଶ ି n = 1: = n = 2: = n = 3: ଵ (ିଵ)! ሺାଶሻ(ାଵ) భ బ ଶ = ሺାଶሻ(ାଵ) = n = 0: »» cn = ଶሺାଵሻశభ ି ଶభ ଶ∗ଵ , n≥0 =1 = ଶమ ିଵ ଵ ଵ 22 ܿ2 −ܿ1 = = = 3∗2 ଷ∗ଶ ଶ ଶ! ଷିଵ ଶ ଵ 23ܿ3 −ܿ2 = = = 4∗3 ସ∗ଷ ଵଶ ଷ∗ଶ 2(4)ܿ4−ܿ3 ଵ 5∗4 = = ଵ ଷ! ସ! and we have the series ଵ ଵ ଵ ଵ ଵ ଵ ଶ! ଷ! ସ! ଶ! ଷ! ସ! y = x + x2 + x3 + x4 + x5 + ... = x[ 1 + x + x2 + x3 + x4 + ...] = xex