MATH 286 Section D1 & X1: Assignment 3Solutions for the graded problems.Section 3.1, #26 [2 points] For f (x) = 2 cos x + 3 sin x, g(x) = 3 cos x − 2 sin x, we cancheck linear (in)dependence directly by considering solutions to the equation c1f + c2g =0. If c1and c2are such that c1f + c2g = 0, then also0 = c1f (0) + c2g(0) = 2c1+ 3c2,0 = c1f0(0) + c2g0(0) = 3c1− 2c2.This system of equations clearly has only one solution c1= c2= 0. Therefore f and gare linearly independent.Note: We could have also computed the Wronskian W( f , g), to determine linear in-dependence using the Wronskians of solutions theorem. This is because, f , g are twosolutions to the ODEy00+ y = 0.Section 3.1, #42 [2 points] For 35y00− y0− 12y = 0, the characteristic polynomial isP(r) = 35r2− r − 12, which by the quadratic formula has rootsr0=1 ±p1 − 4(35)(−12)2(35)=35,−47.Therefore by Theorem 5 in Section 3.1, the general solution isy(x) = c1e3x/5+ c2e−4x/7.Section 3.1, #46 [1 point] Given the general solution y(x) = c1e10x+ c2e100x, we knowthat the characteristic polynomial must be of degree two and have roots 10, 100. There-foreP(r) = (r − 10)(r − 100) = r2− 110r + 1000 =⇒ y00− 110y0+ 1000y = 0is a homogeneous linear ODE with constant coefficients with this general solution.Section 3.2, #30 [1 point] We are given that y1(x) = x and y2(x) = x2are solutions tothe ODEx2y00− 2xy0+ 2y = 0.These functions are linearly independent on the whole real line because the only linearcombinationc1y1(x) + c2y2(x) = c1x + c2x2= 0,1which is valid for all x ∈ R is the trivial one where c1= c2= 0. (Indeed, if x = 1, thenwe havec1+ c2= 0,and when x = 2, we have2c1+ 4c2= 0,and these two equations give c1= c2= 0.) Next we complute their Wronskian:W(x) = W(y1, y2)(x) = y1(x)y02(x) − y2(x)y01(x) = x · 2x − x2· 1 = x2,which vanishes at x = 0. The reason that the condition W(0) = 0 does not contradictpart (b) of Theorem 3 in Section 3.2 is that in order for Theorem 3 to apply we must firstexpress our original equation in the standard formy(2)+ p(x)y0+ q(x)y = 0.In our case p(x) = −2/x and q(x) = 2/x2, and these functions are not continuous atx = 0. Since Theorem 3 only says something about the non-vanishing of W(y1, y2) onan interval I where p, q are continuous, Theorem 3 does not apply to the interval I = R,since it contains x = 0.Section 3.2, #12 [1 point] The chracteristic polynomial associated to the equationy(4)− 3y000+ 3y00− y0= 0is P(r) = r4− 3r3+ 3r2− r = r(r − 1)3. This gives solutions y1(x) = 1 (corresponfingto the root r = 0) and y2(x) = ex, y3(x) = xex, y4(x) = x2ex(correspoding to themultiplicity 3 root r = 1). These functions are linearly independent on R, so the generalsolution isy(x) =4∑i=1ciyi(x) = c1+ c2ex+ c2xex+ c4x2ex.Section 3.3, #18 [1 point] The equationy(4)= 16y ⇐⇒ y(4)− 16y = 0has characteristic polynomial P(r) = r4− 16 = (r − 2)(r + 2)(r − 2i)(r + 2i). So the rootsare ±2, ±2i and the general solution isy(x) = c1e2x+ c2e−2x+ c3cos(2x) + c4sin(2x).2Section 3.3, #24 [2 points] For the IVP2y(3)− 3y00− 2y0= 0; y(0) = 1, y0(0) = −1, y00(0) = 3The characteristic polynomial isP(r) = 2r3− 3r2− 2r = 2r(r2−32r − 1) = 2r(r − 2)(r +12),which has roots 0, 2, −1/2. Thus the general solution isy(x) = c1+ c2e2x+ c3e−x/2.To solve for the initial conditions, we get a system of equationsy(0) = 1 = c1+ c2+ c3y0(0) = −1 = 2c2−12c3y00(0) = 3 = 4c2+14c3.Solving this system of three equations and three unknowns gives c1= −7/2, c2=1/2, c3= 4. Thus the particular solution isy(x) = −7/2 + (1/2)e2x+
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