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IUPUI MATH 22100 - Exam 4 Review

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Exam #4 Review(1) # 9.1.7) Solve 2y00− 5y0+ 2y = 02r2− 5r + 2 = 0 ⇒ (2r − 1)(r − 2) + 0 ⇒ r =12, 2y = C1ex/2+ C2e2x(2) # 9.1.11) Solve y00+ 10y0+ 25y = 0r2+ 10r + 25 = 0 ⇒ (r + 5)2= 0 ⇒ r = {−5, −5}y = C1e−5x+ C2xe−5x(3) # 9.1.21) Solve y(4)− 5y000+ 4y00= 0r4− 5r3+ 4r2= 0 ⇒ r2(r2− 5r + 4) = 0 ⇒ r2(r − 1)(r − 4) = 0 ⇒r = {0, 0, 1, 4}y = C1e0x+ C2xe0x+ C3ex+ C4e4x= C1+ C2x + C3ex+ C4e4x(4) # 9.1.25) Solve y00+ 8y0+ 41y = 0r2+ 8r + 41 = 0 ⇒ r2+ 8r + 16 = −41 + 16 = −25 ⇒(r + 4)2= −25 ⇒ r = −4 ± 5iy = e−4x(C1sin 5x + C2cos 5x)(5) # 9.1.39) Solve the IVP y00+ 4y0+ 4y = 0, y(0) = 3, y0(0) = 2r2+ 4r + 4 = 0 ⇒ (r + 2)2= 0 ⇒ r = {−2, −2}y = C1e−2x+ C2xe−2xy0= −2C1e−2x+ C2(−2xe−2x+ e−2x) = −2C1e−2x− 2C2xe−2x+ C2e−2x3 = C12 = −2C1+ C2= −6 + C2⇒ C2= 8y = 3e−2x+ 8xe−2x(6) # 9.2.3) Solve y00− 2y0+ y = 3r2− 2r + 1 = 0 ⇒ (r − 1)2⇒ r = {1, 1}yh= C1ex+ C2xexyp= C, y0p= 0, y00p= 0y00p− 2y0p+ yp= 0 − 2 · 0 + A = 3 ⇒ A = 3yp= 3y = yh+ yp= C1ex+ C2xex+ 312(7) #9.2.13) Solve y000+ y00= −2r3+ r2= 0 ⇒ r2(r + 1) = 0 ⇒ r = {0, 0, −1}yh= C1e0x+ C2xe0x+ C3e−x= C1+ C2x + C3e−xyp= Cx2, y0p= 2Cx, y00p= 2C, y000p= 0y000p+ y00p= 0 + 2C = −2 ⇒ C = −1 ⇒ yp= −x2y = yh+ yp= C1+ C2x + C3e−x− x2(8) # 9.2.5) Solve y00+ 4y0+ 4y = 8 − 12xr2+ 4r + 4 = 0 ⇒ (r + 2)2= 0 ⇒ r = {−2, −2}yh= C1e−2x+ C2xe−2xyp= Ax + B y0p= A y00p= 0y00p+ 4y0p+ 4yp= 0 + 4A + 4Ax + 4B = −12x + 84Ax + (4A + 4B) = −12x + 84A = −12 ⇒ A = −3; −12 + 4B = 8 ⇒ B = 5yp= −3x + 5y = yh+ yp= C1e−2x+ C2xe−2x− 3x + 5(9) # 9.2.7) Solve y00+ 4y0+ 3y = 6x2− 4r2+ 4r + 3 = 0 ⇒ (r + 3)(r + 1) = 0 ⇒ r = {−3, −1}yh= C1e−3x+ C2e−xyp= Ax2+ Bx + C, y0p= 2Ax + b, y00p= 2Ay00p+ 4y0p+ 3yp= 2A + 8Ax + 4B + 3Ax2+ 3Bx + 3C = 6x2− 43Ax2+(8A + 3B)x + (2A + 4B + 3C) = 6x2+ 0x − 43A = 6 ⇒ A = 28(2) + 3B = 0 ⇒ B = −1632(2) + 4−163+ 3C = −4 ⇒ C =409yp= 2x2−163x +409y = yh+ yp= C1e−3x+ C2e−x+ 2x2−163x +4093(10) #9.2.19) Solve the IVP y00+ 2y0+ y = x2, y(0) = 1, y0(0) = 0r2+ 2r + 1 = 0 ⇒ (r + 1)2= 0 ⇒ r = {−1, −1}yh= C1e−x+ C2xe−xyp= Ax2+ Bx + C, y0p= 2Ax + B, y00p= 2Ay00+ 2y0+ y = 2A + 2(2Ax + B) + Ax2+ Bx + C = x22A + 4Ax + 2B + Ax2+ Bx + C = Ax2+ (4A + B)x + (2A + 2B + C) = x2+ 0x + 0A = 14(1) + B = 0 ⇒ B = −42(1) + 2(−4) + C = 0 ⇒ C = 6yp= x2− 4x + 6y = yh+ yp= C1e−x+ C2xe−x+ x2− 4x + 6y0= −C1e−x+ C2(x · −e−x+ e−x+ 2x − 4)y0= −C1e−x− C2xe−x+ C2e−x+ 2x − 41 = C1+ 0 · C2+ 60 = −C1+ C2− 4⇒1 0 −5−1 1 4⇒1 0 −50 1 −1⇒{C1, C2} = {−5, −1}y = −5e−x− xe−x+ x2− 4x + 6(11) # 9.3.7)x0= 4x + y ⇒ y = x0− 4xy0= −x + 2y(1) x00= 4x0+ y0(2) x00= 4x0+ (−x + 2y) = 4x0− x + 2y(3) x00= 4x0− x + 2(x0− 4x) = 4x0− x + 2x0− 8x = 6x0− 9x(4) x00− 6x0+ 9x = 0r2− 6r + 9 = 0 ⇒ (r − 3)2= 0 ⇒ r = {3, 3}x(t) = C1e3t+ C2te3t(5)x0(t) = 3C1e3t+ C2(3te3t+ e3t) = (3C1+ C2)e3t+ 3tC2e3ty(t) = (3C1+ C2)e3t+ 3tC2e3t− 4(C1e3t+ C2te3t)= 3C1e3t+ C2e3t+ 3C2te3t− 4C1e3t− 4C2te3ty(t) = (−C1+ C2)e3t− C2te3t4(6)x(t) = C1e3t+ C2te3ty(t) = (−C1+ C2)e3t− C2te3tORx0= 4x + yy0= −x + 2y⇒xy0=4 1−1 2xyr2−tr(A)r + det(A) = 0 ⇒ r2+ −6 · r + 9 = 0 ⇒ (r − 3)2= 0 ⇒ r = {3, 3}x(t) = C1e3t+ C2te3tThen Step (5) as above.(12) #9.3.15) Solve the nonhomogeneous system.x0= y + 2t + 3 ⇒ y = x0− 2t − 3y0= −x + 4t − 2(1) x00= y0+ 2 (Remember independent variable is t)(2) x00= −x + 4t − 2 + 2 = −x + 4t(3) (actually 4) x00+ x = 4tr2+ 1 = 0 ⇒ r = ±ixh= e0x(C1sin t + C2cos t) = C1sin t + C2cos txp= At + B; x0p= A; x00p= 0x00+ x = 0 + At + b = 4t ⇒ A = 4, B = 0xp= 4t + 0 = 4tx(t) = xh+ xp= C1sin t + C2cos t + 4t(4) (actually 5)x0(t) = C1cos t − C2sin t + 4y(t) = x0− 2t − 3 = C1cos t − C2sin t + 4 − 2t − 3= C1cos t − C2sin t − 2t + 1(5)x(t) = C1sin t + C2cos t + 4ty(t) = C1cos t − C2sin t − 2t + 1(13) # 9.4.17) Solve the system.x0= −2x + 4yy0= −x − 7y⇒xy0=−2 4−1 −7xy5r2− tr(A) + det(A) = 0 ⇒r2+ 9r + 18 = 0 ⇒ (r + 3)(r + 6) = 0 ⇒r = {−3, −6} ⇒ r1= −3 ⇒ v1=br1− a=4−1r2= −6 ⇒ v2=br2− a=4−4= t1−1z = C1e−3t4−1+ C2e−6t1−1⇒xy=4C1e−3t+ C2e−6t−C1e−3t− C2e−6t(14) # 9.4.33) For exercise # 9.4.17, classify the origin and assess itsstability.Eigenvalues are real, and have the same sign. So the origin is a node.The eigenvalues are < 0, so the node is asymptotically stable.(15) # 9.3.17) Solve the IVP for x(0) = 3, y(0) = −5.x0= 2x − yy0= 2x + 5y⇒xy0=2 −12 5xyr2− tr(A) + det(A) = 0 ⇒r2− 7r + 12 = 0 ⇒ (r − 3)(r − 4) = 0 ⇒r = {3, 4} ⇒ r1= 3 ⇒ v1=br1− a=−11r2= 4 ⇒ v2=br2− a=−12z = C1e3t−11+ C2e4t−12⇒xy=−C1e3t− C2e4tC1e3t+ 2C2e4t3 = −C1− C2−5 = C1+ 2C2⇒−1 −1 31 2 −5⇒−1 −1 30 1 −2⇒−1 0 10 1 −2⇒1 0 −10 1 −2; {C1, C2} = {−1, −2}x(t) = e3t+ 2e4ty(t) = −e3t− 4e4t(16) # 9.4.19) Classify the origin (0,0) of the following system of dif-ferential equations and assess its stability. In particular, state (a) ifthe origin is a node, a spiral point, a center, or a saddle point, and (b)if the origin is asymptotically stable, stable, or unstable.dxdt= −5x + 10ydydt= −4x + 7y6A =−5 10−4 7r2− tr(A) + det(A) = 0 ⇒ r2− 2r + 5 = 0 ⇒ r2− 2r + 1 = −5 + 1 = −4(r − 1)2= −4 ⇒ r = 1 ± 2iEigenvalues complex with positive real part ⇒ origin is an unstable spiral point.(17) # …


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