#set document(title: "3.10 Exercises", author: "Jiří Lebl") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 3.10#h(0.6em)Exercises These are homework exercises to accompany Libl's "#link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)")[Differential Equations for Engineering]" Textmap. This is a textbook targeted for a one semester first course on differential equations, aimed at engineering students. Prerequisite for the course is the basic calculus sequence. === #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/3%3A_Systems_of_ODEs/3.1%3A_Introduction_to_Systems_of_ODEs")[3.1 Introduction to Systems of ODEs] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the general solution of #math.equation(block: false, alt: "x 1 ′ equals x sub 2 minus x sub 1 plus t , x 2 ′ equals x sub 2")[$x_(1)^(′) = x_(2) − x_(1) + t , x_(2)^(′) = x_(2)$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the general solution of #math.equation(block: false, alt: "x 1 ′ equals 3 x sub 1 minus x sub 2 plus e to the power t , x 2 ′ equals x sub 1")[$x_(1)^(′) = 3 x_(1) − x_(2) + e^(t) , x_(2)^(′) = x_(1)$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Write #math.equation(block: false, alt: "a y double prime plus b y prime plus c y equals f open parenthesis x close parenthesis")[$a y^(″) + b y^(′) + c y = f ( x )$] as a first order system of ODEs. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Write #math.equation(block: false, alt: "x double prime plus y squared y prime minus x cubed equals sin open parenthesis t close parenthesis , y double prime plus open parenthesis x prime plus y prime close parenthesis squared minus x equals 0")[$x^(″) + y^(2) y^(′) − x^(3) = sin ( t ) , y^(″) + attach(( x^(′) + y^(′) ), t: 2) − x = 0$] as a first order system of ODEs. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Suppose two masses on carts on frictionless surface are at displacements #math.equation(block: false, alt: "x sub 1")[$x_(1)$] and #math.equation(block: false, alt: "x sub 2")[$x_(2)$] as in Example 3.1.3. Suppose that a rocket applies force #math.equation(block: false, alt: "F")[$F$] in the positive direction on cart #math.equation(block: false, alt: "x sub 1")[$x_(1)$]. Set up the system of equations. ] #examplebox("Example 1")[][ Suppose the tanks are as in Example 3.1.2, starting both at volume #math.equation(block: false, alt: "V")[$V$], but now the rate of flow from tank 1 to tank 2 is #math.equation(block: false, alt: "r sub 1")[$r_(1)$], and rate of flow from tank 2 to tank one is #math.equation(block: false, alt: "r sub 2")[$r_(2)$]. Notice that the volumes are now not constant. Set up the system of equations. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the general solution to #math.equation(block: false, alt: "y 1 ′ equals 3 y sub 1 , y 2 ′ equals y sub 1 plus y sub 2 , y 3 ′ equals y sub 1 plus y sub 3")[$y_(1)^(′) = 3 y_(1) , y_(2)^(′) = y_(1) + y_(2) , y_(3)^(′) = y_(1) + y_(3)$]. Answer #math.equation(block: true, alt: "y sub 1 equals C sub 1 e to the power 3 x")[$y_(1) = C_(1) e^(3 x)$], #math.equation(block: true, alt: "y sub 2 equals y open parenthesis x close parenthesis equals C sub 2 e to the power x plus the fraction C sub 1 over 2 e to the power 3 x")[$y_(2) = y ( x ) = C_(2) e^(x) + frac(C_(1), 2) e^(3 x)$], #math.equation(block: true, alt: "y sub 3 equals y open parenthesis x close parenthesis equals C sub 3 e to the power x plus the fraction C sub 1 over 2 e to the power 3 x")[$y_(3) = y ( x ) = C_(3) e^(x) + frac(C_(1), 2) e^(3 x)$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "y prime equals 2 x , x prime equals x plus y , x open parenthesis 0 close parenthesis equals 1 , y open parenthesis 0 close parenthesis equals 3")[$y^(′) = 2 x , x^(′) = x + y , x ( 0 ) = 1 , y ( 0 ) = 3$]. Answer #math.equation(block: true, alt: "x equals the fraction 5 over 3 e to the power 2 t minus the fraction 2 over 3 e to the power minus t")[$x = frac(5, 3) e^(2 t) − frac(2, 3) e^(− t)$], #math.equation(block: true, alt: "y equals the fraction 5 over 3 e to the power 2 t plus the fraction 4 over 3 e to the power minus t")[$y = frac(5, 3) e^(2 t) + frac(4, 3) e^(− t)$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Write #math.equation(block: false, alt: "x to the power ‴ equals x plus t")[$x^(‴) = x + t$] as a first order system. Answer #math.equation(block: true, alt: "x 1 ′ equals x sub 2")[$x_(1)^(′) = x_(2)$], #math.equation(block: true, alt: "x 2 ′ equals x sub 3")[$x_(2)^(′) = x_(3)$], #math.equation(block: true, alt: "x 3 ′ equals x sub 1 plus t")[$x_(3)^(′) = x_(1) + t$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Write #math.equation(block: false, alt: "y 1 ″ plus y sub 1 plus y sub 2 equals t y 2 ″ plus y sub 1 minus y sub 2 equals t squared")[$y_(1)^(″) + y_(1) + y_(2) = t y_(2)^(″) + y_(1) − y_(2) = t^(2)$] as a first order system. Answer #math.equation(block: true, alt: "y 3 ′ plus y sub 1 plus y sub 2 equals t")[$y_(3)^(′) + y_(1) + y_(2) = t$], #math.equation(block: true, alt: "y 4 ′ plus y sub 1 minus y sub 2 equals t squared")[$y_(4)^(′) + y_(1) − y_(2) = t^(2)$], #math.equation(block: true, alt: "y 1 ′ equals y sub 3")[$y_(1)^(′) = y_(3)$], #math.equation(block: true, alt: "y 2 ′ equals y sub 4")[$y_(2)^(′) = y_(4)$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Suppose two masses on carts on frictionless surface are at displacements #math.equation(block: false, alt: "x sub 1")[$x_(1)$] and #math.equation(block: false, alt: "x sub 2")[$x_(2)$] as in Example 3.1.3. Suppose initial displacement is #math.equation(block: false, alt: "x sub 1 open parenthesis 0 close parenthesis equals x sub 2 open parenthesis 0 close parenthesis equals 0")[$x_(1) ( 0 ) = x_(2) ( 0 ) = 0$], and initial velocity is #math.equation(block: false, alt: "x 1 ′ open parenthesis 0 close parenthesis equals x 2 ′ open parenthesis 0 close parenthesis equals a")[$x_(1)^(′) ( 0 ) = x_(2)^(′) ( 0 ) = a$] for some number #math.equation(block: false, alt: "a")[$a$]. Use your intuition to solve the system, explain your reasoning. Answer #math.equation(block: false, alt: "x sub 1 equals x sub 2 equals a t")[$x_(1) = x_(2) = a t$]. Explanation of the intuition is left to reader. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Suppose the tanks are as in Example 3.1.2 except that clean water flows in at the rate #math.equation(block: false, alt: "s")[$s$] liters per second into tank 1, and brine flows out of tank 2 and into the sewer also at the rate of #math.equation(block: false, alt: "s")[$s$] liters per second. + Draw the picture. + Set up the system of equations. + Intuitively, what happens as #math.equation(block: false, alt: "t")[$t$] goes to infinity, explain. Answer + Left to reader + #math.equation(block: false, alt: "x 1 ′ equals the fraction r over V open parenthesis x sub 2 minus x sub 1 close parenthesis")[$x_(1)^(′) = frac(r, V) ( x_(2) − x_(1) )$], #math.equation(block: false, alt: "x 2 ′ equals the fraction r over V x sub 1 minus the fraction r minus s over V x sub 2")[$x_(2)^(′) = frac(r, V) x_(1) − frac(r − s, V) x_(2)$]. + As #math.equation(block: false, alt: "t")[$t$] goes to infinity, both #math.equation(block: false, alt: "x sub 1")[$x_(1)$] and #math.equation(block: false, alt: "x sub 2")[$x_(2)$] go to zero, explanation is left to reader. ] === #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/3%3A_Systems_of_ODEs/3.2%3A_Matrices_and_linear_systems")[3.2: Matrices and linear systems] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "[ 1, 2; 3, 4 ] x → equals [ 5; 6 ]")[$[ 1 & 2 \ 3 & 4 ] arrow(x) = [ 5 \ 6 ]$] by using matrix inverse. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Compute determinant of #math.equation(block: false, alt: "[ 9, minus 2, minus 6; minus 8, 3, 6; 10, minus 2, minus 6 ]")[$[ 9 & − 2 & − 6 \ − 8 & 3 & 6 \ 10 & − 2 & − 6 ]$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Compute determinant of #math.equation(block: false, alt: "[ 1, 2, 3, 1; 4, 0, 5, 0; 6, 0, 7, 0; 8, 0, 10, 1 ]")[$[ 1 & 2 & 3 & 1 \ 4 & 0 & 5 & 0 \ 6 & 0 & 7 & 0 \ 8 & 0 & 10 & 1 ]$]. Hint: Expand along the proper row or column to make the calculations simpler. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Compute inverse of #math.equation(block: false, alt: "[ 1, 2, 3; 1, 1, 1; 0, 1, 0 ]")[$[ 1 & 2 & 3 \ 1 & 1 & 1 \ 0 & 1 & 0 ]$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ For which #math.equation(block: false, alt: "h")[$h$] is #math.equation(block: false, alt: "[ 1, 2, 3; 4, 5, 6; 7, 8, h ]")[$[ 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & h ]$] not invertible? Is there only one such #math.equation(block: false, alt: "h")[$h$]? Are there several? Infinitely many? ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ For which #math.equation(block: false, alt: "h")[$h$] is #math.equation(block: false, alt: "[ h, 1, 1; 0, h, 0; 1, 1, h ]")[$[ h & 1 & 1 \ 0 & h & 0 \ 1 & 1 & h ]$] not invertible? Find all such #math.equation(block: false, alt: "h")[$h$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "[ 9, minus 2, minus 6; minus 8, 3, 6; 10, minus 2, minus 6 ] x → equals [ 1; 2; 3 ]")[$[ 9 & − 2 & − 6 \ − 8 & 3 & 6 \ 10 & − 2 & − 6 ] arrow(x) = [ 1 \ 2 \ 3 ]$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "[ 5, 3, 7; 8, 4, 4; 6, 3, 3 ] x → equals [ 2; 0; 0 ]")[$[ 5 & 3 & 7 \ 8 & 4 & 4 \ 6 & 3 & 3 ] arrow(x) = [ 2 \ 0 \ 0 ]$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "[ 3, 2, 3, 0; 3, 3, 3, 3; 0, 2, 4, 2; 2, 3, 4, 3 ] x → equals [ 2; 0; 4; 1 ]")[$[ 3 & 2 & 3 & 0 \ 3 & 3 & 3 & 3 \ 0 & 2 & 4 & 2 \ 2 & 3 & 4 & 3 ] arrow(x) = [ 2 \ 0 \ 4 \ 1 ]$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find 3 nonzero #math.equation(block: false, alt: "2 times 2")[$2 × 2$] matrices #math.equation(block: false, alt: "A , B ,")[$A , B ,$] and #math.equation(block: false, alt: "C")[$C$] such that #math.equation(block: false, alt: "A B equals A C")[$A B = A C$] but #math.equation(block: false, alt: "B not equal to C")[$B ≠ C$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Compute determinant of #math.equation(block: false, alt: "[ 1, 1, 1; 2, 3, minus 5; 1, minus 1, 0 ]")[$[ 1 & 1 & 1 \ 2 & 3 & − 5 \ 1 & − 1 & 0 ]$] Answer #math.equation(block: true, alt: "minus 15")[$− 15$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find #math.equation(block: false, alt: "t")[$t$] such that #math.equation(block: false, alt: "[ 1, t; minus 1, 2 ]")[$[ 1 & t \ − 1 & 2 ]$] is not invertible. Answer #math.equation(block: true, alt: "minus 2")[$− 2$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "[ 1, 1; 1, minus 1 ] x → equals [ 10; 20 ]")[$[ 1 & 1 \ 1 & − 1 ] arrow(x) = [ 10 \ 20 ]$]. Answer #math.equation(block: true, alt: "15; minus 5")[$15 \ − 5$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Suppose #math.equation(block: false, alt: "a , b , c")[$a , b , c$] are nonzero numbers. Let #math.equation(block: false, alt: "M equals [ a, 0; 0, b ] , N equals [ a, 0, 0; 0, b, 0; 0, 0, c ]")[$M = [ a & 0 \ 0 & b ] , N = [ a & 0 & 0 \ 0 & b & 0 \ 0 & 0 & c ]$]. + Compute #math.equation(block: false, alt: "M to the power minus 1")[$M^(− 1)$]. + Compute #math.equation(block: false, alt: "N to the power minus 1")[$N^(− 1)$]. Answer + #math.equation(block: false, alt: "the fraction 1 over a 0; 0 the fraction 1 over b")[$frac(1, a) 0 \ 0 frac(1, b)$] + #math.equation(block: false, alt: "the fraction 1 over a 0 0; 0 the fraction 1 over b 0; 0 0 the fraction 1 over c")[$frac(1, a) 0 0 \ 0 frac(1, b) 0 \ 0 0 frac(1, c)$] ] === #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/3%3A_Systems_of_ODEs/3.3%3A_Linear_systems_of_ODEs")[3.3: Linear systems of ODEs] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Write the system #math.equation(block: false, alt: "x 1 ′ equals 2 x sub 1 minus 3 t x sub 2 plus sin t")[$x_(1)^(′) = 2 x_(1) − 3 t x_(2) + sin t$] and #math.equation(block: false, alt: "x 2 ′ equals e to the power t x sub 1 minus 3 x sub 2 plus cos t")[$x_(2)^(′) = e^(t) x_(1) − 3 x_(2) + cos t$] in the form #math.equation(block: false, alt: "x → prime equals P open parenthesis t close parenthesis x → plus f → open parenthesis t close parenthesis")[$arrow(x)^(′) = P ( t ) arrow(x) + arrow(f) ( t )$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ + Verify that the system #math.equation(block: false, alt: "x → prime equals [ 1, 3; 3, 1 ] x →")[$arrow(x)^(′) = [ 1 & 3 \ 3 & 1 ] arrow(x)$] has the two solutions #math.equation(block: false, alt: "[ 1; 1 ] e to the power 4 t")[$[ 1 \ 1 ] e^(4 t)$] and #math.equation(block: false, alt: "[ 1; minus 1 ] e to the power minus 2 t")[$[ 1 \ − 1 ] e^(− 2 t)$]. + Write down the general solution. + Write down the general solution in the form #math.equation(block: false, alt: "x sub 1 equals ?")[$x_(1) = ?$],#math.equation(block: false, alt: "x sub 2 equals ?")[$x_(2) = ?$] (i.e. write down a formula for each element of the solution). ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Verify that #math.equation(block: false, alt: "[ 1; 1 ] e to the power t")[$[ 1 \ 1 ] e^(t)$] and #math.equation(block: false, alt: "[ 1; minus 1 ] e to the power t")[$[ 1 \ − 1 ] e^(t)$] are linearly independent. Hint: Just plug in #math.equation(block: false, alt: "t equals 0")[$t = 0$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Verify that #math.equation(block: false, alt: "[ 1; 1; 0 ] e to the power t")[$[ 1 \ 1 \ 0 ] e^(t)$] and #math.equation(block: false, alt: "[ 1; minus 1; 1 ] e to the power t")[$[ 1 \ − 1 \ 1 ] e^(t)$] and #math.equation(block: false, alt: "[ 1; minus 1; 1 ] e to the power 2 t")[$[ 1 \ − 1 \ 1 ] e^(2 t)$] are linearly independent. Hint: You must be a bit more tricky than in the previous exercise. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Verify that#math.equation(block: false, alt: "[ t; t squared ]")[$[ t \ t^(2) ]$] and #math.equation(block: false, alt: "[ t cubed; t to the power 4 ]")[$[ t^(3) \ t^(4) ]$] are linearly independent. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Take the system #math.equation(block: false, alt: "x 1 ′ plus x 2 ′ equals x sub 1")[$x_(1)^(′) + x_(2)^(′) = x_(1)$], #math.equation(block: false, alt: "x 1 ′ minus x 2 ′ equals x sub 2")[$x_(1)^(′) − x_(2)^(′) = x_(2)$]. + Write it in the form #math.equation(block: false, alt: "A x → prime equals B x →")[$A arrow(x)^(′) = B arrow(x)$] for matrices #math.equation(block: false, alt: "A")[$A$] and #math.equation(block: false, alt: "B")[$B$]. + Compute #math.equation(block: false, alt: "A to the power minus 1")[$A^(− 1)$] and use that to write the system in the form #math.equation(block: false, alt: "x → prime equals P x →")[$arrow(x)^(′) = P arrow(x)$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Are #math.equation(block: false, alt: "[ e to the power 2 t; e to the power t ]")[$[ e^(2 t) \ e^(t) ]$] and #math.equation(block: false, alt: "[ e to the power t; e to the power 2 t ]")[$[ e^(t) \ e^(2 t) ]$] linearly independent? Justify. Answer Yes. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Are #math.equation(block: false, alt: "[ c o s h open parenthesis t close parenthesis; 1 ]")[$[ c o s h ( t ) \ 1 ]$], #math.equation(block: false, alt: "[ e to the power t; 1 ]")[$[ e^(t) \ 1 ]$] and #math.equation(block: false, alt: "[ e to the power minus t; 1 ]")[$[ e^(− t) \ 1 ]$]linearly independent? Justify. Answer No. #math.equation(block: false, alt: "cosh open parenthesis t close parenthesis 1")[$cosh ( t ) 1$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Write #math.equation(block: false, alt: "x prime equals 3 x minus y plus e to the power t")[$x^(′) = 3 x − y + e^(t)$] and #math.equation(block: false, alt: "y prime equals t x")[$y^(′) = t x$] in matrix notation. Answer #math.equation(block: true, alt: "x; y")[$x \ y$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ + Write #math.equation(block: false, alt: "x 1 ′ equals 2 t x sub 2")[$x_(1)^(′) = 2 t x_(2)$] and #math.equation(block: false, alt: "x 2 ′ equals 2 t x sub 2")[$x_(2)^(′) = 2 t x_(2)$] in matrix notation. + Solve and write the solution in matrix notation.Add exercises text here. Answer + #math.equation(block: false, alt: "0 2 t; 0 2 t")[$0 2 t \ 0 2 t$] + #math.equation(block: false, alt: "C sub 2 e to the power t squared plus C sub 1; C sub 2 e to the power t squared")[$C_(2) e^(t^(2)) + C_(1) \ C_(2) e^(t^(2))$] ] === #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/3%3A_Systems_of_ODEs/3.4%3A_Eigenvalue_Method")[3.4: Eigenvalue Method] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[(easy)] Let #math.equation(block: false, alt: "A")[$A$] be a #math.equation(block: false, alt: "3 times 3")[$3 × 3$] matrix with an eigenvalue of #math.equation(block: false, alt: "3")[$3$] and a corresponding eigenvector #math.equation(block: false, alt: "1; minus 1; 3")[$1 \ − 1 \ 3$]. Find #math.equation(block: false, alt: "A v →")[$A arrow(v)$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ + Find the general solution of #math.equation(block: false, alt: "x 1 ′ equals 2 x sub 1 , x 2 ′ equals 3 x sub 2")[$x_(1)^(′) = 2 x_(1) , x_(2)^(′) = 3 x_(2)$] using the eigenvalue method (first write the system in the form #math.equation(block: false, alt: "x → prime equals A x →")[$arrow(x)^(′) = A arrow(x)$] ). + Solve the system by solving each equation separately and verify you get the same general solution. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the general solution of #math.equation(block: false, alt: "x 1 ′ equals 3 x sub 1 plus x sub 2 , x 2 ′ equals 2 x sub 1 plus 4 x sub 2")[$x_(1)^(′) = 3 x_(1) + x_(2) , x_(2)^(′) = 2 x_(1) + 4 x_(2)$] using the eigenvalue method. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the general solution of #math.equation(block: false, alt: "x 1 ′ equals x sub 1 minus 2 x sub 2 , x 2 ′ equals 2 x sub 1 x sub 2")[$x_(1)^(′) = x_(1) − 2 x_(2) , x_(2)^(′) = 2 x_(1) x_(2)$] using the eigenvalue method. Do not use complex exponentials in your solution. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ + Compute eigenvalues and eigenvectors of #math.equation(block: false, alt: "9 minus 2 minus 6; minus 8 3 6; 10 minus 2 minus 6")[$9 − 2 − 6 \ − 8 3 6 \ 10 − 2 − 6$]. + Find the general solution of #math.equation(block: false, alt: "x → prime equals A x →")[$arrow(x)^(′) = A arrow(x)$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Compute eigenvalues and eigenvectors of #math.equation(block: false, alt: "minus 2 minus 1 minus 1; 3 2 1; minus 3 minus 1 0")[$− 2 − 1 − 1 \ 3 2 1 \ − 3 − 1 0$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Let #math.equation(block: false, alt: "a , b , c , d , e , f")[$a , b , c , d , e , f$] be numbers. Find the eigenvalues of #math.equation(block: false, alt: "a b c; 0 d e; 0 0 f")[$a b c \ 0 d e \ 0 0 f$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ + Compute eigenvalues and eigenvectors of #math.equation(block: false, alt: "1 0 3; minus 1 0 1; 2 0 2")[$1 0 3 \ − 1 0 1 \ 2 0 2$]. + Solve the system #math.equation(block: false, alt: "x → prime equals A x →")[$arrow(x)^(′) = A arrow(x)$]. Answer + Eigenvalues: #math.equation(block: false, alt: "4 , 0 , minus 1")[$4 , #h(0.222em) 0 , #h(0.222em) − 1$] Eigenvectors: #math.equation(block: false, alt: "1; 0; 1")[$1 \ 0 \ 1$], #math.equation(block: false, alt: "0; 1; 0")[$0 \ 1 \ 0$], #math.equation(block: false, alt: "3; 5; minus 2")[$3 \ 5 \ − 2$] + #math.equation(block: false, alt: "1; 0; 1")[$1 \ 0 \ 1$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ + Compute eigenvalues and eigenvectors of #math.equation(block: false, alt: "1 1; minus 1 0")[$1 1 \ − 1 0$] + Solve the system #math.equation(block: false, alt: "x → prime equals A x →")[$arrow(x)^(′) = A arrow(x)$]. Answer + Eigenvalues: #math.equation(block: false, alt: "the fraction 1 plus the square root of 3 i over 2")[$frac(1 + sqrt(3) i, 2)$], #math.equation(block: false, alt: "the fraction 1 minus the square root of 3 i over 2")[$frac(1 − sqrt(3) i, 2)$], Eigenvectors: #math.equation(block: false, alt: "minus 2; 1 minus the square root of 3 i")[$− 2 \ 1 − sqrt(3) i$], #math.equation(block: false, alt: "minus 2; 1 plus the square root of 3 i")[$− 2 \ 1 + sqrt(3) i$] + #math.equation(block: false, alt: "minus 2 cos open parenthesis the fraction the square root of 3 t over 2 close parenthesis; cos open parenthesis the fraction the square root of 3 t over 2 close parenthesis plus the square root of 3 sin open parenthesis the fraction the square root of 3 t over 2 close parenthesis")[$− 2 cos ( frac(sqrt(3) t, 2) ) \ cos ( frac(sqrt(3) t, 2) ) + sqrt(3) sin ( frac(sqrt(3) t, 2) )$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "x 1 ′ equals x sub 2 , x 2 ′ equals x sub 1")[$x_(1)^(′) = x_(2) , x_(2)^(′) = x_(1)$] using the eigenvalue method. Answer #math.equation(block: true, alt: "1; 1")[$1 \ 1$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "x 1 ′ equals x sub 2 , x 2 ′ equals minus x sub 1")[$x_(1)^(′) = x_(2) , x_(2)^(′) = − x_(1)$] using the eigenvalue method. Answer #math.equation(block: true, alt: "cos open parenthesis t close parenthesis; minus sin open parenthesis t close parenthesis")[$cos ( t ) \ − sin ( t )$] ] === #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/3%3A_Systems_of_ODEs/3.5%3A_Two_dimensional_systems_and_their_vector_fields")[3.5: Two dimensional systems and their vector fields] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Take the equation #math.equation(block: false, alt: "m x double prime plus c x prime plus k x equals 0")[$m x^(″) + c x^(′) + k x = 0$], with #math.equation(block: false, alt: "m greater than 0 , c greater than or equal to 0 , k greater than 0")[$m > 0 , c ≥ 0 , k > 0$] for the mass-spring system. + Convert this to a system of first order equations. + Classify for what #math.equation(block: false, alt: "m")[$m$], #math.equation(block: false, alt: "c")[$c$], #math.equation(block: false, alt: "k")[$k$] do you get which behavior. + Can you explain from physical intuition why you do not get all the different kinds of behavior here? ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Can you find what happens in the case when #math.equation(block: false, alt: "P equals [ 1, 1; 0, 1 ]")[$P = [ 1 & 1 \ 0 & 1 ]$]? In this case the eigenvalue is repeated and there is only one eigenvector. What picture does this look like? ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Can you find what happens in the case when #math.equation(block: false, alt: "P equals [ 1, 1; 1, 1 ]")[$P = [ 1 & 1 \ 1 & 1 ]$]? Does this look like any of the pictures we have drawn? ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Which behaviors are possible if #math.equation(block: false, alt: "P")[$P$] is diagonal, that is #math.equation(block: false, alt: "a 0; 0 b")[$a 0 \ 0 b$]? You can assume that #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "b")[$b$] are not zero. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Take the system from Example 3.1.2, #math.equation(block: false, alt: "x 1 ′ equals the fraction r over V open parenthesis x sub 2 minus x sub 1 close parenthesis")[$x_(1)^(′) = frac(r, V) ( x_(2) − x_(1) )$], #math.equation(block: false, alt: "x 2 ′ equals the fraction r over V open parenthesis x sub 1 minus x sub 2 close parenthesis")[$x_(2)^(′) = frac(r, V) ( x_(1) − x_(2) )$]. As we said, one of the eigenvalues is zero. What is the other eigenvalue, how does the picture look like and what happens when #math.equation(block: false, alt: "t")[$t$] goes to infinity. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Describe the behavior of the following systems without solving: + #math.equation(block: false, alt: "x prime equals x plus y , y prime equals x minus y")[$x^(′) = x + y , #h(1em) y^(′) = x − y$] + #math.equation(block: false, alt: "x 1 ′ equals x sub 1 plus x sub 2 , x 2 ′ equals 2 x sub 2")[$x_(1)^(′) = x_(1) + x_(2) , #h(1em) x_(2)^(′) = 2 x_(2)$] + #math.equation(block: false, alt: "x 1 ′ equals minus 2 x sub 2 , x 2 ′ equals 2 x sub 1")[$x_(1)^(′) = − 2 x_(2) , #h(1em) x_(2)^(′) = 2 x_(1)$] + #math.equation(block: false, alt: "x prime equals x plus 3 y , y prime equals minus 2 x minus 4 y")[$x^(′) = x + 3 y , #h(1em) y^(′) = − 2 x − 4 y$] + #math.equation(block: false, alt: "x prime equals x minus 4 y , y prime equals minus 4 x plus y")[$x^(′) = x − 4 y , #h(1em) y^(′) = − 4 x + y$] Answer + Two eigenvalues: #math.equation(block: false, alt: "± the square root of 2")[$± sqrt(2)$] so the behavior is a saddle. + Two eigenvalues: #math.equation(block: false, alt: "1")[$1$] and #math.equation(block: false, alt: "2")[$2$], so the behavior is a source. + Two eigenvalues: #math.equation(block: false, alt: "± 2 i")[$± 2 i$], so the behavior is a center (ellipses). + Two eigenvalues: #math.equation(block: false, alt: "− 1")[$− 1$] and #math.equation(block: false, alt: "− 2")[$− 2$], so the behavior is a sink. + Two eigenvalues: #math.equation(block: false, alt: "5")[$5$] and #math.equation(block: false, alt: "− 3")[$− 3$], so the behavior is a saddle. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Suppose that #math.equation(block: false, alt: "x → equals A x →")[$arrow(x) = A arrow(x)$] where #math.equation(block: false, alt: "A")[$A$] is a #math.equation(block: false, alt: "2 times 2")[$2 × 2$] matrix with eigenvalues #math.equation(block: false, alt: "2 ± i")[$2 ± i$]. Describe the behavior. Answer Spiral source. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Take #math.equation(block: false, alt: "[ to the power x; y equals [ 0, 1; 0, 0 ] [ x; y ]")[$\[^(x \ y) = \[ 0 & 1 \ 0 & 0 \] \[ x \ y \]$]. Draw the vector field and describe the behavior. Is it one of the behaviours that we have seen before? Answer #figure(figph[Vector field of purely horizontal blue arrows: they point right above the central horizontal line and left below it, growing longer with distance from it. The axes have tick marks but no numbers.], alt: "Vector field of purely horizontal blue arrows: they point right above the central horizontal line and left below it, growing longer with distance from it. The axes have tick marks but no numbers.", caption: none) The solution does not move anywhere if #math.equation(block: false, alt: "y equals 0")[$y = 0$]. When #math.equation(block: false, alt: "y")[$y$] is positive, the solution moves (with constant speed) in the positive #math.equation(block: false, alt: "x")[$x$] direction. When #math.equation(block: false, alt: "y")[$y$] is negative, the solution moves (with constant speed) in the negative #math.equation(block: false, alt: "x")[$x$] direction. It is not one of the behaviors we saw. Note that the matrix has a double eigenvalue #math.equation(block: false, alt: "0")[$0$] and the general solution is #math.equation(block: false, alt: "x equals C sub 1 t plus C sub 2")[$x = C_(1) t + C_(2)$] and #math.equation(block: false, alt: "y equals C sub 1")[$y = C_(1)$], which agrees with the description ] === #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/3%3A_Systems_of_ODEs/3.6%3A_Second_order_systems_and_applications")[3.6: Second order systems and applications] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find a particular solution to #math.equation(block: true, alt: "minus 3 1; 2 minus 2")[$− 3 1 \ 2 − 2$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[challenging] Let us take the example in Figure 3.6.3 with the same parameters as before: #math.equation(block: false, alt: "m sub 1 equals 2 , k sub 1 equals 4 ,")[$m_(1) = 2 , k_(1) = 4 ,$] and #math.equation(block: false, alt: "k sub 2 equals 2 ,")[$k_(2) = 2 ,$] except for #math.equation(block: false, alt: "m sub 2")[$m_(2)$], which is unknown. Suppose that there is a force #math.equation(block: false, alt: "cos open parenthesis 5 t close parenthesis")[$cos ( 5 t )$] acting on the first mass. Find an #math.equation(block: false, alt: "m sub 2")[$m_(2)$] such that there exists a particular solution where the first mass does not move. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Note] This idea is called dynamic damping. In practice there will be a small amount of damping and so any transient solution will disappear and after long enough time, the first mass will always come to a stop. ] ] #examplebox("Example 2")[][ Let us take the Example 3.6.2, but that at time of impact, cart 2 is moving to the left at the speed of #math.equation(block: false, alt: "3 the fraction m over s")[$3 #h(0.222em) frac("m", "s")$]. + Find the behavior of the system after linkup. + Will the second car hit the wall, or will it be moving away from the wall as time goes on? + At what speed would the first car have to be traveling for the system to essentially stay in place after linkup? ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Let us take the example in Figure 3.6.2 with parameters #math.equation(block: false, alt: "m sub 1 equals m sub 2 equals 1 , k sub 1 equals k sub 2 equals 1")[$m_(1) = m_(2) = 1 , k_(1) = k_(2) = 1$]. Does there exist a set of initial conditions for which the first cart moves but the second cart does not? If so, find those conditions. If not, argue why not. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the general solution to #math.equation(block: true, alt: "1 0 0; 0 2 0; 0 0 3")[$1 0 0 \ 0 2 0 \ 0 0 3$] Answer #math.equation(block: true, alt: "1; minus 1; 1")[$1 \ − 1 \ 1$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Suppose there are three carts of equal mass #math.equation(block: false, alt: "m")[$m$] and connected by two springs of constant #math.equation(block: false, alt: "k")[$k$] (and no connections to walls). Set up the system and find its general solution. Answer #math.equation(block: true, alt: "m 0 0; 0 m 0; 0 0 m")[$m 0 0 \ 0 m 0 \ 0 0 m$]. Solution: #math.equation(block: false, alt: "1; minus 2; 1")[$1 \ − 2 \ 1$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Suppose a cart of mass #math.equation(block: false, alt: "2 kg")[$2 " kg"$] is attached by a spring of constant #math.equation(block: false, alt: "k equals 1")[$k = 1$] to a cart of mass #math.equation(block: false, alt: "3 kg")[$3 " kg"$], which is attached to the wall by a spring also of constant #math.equation(block: false, alt: "k equals 1")[$k = 1$]. Suppose that the initial position of the first cart is #math.equation(block: false, alt: "1")[$1$] meter in the positive direction from the rest position, and the second mass starts at the rest position. The masses are not moving and are let go. Find the position of the second mass as a function of time. Answer #math.equation(block: true, alt: "x sub 2 equals open parenthesis the fraction 2 over 5 close parenthesis cos open parenthesis the square root of the fraction 1 over 6 t close parenthesis minus open parenthesis the fraction 2 over 5 close parenthesis cos open parenthesis t close parenthesis")[$x_(2) = ( frac(2, 5) ) cos ( sqrt(frac(1, 6)) t ) − ( frac(2, 5) ) cos ( t )$] ] === #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/3%3A_Systems_of_ODEs/3.7%3A_Multiple_Eigenvalues")[3.7: Multiple Eigenvalues] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Let #math.equation(block: false, alt: "A equals [ 5, minus 3; 3, minus 1 ]")[$A = [ 5 & − 3 \ 3 & − 1 ]$]. Find the general solution of #math.equation(block: false, alt: "x → prime equals A x →")[$arrow(x)^(′) = A arrow(x)$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Let #math.equation(block: false, alt: "A equals [ 5, minus 4, 4; 0, 3, 0; minus 2, 4, minus 1 ] .")[$A = [ 5 & − 4 & 4 \ 0 & 3 & 0 \ − 2 & 4 & − 1 ] .$] + What are the eigenvalues? + What is/are the defect(s) of the eigenvalue(s)? + Find the general solution of #math.equation(block: false, alt: "x → prime equals A x →")[$arrow(x)^(′) = A arrow(x)$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Let #math.equation(block: false, alt: "A equals [ 2, 1, 0; 0, 2, 0; 0, 0, 2 ]")[$A = [ 2 & 1 & 0 \ 0 & 2 & 0 \ 0 & 0 & 2 ]$]. + What are the eigenvalues? + What is/are the defect(s) of the eigenvalue(s)? + Find the general solution of #math.equation(block: false, alt: "x → prime equals A x →")[$arrow(x)^(′) = A arrow(x)$] in two different ways and verify you get the same answer. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Let #math.equation(block: false, alt: "A equals [ 0, 1, 2; minus 1, minus 2, minus 2; minus 4, 4, 7 ]")[$A = [ 0 & 1 & 2 \ − 1 & − 2 & − 2 \ − 4 & 4 & 7 ]$]. + What are the eigenvalues? + What is/are the defect(s) of the eigenvalue(s)? + Find the general solution of #math.equation(block: false, alt: "x → prime equals A x →")[$arrow(x)^(′) = A arrow(x)$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Let #math.equation(block: false, alt: "A equals [ 0, 4, minus 2; minus 1, minus 4, 1; 0, 0, minus 2 ]")[$A = [ 0 & 4 & − 2 \ − 1 & − 4 & 1 \ 0 & 0 & − 2 ]$]. + What are the eigenvalues? + What is/are the defect(s) of the eigenvalue(s)? + Find the general solution of #math.equation(block: false, alt: "x → prime equals A x →")[$arrow(x)^(′) = A arrow(x)$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Let #math.equation(block: false, alt: "[ 2, 1, minus 1; minus 1, 0, 2; minus 1, minus 2, 4 ]")[$[ 2 & 1 & − 1 \ − 1 & 0 & 2 \ − 1 & − 2 & 4 ]$]. + What are the eigenvalues? + What is/are the defect(s) of the eigenvalue(s)? + Find the general solution of #math.equation(block: false, alt: "x → prime equals A x →")[$arrow(x)^(′) = A arrow(x)$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Suppose that A is a #math.equation(block: false, alt: "2 times 2")[$2 × 2$] matrix with a repeated eigenvalue #math.equation(block: false, alt: "λ")[$λ$]. Suppose that there are two linearly independent eigenvectors. Show that #math.equation(block: false, alt: "A equals λ I")[$A = λ I$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Let #math.equation(block: false, alt: "A equals [ 1, 1, 1; 1, 1, 1; 1, 1, 1 ]")[$A = [ 1 & 1 & 1 \ 1 & 1 & 1 \ 1 & 1 & 1 ]$]. + What are the eigenvalues? + What is/are the defect(s) of the eigenvalue(s)? + Find the general solution of #math.equation(block: false, alt: "x → prime equals A x →")[$arrow(x)^(′) = A arrow(x)$]. Answer + #math.equation(block: false, alt: "3 , 0 , 0")[$3 , #h(0.222em) 0 , #h(0.222em) 0$] + No defects. + #math.equation(block: false, alt: "1; 1; 1")[$1 \ 1 \ 1$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Let #math.equation(block: false, alt: "A equals [ 1, 3, 3; 1, 1, 0; minus 1, 1, 2 ]")[$A = [ 1 & 3 & 3 \ 1 & 1 & 0 \ − 1 & 1 & 2 ]$]. + What are the eigenvalues? + What is/are the defect(s) of the eigenvalue(s)? + Find the general solution of #math.equation(block: false, alt: "x → prime equals A x →")[$arrow(x)^(′) = A arrow(x)$]. Answer + #math.equation(block: false, alt: "1 , 1 , 2")[$1 , #h(0.222em) 1 , #h(0.222em) 2$] + Eigenvalue #math.equation(block: false, alt: "1")[$1$] has a defect of #math.equation(block: false, alt: "1")[$1$] + #math.equation(block: false, alt: "0; 1; minus 1")[$0 \ 1 \ − 1$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Let #math.equation(block: false, alt: "A equals [ 2, 0, 0; minus 1, minus 1, 9; 0, minus 1, 5 ]")[$A = [ 2 & 0 & 0 \ − 1 & − 1 & 9 \ 0 & − 1 & 5 ]$]. + What are the eigenvalues? + What is/are the defect(s) of the eigenvalue(s)? + Find the general solution of #math.equation(block: false, alt: "x → prime equals A x →")[$arrow(x)^(′) = A arrow(x)$]. Answer + #math.equation(block: false, alt: "2 , 2 , 2")[$2 , #h(0.222em) 2 , #h(0.222em) 2$] + Eigenvalue #math.equation(block: false, alt: "2")[$2$] has a defect of #math.equation(block: false, alt: "2")[$2$] + #math.equation(block: false, alt: "0; 3; 1")[$0 \ 3 \ 1$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Let #math.equation(block: false, alt: "A equals [ a, a; b, c ]")[$A = [ a & a \ b & c ]$], where #math.equation(block: false, alt: "a")[$a$], #math.equation(block: false, alt: "b")[$b$], and #math.equation(block: false, alt: "c")[$c$] are unknowns. Suppose that 5 is a doubled eigenvalue of defect 1, and suppose that #math.equation(block: false, alt: "[ 1; 0 ]")[$[ 1 \ 0 ]$] is the eigenvector. Find #math.equation(block: false, alt: "A")[$A$] and show that there is only one solution. Answer #math.equation(block: true, alt: "5 5; 0 5")[$5 5 \ 0 5$] ] === #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/3%3A_Systems_of_ODEs/3.8%3A_Matrix_exponentials")[3.8: Matrix Exponentials] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Using the matrix exponential, find a fundamental matrix solution for the system , #math.equation(block: false, alt: "x prime equals 3 x plus y , y prime equals x plus 3 y")[$x^(′) = 3 x + y , y^(′) = x + 3 y$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find #math.equation(block: false, alt: "e to the power t A")[$e^(t A)$] for the matrix #math.equation(block: false, alt: "A equals [ 2, 3; 0, 2 ]")[$A = [ 2 & 3 \ 0 & 2 ]$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find a fundamental matrix solution for the system , #math.equation(block: false, alt: "x 1 ′ equals 7 x sub 1 plus 4 x sub 2 plus 12 x sub 3 , x 2 ′ equals x sub 1 plus 2 x sub 2 plus x sub 3 , x 3 ′ equals minus 3 x sub 1 minus 2 x sub 2 minus 5 x sub 3")[$x_(1)^(′) = 7 x_(1) + 4 x_(2) + 12 x_(3) , " " " " " " x_(2)^(′) = x_(1) + 2 x_(2) + x_(3) , " " " " " " x_(3)^(′) = − 3 x_(1) − 2 x_(2) − 5 x_(3)$]. Then find the solution that satisfies #math.equation(block: false, alt: "x → equals [ 0; 1; minus 2 ]")[$arrow(x) = [ 0 \ 1 \ − 2 ]$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Compute the matrix exponential #math.equation(block: false, alt: "e to the power A")[$e^(A)$] for #math.equation(block: false, alt: "A equals [ 1, 2; 0, 2 ]")[$A = [ 1 & 2 \ 0 & 2 ]$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[(challenging)] Suppose #math.equation(block: false, alt: "A B equals B A")[$A B = B A$] . Show that under this assumption, #math.equation(block: false, alt: "e to the power A plus B equals e to the power A e to the power B")[$e^(A + B) = e^(A) e^(B)$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Use Exercise #math.equation(block: false, alt: "3.8 .5")[$3.8 ".5"$] to show that #math.equation(block: false, alt: "open parenthesis e to the power A close parenthesis to the power minus 1 equals e to the power minus A")[$( e^(A) )^(− 1) = e^(− A)$]. In particular this means that #math.equation(block: false, alt: "e to the power A")[$e^(A)$] is invertible even if A is not. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Suppose #math.equation(block: false, alt: "A")[$A$] is a matrix with eigenvalues #math.equation(block: false, alt: "minus 1")[$− 1$], #math.equation(block: false, alt: "1")[$1$], and corresponding eigenvectors #math.equation(block: false, alt: "[ 1; 1 ]")[$[ 1 \ 1 ]$], #math.equation(block: false, alt: "[ 0; 1 ]")[$[ 0 \ 1 ]$]. + Find matrix #math.equation(block: false, alt: "A")[$A$] with these properties. + Find the fundamental matrix solution to #math.equation(block: false, alt: "x → prime equals A x →")[$arrow(x)^(′) = A arrow(x)$]. + Solve the system in with initial conditions #math.equation(block: false, alt: "x → open parenthesis 0 close parenthesis equals [ 2; 3 ]")[$arrow(x) ( 0 ) = [ 2 \ 3 ]$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Suppose that #math.equation(block: false, alt: "A")[$A$] is an #math.equation(block: false, alt: "n times n")[$n × n$] matrix with a repeated eigenvalue #math.equation(block: false, alt: "λ")[$λ$] of multiplicity n. Suppose that there are n linearly independent eigenvectors. Show that the matrix is diagonal, in particular #math.equation(block: false, alt: "A equals λ 𝐼")[$A = λ 𝐼$]. Hint: Use diagonalization and the fact that the identity matrix commutes with every other matrix. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Let #math.equation(block: false, alt: "A equals [ minus 1, minus 1; 1, minus 3 ]")[$A = [ − 1 & − 1 \ 1 & − 3 ]$]. + Find #math.equation(block: false, alt: "e to the power t A")[$e^(t A)$]. + Solve #math.equation(block: false, alt: "x → prime equals A x →")[$arrow(x)^(′) = A arrow(x)$], #math.equation(block: false, alt: "x → open parenthesis 0 close parenthesis equals [ 1; minus 2 ]")[$arrow(x) ( 0 ) = [ 1 \ − 2 ]$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Let #math.equation(block: false, alt: "A equals [ 1, 2; 3, 4 ]")[$A = [ 1 & 2 \ 3 & 4 ]$]. Approximate #math.equation(block: false, alt: "e to the power t A")[$e^(t A)$] by expanding the power series up to the third order. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ For any positive integer #math.equation(block: false, alt: "n")[$n$], find a formula (or a recipe) for #math.equation(block: false, alt: "A to the power n")[$A^(n)$] for the following matrices: + #math.equation(block: false, alt: "3 0; 0 9")[$3 0 \ 0 9$] + #math.equation(block: false, alt: "5 2; 4 7")[$5 2 \ 4 7$] + #math.equation(block: false, alt: "0 1; 0 0")[$0 1 \ 0 0$] + #math.equation(block: false, alt: "2 1; 0 2")[$2 1 \ 0 2$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Compute #math.equation(block: false, alt: "e to the power t A")[$e^(t A)$] where #math.equation(block: false, alt: "A equals [ 1, minus 2; minus 2, 1 ]")[$A = [ 1 & − 2 \ − 2 & 1 ]$]. Answer #math.equation(block: true, alt: "the fraction e to the power 3 t plus e to the power minus t over 2 the fraction e to the power minus t minus e to the power 3 t over 2; the fraction e to the power minus t minus e to the power 3 t over 2 the fraction e to the power 3 t plus e to the power minus t over 2")[$frac(e^(3 t) + e^(− t), 2) frac(e^(− t) − e^(3 t), 2) \ frac(e^(− t) − e^(3 t), 2) frac(e^(3 t) + e^(− t), 2)$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Compute #math.equation(block: false, alt: "e to the power t A")[$e^(t A)$] where #math.equation(block: false, alt: "A equals [ 1, minus 3, 2; minus 2, 1, 2; minus 1, minus 3, 4 ]")[$A = [ 1 & − 3 & 2 \ − 2 & 1 & 2 \ − 1 & − 3 & 4 ]$]. Answer #math.equation(block: true, alt: "2 e to the power 3 t minus 4 e to the power 2 t plus 3 e to the power t the fraction 3 e to the power t over 2 minus the fraction 3 e to the power 3 t over 2 minus e to the power 3 t plus 4 e to the power 2 t minus 3 e to the power t; 2 e to the power t minus 2 e to the power 2 t e to the power t 2 e to the power 2 t minus 2 e to the power t; 2 e to the power 3 t minus 5 e to the power 2 t plus 3 e to the power t the fraction 3 e to the power t over 2 minus the fraction 3 e to the power 3 t over 2 minus e to the power 3 t plus 5 e to the power 2 t minus 3 e to the power t")[$2 e^(3 t) − 4 e^(2 t) + 3 e^(t) frac(3 e^(t), 2) − frac(3 e^(3 t), 2) − e^(3 t) + 4 e^(2 t) − 3 e^(t) \ 2 e^(t) − 2 e^(2 t) e^(t) 2 e^(2 t) − 2 e^(t) \ 2 e^(3 t) − 5 e^(2 t) + 3 e^(t) frac(3 e^(t), 2) − frac(3 e^(3 t), 2) − e^(3 t) + 5 e^(2 t) − 3 e^(t)$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ + Compute #math.equation(block: false, alt: "e to the power t A")[$e^(t A)$] where #math.equation(block: false, alt: "A equals [ 3, minus 1; 1, 1 ]")[$A = [ 3 & − 1 \ 1 & 1 ]$]. + Solve #math.equation(block: false, alt: "x → prime equals A x →")[$arrow(x)^(′) = A arrow(x)$] for #math.equation(block: false, alt: "x → open parenthesis 0 close parenthesis equals [ 1; 2 ]")[$arrow(x) ( 0 ) = [ 1 \ 2 ]$]. Answer + #math.equation(block: false, alt: "open parenthesis t plus 1 close parenthesis e to the power 2 t minus t e to the power 2 t; t e to the power 2 t open parenthesis 1 minus t close parenthesis e to the power 2 t")[$( t + 1 ) e^(2 t) − t e^(2 t) \ t e^(2 t) ( 1 − t ) e^(2 t)$] + #math.equation(block: false, alt: "open parenthesis 1 minus t close parenthesis e to the power 2 t; open parenthesis 2 minus t close parenthesis e to the power 2 t")[$( 1 − t ) e^(2 t) \ ( 2 − t ) e^(2 t)$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Compute the first 3 terms (up to the second degree) of the Taylor expansion of #math.equation(block: false, alt: "e to the power t A")[$e^(t A)$] where #math.equation(block: false, alt: "A equals [ 2, 3; 2, 2 ]")[$A = [ 2 & 3 \ 2 & 2 ]$] (Write as a single matrix). Then use it to approximate #math.equation(block: false, alt: "e to the power 0.1 A")[$e^(0.1 A)$]. Answer #math.equation(block: true, alt: "1 plus 2 t plus 5 t squared 3 t plus 6 t squared; 2 t plus 4 t squared 1 plus 2 t plus 5 t squared")[$1 + 2 t + 5 t^(2) 3 t + 6 t^(2) \ 2 t + 4 t^(2) 1 + 2 t + 5 t^(2)$] #math.equation(block: true, alt: "1.25 0.36; 0.24 1.25")[$1.25 0.36 \ 0.24 1.25$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ For any positive integer #math.equation(block: false, alt: "n")[$n$], find a formula (or a recipe) for #math.equation(block: false, alt: "A to the power n")[$A^(n)$] for the following matrices: + #math.equation(block: false, alt: "7 4; minus 5 minus 2")[$7 4 \ − 5 − 2$] + #math.equation(block: false, alt: "minus 3 4; minus 6 minus 7")[$− 3 4 \ − 6 − 7$] + #math.equation(block: false, alt: "0 1; 1 0")[$0 1 \ 1 0$] Answer + #math.equation(block: false, alt: "5 open parenthesis 3 to the power n close parenthesis minus 2 to the power n plus 2 4 open parenthesis 3 to the power n close parenthesis minus 2 to the power n plus 2; 5 open parenthesis 2 to the power n close parenthesis minus 5 open parenthesis 3 to the power n close parenthesis 5 open parenthesis 2 to the power n close parenthesis minus 4 open parenthesis 3 to the power n close parenthesis")[$5 ( 3^(n) ) − 2^(n + 2) 4 ( 3^(n) ) − 2^(n + 2) \ 5 ( 2^(n) ) − 5 ( 3^(n) ) 5 ( 2^(n) ) − 4 ( 3^(n) )$] + #math.equation(block: false, alt: "3 minus 2 open parenthesis 3 to the power n close parenthesis 2 open parenthesis 3 to the power n close parenthesis minus 2; 3 minus 3 to the power n plus 1 3 to the power n plus 1 minus 2")[$3 − 2 ( 3^(n) ) 2 ( 3^(n) ) − 2 \ 3 − 3^(n + 1) 3^(n + 1) − 2$] + #math.equation(block: false, alt: "1 0; 0 1")[$1 0 \ 0 1$] if #math.equation(block: false, alt: "n")[$n$] is even, and #math.equation(block: false, alt: "0 1; 1 0")[$0 1 \ 1 0$] if #math.equation(block: false, alt: "n")[$n$] is odd. ] === #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/3%3A_Systems_of_ODEs/3.9%3A_Nonhomogeneous_systems")[3.9: Nonhomogeneous Systems] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find a particular solution to #math.equation(block: false, alt: "x prime equals x plus 2 y plus 2 t , y prime equals 3 x plus 2 y minus 4")[$x^(′) = x + 2 y + 2 t , #h(0.222em) y^(′) = 3 x + 2 y − 4$], + using integrating factor method, + using eigenvector decomposition, + using undetermined coefficients. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the general solution to #math.equation(block: false, alt: "x prime equals 4 x plus y minus 1 , y prime equals x plus 4 y minus e to the power t")[$x^(′) = 4 x + y − 1 , #h(0.222em) y^(′) = x + 4 y − e^(t)$], + using integrating factor method, + using eigenvector decomposition, + using undetermined coefficients. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the general solution to #math.equation(block: false, alt: "x 1 ′ prime equals minus 6 x sub 1 plus 3 x sub 2 plus cos open parenthesis t close parenthesis , x 2 ″ equals 2 x sub 1 minus 7 x sub 2 plus 3 cos open parenthesis t close parenthesis")[$x_(1)^(′)^(′) = − 6 x_(1) + 3 x_(2) + cos ( t ) , #h(0.222em) x_(2)^(″) = 2 x_(1) − 7 x_(2) + 3 cos ( t )$], + using eigenvector decomposition, + using undetermined coefficients. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the general solution to #math.equation(block: false, alt: "x 1 ″ equals minus 6 x sub 1 plus 3 x sub 2 plus cos open parenthesis 2 t close parenthesis , x 2 ″ equals 2 x sub 1 minus 7 x sub 2 plus cos open parenthesis 2 t close parenthesis")[$x_(1)^(″) = − 6 x_(1) + 3 x_(2) + cos ( 2 t ) , #h(0.222em) x_(2)^(″) = 2 x_(1) − 7 x_(2) + cos ( 2 t )$], + using eigenvector decomposition, + using undetermined coefficients. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Take the equation #math.equation(block: true, alt: "the fraction 1 over t minus 1; 1 the fraction 1 over t")[$frac(1, t) − 1 \ 1 frac(1, t)$] + Check that #math.equation(block: true, alt: "t sin t; minus t cos t")[$t sin t \ − t cos t$] is the complementary solution. + Use variation of parameters to find a particular solution. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find a particular solution to #math.equation(block: false, alt: "x prime equals 5 x plus 4 y plus t , y prime equals x plus 8 y minus t")[$x^(′) = 5 x + 4 y + t , #h(0.222em) y^(′) = x + 8 y − t$], + using integrating factor method, + using eigenvector decomposition, + using undetermined coefficients. Answer The general solution is (particular solutions should agree with one of these): #math.equation(block: false, alt: "x open parenthesis t close parenthesis equals C sub 1 e to the power 9 t plus 4 C sub 2 e to the power 4 t minus the fraction t over 3 minus the fraction 5 over 54")[$x ( t ) = C_(1) e^(9 t) + 4 C_(2) e^(4 t) − frac(t, 3) − frac(5, 54)$], #math.equation(block: false, alt: "y open parenthesis t close parenthesis equals C sub 1 e to the power 9 t minus C sub 2 e to the power 4 t plus the fraction t over 6 plus the fraction 7 over 216")[$y ( t ) = C_(1) e^(9 t) − C_(2) e^(4 t) + frac(t, 6) + frac(7, 216)$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find a particular solution to #math.equation(block: false, alt: "x prime equals y plus e to the power t , y prime equals x plus e to the power t")[$x^(′) = y + e^(t) , #h(0.222em) y^(′) = x + e^(t)$], + using integrating factor method, + using eigenvector decomposition, + using undetermined coefficients. Answer The general solution is (particular solutions should agree with one of these): #math.equation(block: false, alt: "x open parenthesis t close parenthesis equals C sub 1 e to the power t plus C sub 2 e to the power minus t plus t e to the power t")[$x ( t ) = C_(1) e^(t) + C_(2) e^(− t) + t e^(t)$], #math.equation(block: false, alt: "y open parenthesis t close parenthesis equals C sub 1 e to the power t minus C sub 2 e to the power minus t plus t e to the power t")[$y ( t ) = C_(1) e^(t) − C_(2) e^(− t) + t e^(t)$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "x 1 ′ equals x sub 2 plus t , x 2 ′ equals x sub 1 plus t")[$x_(1)^(′) = x_(2) + t , x_(2)^(′) = x_(1) + t$] with initial conditions #math.equation(block: false, alt: "x sub 1 open parenthesis 0 close parenthesis equals 1 , x sub 2 open parenthesis 0 close parenthesis equals 2")[$x_(1) ( 0 ) = 1 , x_(2) ( 0 ) = 2$], using eigenvector decomposition. Answer #math.equation(block: true, alt: "1; 1")[$1 \ 1$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "x 1 ″ equals minus 3 x sub 1 plus x sub 2 plus t , x 2 ″ equals 9 x sub 1 plus 5 x sub 2 plus cos open parenthesis t close parenthesis")[$x_(1)^(″) = − 3 x_(1) + x_(2) + t , x_(2)^(″) = 9 x_(1) + 5 x_(2) + cos ( t )$] with initial conditions #math.equation(block: false, alt: "x sub 1 open parenthesis 0 close parenthesis equals 0 , x sub 2 open parenthesis 0 close parenthesis equals 0 , x 1 ′ open parenthesis 0 close parenthesis equals 0 , x 2 ′ open parenthesis 0 close parenthesis equals 0")[$x_(1) ( 0 ) = 0 , x_(2) ( 0 ) = 0 , x_(1)^(′) ( 0 ) = 0 , x_(2)^(′) ( 0 ) = 0$], using eigenvector decomposition. Answer #math.equation(block: true, alt: "1; 9")[$1 \ 9$] ]