#set document(title: "2.7 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")) == 2.7#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)/2%3A_Higher_order_linear_ODEs/2.1%3A_Second_order_linear_ODEs")[2.1: Second order linear ODEs] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Show that #math.equation(block: false, alt: "y equals e to the power x")[$y = e^(x)$] and #math.equation(block: false, alt: "y equals e to the power 2 x")[$y = e^(2 x)$] are linearly independent. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Take #math.equation(block: false, alt: "y double prime plus 5 y equals 10 x plus 5")[$y^(″) + 5 y = 10 x + 5$]. Find (guess!) a solution. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Prove the superposition principle for nonhomogeneous equations. Suppose that #math.equation(block: false, alt: "y sub 1")[$y_(1)$] is a solution to #math.equation(block: false, alt: "L y sub 1 equals f open parenthesis x close parenthesis")[$L y_(1) = f ( x )$] and #math.equation(block: false, alt: "y sub 2")[$y_(2)$] is a solution to #math.equation(block: false, alt: "L y sub 2 equals g open parenthesis x close parenthesis")[$L y_(2) = g ( x )$] (same linear operator #math.equation(block: false, alt: "L")[$L$]). Show that #math.equation(block: false, alt: "y equals y sub 1 plus y sub 2")[$y = y_(1) + y_(2)$] solves #math.equation(block: false, alt: "L y equals f open parenthesis x close parenthesis plus g open parenthesis x close parenthesis")[$L y = f ( x ) + g ( x )$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ For the equation #math.equation(block: false, alt: "x squared y double prime minus x y prime equals 0")[$x^(2) y^(″) − x y^(′) = 0$], find two solutions, show that they are linearly independent and find the general solution. Hint: Try #math.equation(block: false, alt: "y equals x prime")[$y = x^(′)$]. ] Equations of the form #math.equation(block: false, alt: "a x squared y double prime plus b x y prime plus c y equals 0")[$a x^(2) y^(″) + b x y^(′) + c y = 0$] are called #emph[Euler’s equations] or #emph[Cauchy-Euler equations]. They are solved by trying #math.equation(block: false, alt: "y equals x to the power r")[$y = x^(r)$] and solving for #math.equation(block: false, alt: "r")[$r$] (we can assume that #math.equation(block: false, alt: "x ⩾ 0")[$x ⩾ 0$] for simplicity). #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Suppose that #math.equation(block: false, alt: "open parenthesis b minus a close parenthesis squared minus 4 a c greater than 0")[$attach(( b − a ), t: 2) − 4 a c > 0$]. + Find a formula for the general solution of #math.equation(block: false, alt: "a x squared y double prime plus b x y prime plus c y equals 0")[$a x^(2) y^(″) + b x y^(′) + c y = 0$]. Hint: Try #math.equation(block: false, alt: "y equals x to the power r")[$y = x^(r)$] and find a formula for #math.equation(block: false, alt: "r")[$r$]. + What happens when #math.equation(block: false, alt: "open parenthesis b minus a close parenthesis squared minus 4 a c equals 0")[$attach(( b − a ), t: 2) − 4 a c = 0$] or #math.equation(block: false, alt: "open parenthesis b minus a close parenthesis squared minus 4 a c less than 0")[$attach(( b − a ), t: 2) − 4 a c < 0$]? We will revisit the case when #math.equation(block: false, alt: "open parenthesis b minus a close parenthesis squared minus 4 a c less than 0")[$( b − a )^(2) − 4 a c < 0$] later. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Same equation as in Exercise #math.equation(block: false, alt: "2.1 .5")[$2.1 ".5"$]. Suppose #math.equation(block: false, alt: "open parenthesis b minus a close parenthesis squared minus 4 a c equals 0")[$attach(( b − a ), t: 2) − 4 a c = 0$]. Find a formula for the general solution of #math.equation(block: false, alt: "a x squared y double prime plus b x y prime plus c y equals 0")[$a x^(2) y^(″) + b x y^(′) + c y = 0$]. Hint: Try #math.equation(block: false, alt: "y equals x to the power r ln x")[$y = x^(r) ln x$] for the second solution. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[reduction of order] Suppose #math.equation(block: false, alt: "y sub 1")[$y_(1)$] is a solution to #math.equation(block: false, alt: "y double prime plus p open parenthesis x close parenthesis y prime plus q open parenthesis x close parenthesis y equals 0")[$y^(″) + p ( x ) y^(′) + q ( x ) y = 0$]. Show that #math.equation(block: true, alt: "y sub 2 open parenthesis x close parenthesis equals y sub 1 open parenthesis x close parenthesis ∫ the fraction e to the power minus ∫ p open parenthesis x close parenthesis d x over open parenthesis y sub 1 open parenthesis x close parenthesis close parenthesis squared d x")[$y_(2) ( x ) = y_(1) ( x ) ∫ frac(e^(− ∫ p ( x ) d x), attach(( y_(1) ( x ) ), t: 2)) d x$] is also a solution. Note: If you wish to come up with the formula for reduction of order yourself, start by trying #math.equation(block: false, alt: "y sub 2 open parenthesis x close parenthesis equals y sub 1 open parenthesis x close parenthesis v open parenthesis x close parenthesis")[$y_(2) ( x ) = y_(1) ( x ) v ( x )$]. Then plug #math.equation(block: false, alt: "y sub 2")[$y_(2)$] into the equation, use the fact that #math.equation(block: false, alt: "y sub 1")[$y_(1)$] is a solution, substitute #math.equation(block: false, alt: "w equals v prime")[$w = v^(′)$], and you have a first order linear equation in #math.equation(block: false, alt: "w")[$w$]. Solve for #math.equation(block: false, alt: "w")[$w$] and then for #math.equation(block: false, alt: "v")[$v$]. When solving for #math.equation(block: false, alt: "w")[$w$], make sure to include a constant of integration. Let us solve some famous equations using the method. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Chebyshev’s equation of order 1] Take #math.equation(block: false, alt: "open parenthesis 1 minus x squared close parenthesis y double prime minus x y prime plus y equals 0")[$( 1 − x^(2) ) y^(″) − x y^(′) + y = 0$]. + Show that #math.equation(block: false, alt: "y equals x")[$y = x$] is a solution. + Use reduction of order to find a second linearly independent solution. + Write down the general solution. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Hermite’s equation of order 2] Take #math.equation(block: false, alt: "y double prime minus 2 x y prime plus 4 y equals 0")[$y^(″) − 2 x y^(′) + 4 y = 0$]. + Show that #math.equation(block: false, alt: "y equals 1 minus 2 x squared")[$y = 1 − 2 x^(2)$] is a solution. + Use reduction of order to find a second linearly independent solution. + Write down the general solution. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Are #math.equation(block: false, alt: "sin open parenthesis x close parenthesis")[$sin ( x )$] and #math.equation(block: false, alt: "e to the power x")[$e^(x)$] linearly independent? Justify. Answer Yes. To justify try to find a constant #math.equation(block: false, alt: "A")[$A$] such that #math.equation(block: false, alt: "sin open parenthesis x close parenthesis equals A e to the power x")[$sin ( x ) = A e^(x)$] for all #math.equation(block: false, alt: "x")[$x$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Are #math.equation(block: false, alt: "e to the power x")[$e^(x)$] and #math.equation(block: false, alt: "e to the power x plus 2")[$e^(x + 2)$] linearly independent? Justify. Answer No. #math.equation(block: false, alt: "e to the power x plus 2 equals e squared e to the power x")[$e^(x + 2) = e^(2) e^(x)$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Guess a solution to #math.equation(block: false, alt: "y double prime plus y prime plus y equals 5")[$y^(″) + y^(′) + y = 5$]. Answer #math.equation(block: true, alt: "y equals 5")[$y = 5$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the general solution to #math.equation(block: false, alt: "x y double prime plus y prime equals 0")[$x y^(″) + y^(′) = 0$]. Hint: Notice that it is a first order ODE in #math.equation(block: false, alt: "y prime")[$y^(′)$]. Answer #math.equation(block: true, alt: "y equals C sub 1 ln open parenthesis x close parenthesis plus C sub 2")[$y = C_(1) ln ( x ) + C_(2)$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Write down an equation (guess) for which we have the solutions #math.equation(block: false, alt: "e to the power x")[$e^(x)$] and #math.equation(block: false, alt: "e to the power 2 x")[$e^(2 x)$]. Hint: Try an equation of the form #math.equation(block: false, alt: "y double prime plus A y prime plus B y equals 0")[$y^(″) + A y^(′) + B y = 0$] for constants #math.equation(block: false, alt: "A")[$A$] and #math.equation(block: false, alt: "B")[$B$], plug in both #math.equation(block: false, alt: "e to the power x")[$e^(x)$] and #math.equation(block: false, alt: "e to the power 2 x")[$e^(2 x)$] and solve for #math.equation(block: false, alt: "A")[$A$] and #math.equation(block: false, alt: "B")[$B$]. Answer #math.equation(block: true, alt: "y double prime minus 3 y prime plus 2 y equals 0")[$y^(″) − 3 y^(′) + 2 y = 0$] ] === #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/2%3A_Higher_order_linear_ODEs/2.2%3A_Constant_coefficient_second_order_linear_ODEs")[2.2: Constant coefficient second order linear ODEs] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the general solution of #math.equation(block: false, alt: "2 y double prime plus 2 y prime minus 4 y equals 0")[$2 y^(″) + 2 y^(′) − 4 y = 0$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the general solution of #math.equation(block: false, alt: "y double prime plus 9 y prime minus 10 y equals 0")[$y^(″) + 9 y^(′) − 10 y = 0$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "y double prime minus 8 y prime plus 16 y equals 0")[$y^(″) − 8 y^(′) + 16 y = 0$] for #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals 2 , y prime open parenthesis 0 close parenthesis equals 0")[$y ( 0 ) = 2 , y^(′) ( 0 ) = 0$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "y double prime plus 9 y prime equals 0")[$y^(″) + 9 y^(′) = 0$] for #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals 1 , y prime open parenthesis 0 close parenthesis equals 1")[$y ( 0 ) = 1 , y^(′) ( 0 ) = 1$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the general solution of #math.equation(block: false, alt: "2 y double prime plus 50 y equals 0")[$2 y^(″) + 50 y = 0$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the general solution of #math.equation(block: false, alt: "y double prime plus 6 y prime plus 13 y equals 0")[$y^(″) + 6 y^(′) + 13 y = 0$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the general solution of #math.equation(block: false, alt: "y double prime equals 0")[$y^(″) = 0$] using the methods of this section. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ The method of this section applies to equations of other orders than two. We will see higher orders later. Try to solve the first order equation #math.equation(block: false, alt: "2 y prime plus 3 y equals 0")[$2 y^(′) + 3 y = 0$] using the methods of this section. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Let us revisit Euler’s equations of Exercise #math.equation(block: false, alt: "1")[$1$]. Suppose now that #math.equation(block: false, alt: "open parenthesis b minus a close parenthesis squared minus 4 a c less than 0")[$attach(( b − a ), t: 2) − 4 a c < 0$]. Find a formula for the general solution of #math.equation(block: false, alt: "a x squared y double prime plus b x y prime plus c y equals 0")[$a x^(2) y^(″) + b x y^(′) + c y = 0$]. Hint: Note that #math.equation(block: false, alt: "x to the power r equals e to the power r ln x")[$x^(r) = e^(r ln x)$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the solution to #math.equation(block: false, alt: "y double prime minus open parenthesis 2 α close parenthesis y prime plus α squared y equals 0")[$y^(″) − ( 2 α ) y^(′) + α^(2) y = 0$], #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals a")[$y ( 0 ) = a$], #math.equation(block: false, alt: "y prime open parenthesis 0 close parenthesis equals b")[$y^(′) ( 0 ) = b$], where #math.equation(block: false, alt: "α")[$α$], #math.equation(block: false, alt: "a")[$a$], and #math.equation(block: false, alt: "b")[$b$] are real numbers. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Construct an equation such that #math.equation(block: false, alt: "y equals C sub 1 e to the power minus 2 x cos open parenthesis 3 x close parenthesis plus C sub 2 e to the power minus 2 x sin open parenthesis 3 x close parenthesis")[$y = C_(1) e^(− 2 x) cos ( 3 x ) + C_(2) e^(− 2 x) sin ( 3 x )$] is the general solution. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the general solution to #math.equation(block: false, alt: "y double prime plus 4 y prime plus 2 y equals 0")[$y^(″) + 4 y^(′) + 2 y = 0$]. Answer #math.equation(block: true, alt: "y equals C sub 1 e to the power open parenthesis minus 2 plus the square root of 2 close parenthesis x plus C sub 2 e to the power open parenthesis minus 2 minus the square root of 2 close parenthesis x")[$y = C_(1) e^(( − 2 + sqrt(2) ) x) + C_(2) e^(( − 2 − sqrt(2) ) x)$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the general solution to #math.equation(block: false, alt: "y double prime minus 6 y prime plus 9 y equals 0")[$y^(″) − 6 y^(′) + 9 y = 0$]. Answer #math.equation(block: true, alt: "y equals C sub 1 e to the power 3 x plus C sub 2 x e to the power 3 x")[$y = C_(1) e^(3 x) + C_(2) x e^(3 x)$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the solution to #math.equation(block: false, alt: "2 y double prime plus y prime plus y equals 0 , y open parenthesis 0 close parenthesis equals 1 , y prime open parenthesis 0 close parenthesis equals minus 2")[$2 y^(″) + y^(′) + y = 0 , y ( 0 ) = 1 , y^(′) ( 0 ) = − 2$]. Answer #math.equation(block: true, alt: "y equals e to the power minus x / 4 cos open parenthesis open parenthesis the fraction the square root of 7 over 4 close parenthesis x close parenthesis minus the square root of 7 e to the power minus x / 4 sin open parenthesis open parenthesis the fraction the square root of 7 over 4 close parenthesis x close parenthesis")[$y = e^(− x / 4) cos ( ( frac(sqrt(7), 4) ) x ) − sqrt(7) e^(− x / 4) sin ( ( frac(sqrt(7), 4) ) x )$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the solution to #math.equation(block: false, alt: "2 y double prime plus y prime minus 3 y equals 0 , y open parenthesis 0 close parenthesis equals a , y prime open parenthesis 0 close parenthesis equals b")[$2 y^(″) + y^(′) − 3 y = 0 , y ( 0 ) = a , y^(′) ( 0 ) = b$]. Answer #math.equation(block: true, alt: "y equals the fraction 2 open parenthesis a minus b close parenthesis over 5 e to the power minus 3 x / 2 plus the fraction 3 a plus 2 b over 5 e to the power x")[$y = frac(2 ( a − b ), 5) e^(− 3 x / 2) + frac(3 a + 2 b, 5) e^(x)$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the solution to #math.equation(block: false, alt: "z double prime open parenthesis t close parenthesis equals minus 2 z prime open parenthesis t close parenthesis minus 2 z open parenthesis t close parenthesis , z open parenthesis 0 close parenthesis equals 2 , z prime open parenthesis 0 close parenthesis equals minus 2")[$z^(″) ( t ) = − 2 z^(′) ( t ) − 2 z ( t ) , z ( 0 ) = 2 , z^(′) ( 0 ) = − 2$]. Answer #math.equation(block: true, alt: "z open parenthesis t close parenthesis equals 2 e to the power minus t cos open parenthesis t close parenthesis")[$z ( t ) = 2 e^(− t) cos ( t )$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the solution to #math.equation(block: false, alt: "y double prime minus open parenthesis α plus β close parenthesis y prime plus α β y equals 0")[$y^(″) − ( α + β ) y^(′) + α β y = 0$], #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals a")[$y ( 0 ) = a$], #math.equation(block: false, alt: "y prime open parenthesis 0 close parenthesis equals b")[$y^(′) ( 0 ) = b$], where #math.equation(block: false, alt: "α")[$α$], (\\beta\\), #math.equation(block: false, alt: "a")[$a$], and #math.equation(block: false, alt: "b")[$b$] are real numbers, and #math.equation(block: false, alt: "α not equal to β")[$α ≠ β$]. Answer #math.equation(block: true, alt: "y equals the fraction α β minus b over β minus α e to the power α x plus the fraction b minus a α over β minus α e to the power β x")[$y = frac(α β − b, β − α) e^(α x) + frac(b − a α, β − α) e^(β x)$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Construct an equation such that #math.equation(block: false, alt: "y equals C sub 1 e to the power 3 x plus C sub 2 e to the power minus 2 x")[$y = C_(1) e^(3 x) + C_(2) e^(− 2 x)$] is the general solution. Answer #math.equation(block: true, alt: "y double prime minus y prime minus 6 y equals 0")[$y^(″) − y^(′) − 6 y = 0$] ] === #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/2%3A_Higher_order_linear_ODEs/2.3%3A_Higher_order_linear_ODEs")[2.3: Higher order linear ODEs] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the general solution for #math.equation(block: false, alt: "y to the power ‴ minus y double prime plus y prime minus y equals 0")[$y^(‴) − y^(″) + y^(′) − y = 0$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the general solution for #math.equation(block: false, alt: "y to the power open parenthesis 4 close parenthesis minus 5 y to the power ‴ plus 6 y double prime equals 0")[$y^(( 4 )) − 5 y^(‴) + 6 y^(″) = 0$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the general solution for #math.equation(block: false, alt: "y to the power ‴ plus 2 y double prime plus 2 y prime equals 0")[$y^(‴) + 2 y^(″) + 2 y^(′) = 0$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Suppose the characteristic equation for a differential equation is #math.equation(block: false, alt: "open parenthesis r minus 1 close parenthesis squared open parenthesis r minus 2 close parenthesis squared equals 0")[$attach(( r − 1 ), t: 2) attach(( r − 2 ), t: 2) = 0$]. + Find such a differential equation. + Find its general solution. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Suppose that a fourth order equation has a solution #math.equation(block: false, alt: "y equals 2 e to the power 4 x x cos x")[$y = 2 e^(4 x) x cos x$]. + Find such an equation. + Find the initial conditions that the given solution satisfies. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the general solution for the equation of Exercise #math.equation(block: false, alt: "2.3 .5")[$2.3 ".5"$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Let #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals e to the power x minus cos x , g open parenthesis x close parenthesis equals e to the power x plus cos x")[$f ( x ) = e^(x) − cos x , g ( x ) = e^(x) + cos x$] and #math.equation(block: false, alt: "h open parenthesis x close parenthesis equals cos x")[$h ( x ) = cos x$]. Are #math.equation(block: false, alt: "f open parenthesis x close parenthesis , g open parenthesis x close parenthesis ,")[$f ( x ) , g ( x ) ,$] and #math.equation(block: false, alt: "h open parenthesis x close parenthesis")[$h ( x )$] and linearly independent? If so, show it, if not, find a linear combination that works. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Let #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 0 , g open parenthesis x close parenthesis equals cos x")[$f ( x ) = 0 , g ( x ) = cos x$], and #math.equation(block: false, alt: "h open parenthesis x close parenthesis equals sin x")[$h ( x ) = sin x$]. Are #math.equation(block: false, alt: "f open parenthesis x close parenthesis , g open parenthesis x close parenthesis ,")[$f ( x ) , g ( x ) ,$] and #math.equation(block: false, alt: "h open parenthesis x close parenthesis")[$h ( x )$] and linearly independent? If so, show it, if not, find a linear combination that works. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Are #math.equation(block: false, alt: "x , x squared , a n d x to the power 4")[$x , x^(2) , a n d x^(4)$] linearly independent? If so, show it, if not, find a linear combination that works. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Are #math.equation(block: false, alt: "e to the power x , x e to the power x , a n d x squared e to the power x")[$e^(x) , x e^(x) , a n d x^(2) e^(x)$] linearly independent? If so, show it, if not, find a linear combination that works. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find an equation such that #math.equation(block: false, alt: "y equals x e to the power minus 2 x sin open parenthesis 3 x close parenthesis")[$y = x e^(− 2 x) sin ( 3 x )$] is a solution. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the general solution of #math.equation(block: false, alt: "y to the power open parenthesis 5 close parenthesis minus y to the power open parenthesis 4 close parenthesis equals 0")[$y^(( 5 )) − y^(( 4 )) = 0$]. Answer #math.equation(block: true, alt: "y equals C sub 1 e to the power x plus C sub 2 x cubed plus C sub 3 x squared plus C sub 4 x plus C sub 5")[$y = C_(1) e^(x) + C_(2) x^(3) + C_(3) x^(2) + C_(4) x + C_(5)$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Suppose that the characteristic equation of a third order differential equation has roots #math.equation(block: false, alt: "3 ± 2 i")[$3 ± 2 i$]. + What is the characteristic equation? + Find the corresponding differential equation. + Find the general solution. Answer + #math.equation(block: false, alt: "r cubed minus 3 r squared plus 4 r minus 12 equals 0")[$r^(3) − 3 r^(2) + 4 r − 12 = 0$] + #math.equation(block: false, alt: "y to the power ‴ minus 3 y double prime plus 4 y prime minus 12 y equals 0")[$y^(‴) − 3 y^(″) + 4 y^(′) − 12 y = 0$] + #math.equation(block: false, alt: "y equals C sub 1 e to the power 3 x plus C sub 2 sin open parenthesis 2 x close parenthesis plus C sub 3 cos open parenthesis 2 x close parenthesis")[$y = C_(1) e^(3 x) + C_(2) sin ( 2 x ) + C_(3) cos ( 2 x )$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "1001 y to the power ‴ plus 3.2 y double prime plus π y prime minus the square root of 4 y equals 0 , y open parenthesis 0 close parenthesis equals 0 , y prime open parenthesis 0 close parenthesis equals 0 , y double prime open parenthesis 0 close parenthesis equals 0")[$1001 y^(‴) + 3.2 y^(″) + π y^(′) − sqrt(4) y = 0 , y ( 0 ) = 0 , y^(′) ( 0 ) = 0 , y^(″) ( 0 ) = 0$]. Answer #math.equation(block: true, alt: "y equals 0")[$y = 0$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Are #math.equation(block: false, alt: "e to the power x , e to the power x plus 1 , e to the power 2 x , sin open parenthesis x close parenthesis")[$e^(x) , e^(x + 1) , e^(2 x) , sin ( x )$] linearly independent? If so, show it, if not find a linear combination that works. Answer No. #math.equation(block: false, alt: "e to the power 1 e to the power x minus e to the power x plus 1 equals 0")[$e^(1) e^(x) − e^(x + 1) = 0$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Are #math.equation(block: false, alt: "sin open parenthesis x close parenthesis , x , x sin open parenthesis x close parenthesis")[$sin ( x ) , x , x sin ( x )$] linearly independent? If so, show it, if not find a linear combination that works. Answer Yes. (Hint: First note that #math.equation(block: false, alt: "sin open parenthesis x close parenthesis")[$sin ( x )$] is bounded. Then note that #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "x sin open parenthesis x close parenthesis")[$x sin ( x )$] cannot be multiples of each other.) ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find an equation such that #math.equation(block: false, alt: "y equals cos open parenthesis x close parenthesis")[$y = cos ( x )$], #math.equation(block: false, alt: "y equals sin open parenthesis x close parenthesis")[$y = sin ( x )$], #math.equation(block: false, alt: "y equals e to the power x")[$y = e^(x)$] are solutions. Answer #math.equation(block: true, alt: "y to the power ‴ minus y double prime plus y prime minus y equals 0")[$y^(‴) − y^(″) + y^(′) − y = 0$] ] === #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/2%3A_Higher_order_linear_ODEs/2.4%3A_Mechanical_Vibrations")[2.4: Mechanical Vibrations] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Consider a mass and spring system with a mass #math.equation(block: false, alt: "m equals 2")[$m = 2$], spring constant #math.equation(block: false, alt: "k equals 3")[$k = 3$], and damping constant #math.equation(block: false, alt: "c equals 1")[$c = 1$]. + Set up and find the general solution of the system. + Is the system underdamped, overdamped or critically damped? + If the system is not critically damped, find a #math.equation(block: false, alt: "c")[$c$] that makes the system critically damped. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Do Exercise #math.equation(block: false, alt: "2.4 .1")[$2.4 ".1"$] for #math.equation(block: false, alt: "m equals 3 , k equals 12 , a n d c equals 12")[$m = 3 , k = 12 , a n d c = 12$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Using the mks units (meters-kilograms-seconds), suppose you have a spring with spring constant #math.equation(block: false, alt: "4 the fraction N over m")[$4 frac(N, m)$]. You want to use it to weigh items. Assume no friction. You place the mass on the spring and put it in motion. + You count and find that the frequency is #math.equation(block: false, alt: "0.8 Hz")[$0.8 " Hz"$] (cycles per second). What is the mass? + Find a formula for the mass #math.equation(block: false, alt: "m")[$m$] given the frequency #math.equation(block: false, alt: "w")[$w$] in #math.equation(block: false, alt: "Hz")[$"Hz"$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Suppose we add possible friction to Exercise #math.equation(block: false, alt: "2.4 .3")[$2.4 ".3"$]. Further, suppose you do not know the spring constant, but you have two reference weights #math.equation(block: false, alt: "1 kg")[$1 " kg"$] and #math.equation(block: false, alt: "2 kg")[$2 " kg"$] to calibrate your setup. You put each in motion on your spring and measure the frequency. For the #math.equation(block: false, alt: "1 kg")[$1 " kg"$] weight you measured #math.equation(block: false, alt: "1.1 Hz")[$1.1 " Hz"$], for the #math.equation(block: false, alt: "2 kg")[$2 " kg"$] weight you measured #math.equation(block: false, alt: "0.8 Hz")[$0.8 " Hz"$]. + Find #math.equation(block: false, alt: "k")[$k$] (spring constant) and #math.equation(block: false, alt: "c")[$c$] (damping constant). + Find a formula for the mass in terms of the frequency in #math.equation(block: false, alt: "Hz")[$"Hz"$]. Note that there may be more than one possible mass for a given frequency. + For an unknown object you measured \\(0.2\\text{ Hz}\\), what is the mass of the object? Suppose that you know that the mass of the unknown object is more than a kilogram. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Suppose you wish to measure the friction a mass of #math.equation(block: false, alt: "0.1 kg")[$0.1 " kg"$] experiences as it slides along a floor (you wish to find #math.equation(block: false, alt: "c")[$c$]). You have a spring with spring constant #math.equation(block: false, alt: "k equals 5 the fraction N over m")[$k = 5 frac(N, m)$]. You take the spring, you attach it to the mass and fix it to a wall. Then you pull on the spring and let the mass go. You find that the mass oscillates with frequency #math.equation(block: false, alt: "1 Hz")[$1 " Hz"$]. What is the friction? ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ A mass of #math.equation(block: false, alt: "2")[$2$] kilograms is on a spring with spring constant #math.equation(block: false, alt: "k")[$k$] newtons per meter with no damping. Suppose the system is at rest and at time #math.equation(block: false, alt: "t equals 0")[$t = 0$] the mass is kicked and starts traveling at #math.equation(block: false, alt: "2")[$2$] meters per second. How large does #math.equation(block: false, alt: "k")[$k$] have to be to so that the mass does not go further than #math.equation(block: false, alt: "3")[$3$] meters from the rest position? Answer #math.equation(block: false, alt: "k equals the fraction 8 over 9")[$k = frac(8, 9)$] (and larger) ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Suppose we have an RLC circuit with a resistor of #math.equation(block: false, alt: "100")[$100$] miliohms (#math.equation(block: false, alt: "0.1")[$0.1$] ohms), inductor of inductance of #math.equation(block: false, alt: "50")[$50$] millihenries (#math.equation(block: false, alt: "0.05")[$0.05$] henries), and a capacitor of #math.equation(block: false, alt: "5")[$5$] farads, with constant voltage. + Set up the ODE equation for the current #math.equation(block: false, alt: "I")[$I$]. + Find the general solution. + Solve for #math.equation(block: false, alt: "I open parenthesis 0 close parenthesis equals 10")[$I ( 0 ) = 10$] and #math.equation(block: false, alt: "I prime open parenthesis 0 close parenthesis equals 0")[$I^(′) ( 0 ) = 0$]. Answer + #math.equation(block: false, alt: "0.05 I double prime plus 0.1 I prime plus open parenthesis the fraction 1 over 5 close parenthesis I equals 0")[$0.05 I^(″) + 0.1 I^(′) + ( frac(1, 5) ) I = 0$] + #math.equation(block: false, alt: "I equals C e to the power minus t cos open parenthesis the square root of 3 t minus γ close parenthesis")[$I = C e^(− t) cos ( sqrt(3) t − γ )$] + #math.equation(block: false, alt: "I equals 10 e to the power minus t cos open parenthesis the square root of 3 t close parenthesis plus the fraction 10 over the square root of 3 e to the power minus t sin open parenthesis the square root of 3 t close parenthesis")[$I = 10 e^(− t) cos ( sqrt(3) t ) + frac(10, sqrt(3)) e^(− t) sin ( sqrt(3) t )$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ A #math.equation(block: false, alt: "5000 kg")[$5000 " kg"$] railcar hits a bumper (a spring) at #math.equation(block: false, alt: "1 the fraction m over s")[$1 frac(m, s)$], and the spring compresses by #math.equation(block: false, alt: "0.1 m")[$0.1 " m"$]. Assume no damping. + Find #math.equation(block: false, alt: "k")[$k$]. + Find out how far does the spring compress when a #math.equation(block: false, alt: "10000 kg")[$10000 " kg"$] railcar hits the spring at the same speed. + If the spring would break if it compresses further than #math.equation(block: false, alt: "0.3 m")[$0.3 " m"$], what is the maximum mass of a railcar that can hit it at #math.equation(block: false, alt: "1 the fraction m over s")[$1 frac(m, s)$]? + What is the maximum mass of a railcar that can hit the spring without breaking at #math.equation(block: false, alt: "2 the fraction m over s")[$2 frac(m, s)$]? Answer + #math.equation(block: false, alt: "k equals 500000")[$k = 500000$] + #math.equation(block: false, alt: "the fraction 1 over 5 the square root of 2 approximately equals 0.141")[$frac(1, 5 sqrt(2)) ≈ 0.141$] + #math.equation(block: false, alt: "45000 kg")[$45000 " kg"$] + #math.equation(block: false, alt: "11250 kg")[$11250 " kg"$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ A mass of #math.equation(block: false, alt: "m kg")[$m " kg"$] is on a spring with #math.equation(block: false, alt: "k equals 3 the fraction N over m")[$k = 3 frac("N", "m")$] and #math.equation(block: false, alt: "c equals 2 the fraction Ns over m")[$c = 2 frac("Ns", "m")$]. Find the mass #math.equation(block: false, alt: "m sub 0")[$m_(0)$] for which there is critical damping. If #math.equation(block: false, alt: "m less than m sub 0")[$m < m_(0)$], does the system oscillate or not, that is, is it underdamped or overdamped? Answer #math.equation(block: false, alt: "m sub 0 equals the fraction 1 over 2")[$m_(0) = frac(1, 2)$]. If #math.equation(block: false, alt: "m less than m sub 0")[$m < m_(0)$], then the system is overdamped and will not oscillate. ] === #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/2%3A_Higher_order_linear_ODEs/2.5%3A_Nonhomogeneous_Equations")[2.5: Nonhomogeneous Equations] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find a particular solution of #math.equation(block: false, alt: "y double prime minus y prime minus 6 y equals e to the power 2 x")[$y^(″) − y^(′) − 6 y = e^(2 x)$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find a particular solution of #math.equation(block: false, alt: "y double prime minus 4 y prime plus 4 y equals e to the power 2 x")[$y^(″) − 4 y^(′) + 4 y = e^(2 x)$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve the initial value problem #math.equation(block: false, alt: "y double prime plus 9 y equals cos open parenthesis 3 x close parenthesis plus sin open parenthesis 3 x close parenthesis")[$y^(″) + 9 y = cos ( 3 x ) + sin ( 3 x )$] for #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals 2 , y prime open parenthesis 0 close parenthesis equals 1")[$y ( 0 ) = 2 , y^(′) ( 0 ) = 1$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Setup the form of the particular solution but do not solve for the coefficients for #math.equation(block: false, alt: "y to the power open parenthesis 4 close parenthesis minus 2 y to the power ‴ plus y double prime equals e to the power x")[$y^(( 4 )) − 2 y^(‴) + y^(″) = e^(x)$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Setup the form of the particular solution but do not solve for the coefficients for #math.equation(block: false, alt: "y to the power open parenthesis 4 close parenthesis minus 2 y to the power ‴ plus y double prime equals e to the power x plus x plus sin x")[$y^(( 4 )) − 2 y^(‴) + y^(″) = e^(x) + x + sin x$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ + Using variation of parameters find a particular solution of #math.equation(block: false, alt: "y double prime minus 2 y prime plus y equals e to the power x")[$y^(″) − 2 y^(′) + y = e^(x)$]. + Find a particular solution using undetermined coefficients. + Are the two solutions you found the same? What is going on? See also Exercise #math.equation(block: false, alt: "2.5 .9")[$2.5 ".9"$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find a particular solution of #math.equation(block: false, alt: "y double prime minus 2 y prime plus y equals sin open parenthesis x squared close parenthesis")[$y^(″) − 2 y^(′) + y = sin ( x^(2) )$]. It is OK to leave the answer as a definite integral. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ For an arbitrary constant #math.equation(block: false, alt: "c")[$c$] find a particular solution to #math.equation(block: false, alt: "y double prime minus y equals e to the power c x")[$y^(″) − y = e^(c x)$]. Hint: Make sure to handle every possible real #math.equation(block: false, alt: "c")[$c$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ + Using variation of parameters find a particular solution of #math.equation(block: false, alt: "y double prime minus y equals e to the power x")[$y^(″) − y = e^(x)$] + Find a particular solution using undetermined coefficients. + Are the two solutions you found the same? What is going on? ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find a polynomial #math.equation(block: false, alt: "P open parenthesis x close parenthesis")[$P ( x )$], so that #math.equation(block: false, alt: "y equals 2 x squared plus 3 x plus 4")[$y = 2 x^(2) + 3 x + 4$] solves #math.equation(block: false, alt: "y double prime plus 5 y prime plus y equals P open parenthesis x close parenthesis")[$y^(″) + 5 y^(′) + y = P ( x )$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find a particular solution to #math.equation(block: false, alt: "y double prime minus y prime plus y equals 2 sin open parenthesis 3 x close parenthesis")[$y^(″) − y^(′) + y = 2 sin ( 3 x )$] Answer #math.equation(block: true, alt: "y equals the fraction minus 16 sin open parenthesis 3 x close parenthesis plus 6 cos open parenthesis 3 x close parenthesis over 73")[$y = frac(− 16 sin ( 3 x ) + 6 cos ( 3 x ), 73)$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ + Find a particular solution to #math.equation(block: false, alt: "y double prime plus 2 y equals e to the power x plus x cubed")[$y^(″) + 2 y = e^(x) + x^(3)$]. + Find the general solution. Answer + #math.equation(block: false, alt: "y equals the fraction 2 e to the power x plus 3 x cubed minus 9 x over 6")[$y = frac(2 e^(x) + 3 x^(3) − 9 x, 6)$] + #math.equation(block: false, alt: "y equals C sub 1 cos open parenthesis the square root of 2 x close parenthesis plus C sub 2 sin open parenthesis the square root of 2 x close parenthesis plus the fraction 2 e to the power x plus 3 x cubed minus 9 x over 6")[$y = C_(1) cos ( sqrt(2) x ) + C_(2) sin ( sqrt(2) x ) + frac(2 e^(x) + 3 x^(3) − 9 x, 6)$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "y double prime plus 2 y prime plus y equals x squared , y open parenthesis 0 close parenthesis equals 1 , y prime open parenthesis 0 close parenthesis equals 2")[$y^(″) + 2 y^(′) + y = x^(2) , y ( 0 ) = 1 , y^(′) ( 0 ) = 2$]. Answer #math.equation(block: true, alt: "y open parenthesis x close parenthesis equals x squared minus 4 x plus 6 plus e to the power minus x open parenthesis x minus 5 close parenthesis")[$y ( x ) = x^(2) − 4 x + 6 + e^(− x) ( x − 5 )$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Use variation of parameters to find a particular solution of #math.equation(block: false, alt: "y double prime minus y equals the fraction 1 over e to the power x plus e to the power minus x")[$y^(″) − y = frac(1, e^(x) + e^(− x))$]. Answer #math.equation(block: true, alt: "y equals the fraction 2 x e to the power x minus open parenthesis e to the power x plus e to the power minus x close parenthesis log open parenthesis e to the power 2 x plus 1 close parenthesis over 4")[$y = frac(2 x e^(x) − ( e^(x) + e^(− x) ) log ( e^(2 x) + 1 ), 4)$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ For an arbitrary constant #math.equation(block: false, alt: "c")[$c$] find the general solution to #math.equation(block: false, alt: "y double prime minus 2 y equals sin open parenthesis x plus c close parenthesis")[$y^(″) − 2 y = sin ( x + c )$]. Answer #math.equation(block: true, alt: "y equals the fraction minus sin open parenthesis x plus c close parenthesis over 3 plus C sub 1 e to the power the square root of 2 x plus C sub 2 e to the power minus the square root of 2 x")[$y = frac(− sin ( x + c ), 3) + C_(1) e^(sqrt(2) x) + C_(2) e^(− sqrt(2) x)$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Undetermined coefficients can sometimes be used to guess a particular solution to other equations than constant coefficients. Find a polynomial #math.equation(block: false, alt: "y open parenthesis x close parenthesis")[$y ( x )$] that solves #math.equation(block: false, alt: "y prime plus x y equals x cubed plus 2 x squared plus 5 x plus 2")[$y^(′) + x y = x^(3) + 2 x^(2) + 5 x + 2$]. Note: Not every right hand side will allow a polynomial solution, for example, #math.equation(block: false, alt: "y prime plus x y equals 1")[$y^(′) + x y = 1$] does not, but a technique based on undetermined coefficients does work, see #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/7%3A_Power_series_methods")[Chapter 7]. Answer #math.equation(block: true, alt: "y equals x squared plus 2 x plus 3")[$y = x^(2) + 2 x + 3$] ] === #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/2%3A_Higher_order_linear_ODEs/2.6%3A_Forced_Oscillations_and_Resonance")[2.6: Forced Oscillations and Resonance] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Derive a formula for #math.equation(block: false, alt: "x sub s p")[$x_(s p)$] if the equation is #math.equation(block: false, alt: "m x double prime plus c x prime plus k x equals F sub 0 sin open parenthesis ω t close parenthesis")[$m x^(″) + c x^(′) + k x = F_(0) sin ( ω t )$]. Assume #math.equation(block: false, alt: "c greater than 0")[$c > 0$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Derive a formula for #math.equation(block: false, alt: "x sub s p")[$x_(s p)$] if the equation is #math.equation(block: false, alt: "m x double prime plus c x prime plus k x equals F sub 0 cos open parenthesis ω t close parenthesis plus F sub 1 cos open parenthesis 3 ω t close parenthesis")[$m x^(″) + c x^(′) + k x = F_(0) cos ( ω t ) + F_(1) cos ( 3 ω t )$]. Assume #math.equation(block: false, alt: "c greater than 0")[$c > 0$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Take #math.equation(block: false, alt: "m x double prime plus c x prime plus k x equals F sub 0 cos open parenthesis ω t close parenthesis")[$m x^(″) + c x^(′) + k x = F_(0) cos ( ω t )$]. Fix #math.equation(block: false, alt: "m greater than 0")[$m > 0$] and #math.equation(block: false, alt: "k greater than 0")[$k > 0$]. Now think of the function #math.equation(block: false, alt: "C open parenthesis ω close parenthesis")[$C ( ω )$]. For what values of #math.equation(block: false, alt: "c")[$c$] (solve in terms of #math.equation(block: false, alt: "m , k ,")[$m , k ,$] and #math.equation(block: false, alt: "F sub 0")[$F_(0)$] ) will there be no practical resonance (that is, for what values of #math.equation(block: false, alt: "c")[$c$] is there no maximum of #math.equation(block: false, alt: "C open parenthesis ω close parenthesis")[$C ( ω )$] for #math.equation(block: false, alt: "ω greater than 0")[$ω > 0$] )? ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Take #math.equation(block: false, alt: "m x double prime plus c x prime plus k x equals F sub 0 cos open parenthesis ω t close parenthesis")[$m x^(″) + c x^(′) + k x = F_(0) cos ( ω t )$]. Fix #math.equation(block: false, alt: "c greater than 0")[$c > 0$] and #math.equation(block: false, alt: "k greater than 0")[$k > 0$]. Now think of the function #math.equation(block: false, alt: "C open parenthesis ω close parenthesis")[$C ( ω )$]. For what values of #math.equation(block: false, alt: "m")[$m$] (solve in terms of #math.equation(block: false, alt: "c , k ,")[$c , k ,$] and #math.equation(block: false, alt: "F sub 0")[$F_(0)$]) will there be no practical resonance (that is, for what values of #math.equation(block: false, alt: "m")[$m$] is there no maximum of #math.equation(block: false, alt: "C open parenthesis ω close parenthesis")[$C ( ω )$] for #math.equation(block: false, alt: "ω greater than 0")[$ω > 0$] )? ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Suppose a water tower in an earthquake acts as a mass-spring system. Assume that the container on top is full and the water does not move around. The container then acts as a mass and the support acts as the spring, where the induced vibrations are horizontal. Suppose that the container with water has a mass of #math.equation(block: false, alt: "m equals 10 , 000 k g")[$m = 10 , 000 k g$]. It takes a force of #math.equation(block: false, alt: "1000")[$1000$] newtons to displace the container #math.equation(block: false, alt: "1")[$1$] meter. For simplicity assume no friction. When the earthquake hits the water tower is at rest (it is not moving). Suppose that an earthquake induces an external force #math.equation(block: false, alt: "F open parenthesis t close parenthesis equals m A ω squared cos open parenthesis ω t close parenthesis .")[$F ( t ) = m A ω^(2) cos ( ω t ) .$] + What is the natural frequency of the water tower? + If #math.equation(block: false, alt: "ω")[$ω$] is not the natural frequency, find a formula for the maximal amplitude of the resulting oscillations of the water container (the maximal deviation from the rest position). The motion will be a high frequency wave modulated by a low frequency wave, so simply find the constant in front of the sines. + Suppose #math.equation(block: false, alt: "A equals 1")[$A = 1$] and an earthquake with frequency #math.equation(block: false, alt: "0.5")[$0.5$] cycles per second comes. What is the amplitude of the oscillations? Suppose that if the water tower moves more than #math.equation(block: false, alt: "1.5")[$1.5$] meter, the tower collapses. Will the tower collapse? ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ A mass of #math.equation(block: false, alt: "4 kg")[$4 " kg"$] on a spring with #math.equation(block: false, alt: "k equals 4")[$k = 4$] and a damping constant #math.equation(block: false, alt: "c equals 1")[$c = 1$]. Suppose that #math.equation(block: false, alt: "F sub 0 equals 2")[$F_(0) = 2$]. Using forcing of #math.equation(block: false, alt: "F sub 0 cos open parenthesis ω t close parenthesis")[$F_(0) cos ( ω t )$]. Find the #math.equation(block: false, alt: "ω")[$ω$] that causes practical resonance and find the amplitude. Answer #math.equation(block: true, alt: "ω equals the fraction the square root of 31 over 4 the square root of 2 approximately equals 0.984 C open parenthesis ω close parenthesis equals the fraction 16 over 3 the square root of 7 approximately equals 2.016")[$ω = frac(sqrt(31), 4 sqrt(2)) ≈ 0.984 #h(1em) C ( ω ) = frac(16, 3 sqrt(7)) ≈ 2.016$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Derive a formula for #math.equation(block: false, alt: "x sub s p")[$x_(s p)$] for #math.equation(block: false, alt: "m x double prime plus c x prime plus k x equals F sub 0 cos open parenthesis ω t close parenthesis plus A")[$m x^(″) + c x^(′) + k x = F_(0) cos ( ω t ) + A$] where #math.equation(block: false, alt: "A")[$A$] is some constant. Assume #math.equation(block: false, alt: "c greater than 0")[$c > 0$]. Answer #math.equation(block: false, alt: "x sub s p equals the fraction open parenthesis ω 0 2 minus ω squared close parenthesis F sub 0 over m open parenthesis 2 ω p close parenthesis squared plus m open parenthesis ω 0 2 minus ω squared close parenthesis squared cos open parenthesis ω t close parenthesis plus the fraction 2 ω p F sub 0 over m open parenthesis 2 ω p close parenthesis squared plus m open parenthesis ω 0 2 minus ω squared close parenthesis squared sin open parenthesis ω t close parenthesis plus the fraction A over k")[$x_(s p) = frac(( ω_(0)^(2) − ω^(2) ) F_(0), m ( 2 ω p )^(2) + m ( ω_(0)^(2) − ω^(2) )^(2)) cos ( ω t ) + frac(2 ω p F_(0), m ( 2 ω p )^(2) + m ( ω_(0)^(2) − ω^(2) )^(2)) sin ( ω t ) + frac(A, k)$], where #math.equation(block: false, alt: "p equals the fraction c over 2 m")[$p = frac(c, 2 m)$] and #math.equation(block: false, alt: "ω sub 0 equals the square root of the fraction k over m")[$ω_(0) = sqrt(frac(k, m))$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Suppose there is no damping in a mass and spring system with #math.equation(block: false, alt: "m equals 5 , k equals 20 ,")[$m = 5 , k = 20 ,$] and #math.equation(block: false, alt: "F sub 0 equals 5")[$F_(0) = 5$]. Suppose that #math.equation(block: false, alt: "ω")[$ω$] is chosen to be precisely the resonance frequency. + Find #math.equation(block: false, alt: "ω")[$ω$]. + Find the amplitude of the oscillations at time #math.equation(block: false, alt: "t equals 100")[$t = 100$]. Answer + #math.equation(block: false, alt: "ω equals 2")[$ω = 2$] + #math.equation(block: false, alt: "25")[$25$] ]