#set document(title: "0.4 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")) == 0.4#h(0.6em)Exercises === Introduction to Differential Equations #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Show that #math.equation(block: false, alt: "x equals e to the power 4 t")[$x = e^(4 t)$] is a solution to #math.equation(block: false, alt: "x to the power ‴ minus 12 x double prime plus 48 x prime minus 64 x equals 0")[$x^(‴) − 12 x^(″) + 48 x^(′) − 64 x = 0$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Show that #math.equation(block: false, alt: "x equals e to the power t")[$x = e^(t)$] is not a solution to #math.equation(block: false, alt: "x to the power ‴ minus 12 x double prime plus 48 x prime minus 64 x equals 0")[$x^(‴) − 12 x^(″) + 48 x^(′) − 64 x = 0$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Is #math.equation(block: false, alt: "y equals sin t")[$y = sin t$] a solution to #math.equation(block: false, alt: "open parenthesis the fraction d y over d t close parenthesis squared equals 1 minus y squared")[$attach(( frac(d y, d t) ), t: 2) = 1 − y^(2)$]? Justify. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Let #math.equation(block: false, alt: "y double prime plus 2 y prime minus 8 y equals 0")[$y^(″) + 2 y^(′) − 8 y = 0$]. Now try a solution of the form #math.equation(block: false, alt: "y equals e to the power r x")[$y = e^(r x)$] for some (unknown) constant #math.equation(block: false, alt: "r")[$r$]. Is this a solution for some #math.equation(block: false, alt: "r")[$r$]? If so, find all such #math.equation(block: false, alt: "r")[$r$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Verify that #math.equation(block: false, alt: "x equals C e to the power minus 2 t")[$x = C e^(− 2 t)$] is a solution to #math.equation(block: false, alt: "x prime equals minus 2 x")[$x^(′) = − 2 x$]. Find #math.equation(block: false, alt: "C")[$C$] to solve for the initial condition #math.equation(block: false, alt: "x open parenthesis 0 close parenthesis equals 100")[$x ( 0 ) = 100$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Verify that #math.equation(block: false, alt: "x equals C sub 1 e to the power minus t plus C sub 2 e to the power 2 t")[$x = C_(1) e^(− t) + C_(2) e^(2 t)$] is a solution to #math.equation(block: false, alt: "x double prime minus x prime minus 2 x equals 0")[$x^(″) − x^(′) − 2 x = 0$]. Find #math.equation(block: false, alt: "C sub 1")[$C_(1)$] and #math.equation(block: false, alt: "C sub 2")[$C_(2)$] to solve for the initial conditions #math.equation(block: false, alt: "x open parenthesis 0 close parenthesis equals 10")[$x ( 0 ) = 10$] and #math.equation(block: false, alt: "x prime open parenthesis 0 close parenthesis equals 0")[$x^(′) ( 0 ) = 0$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find a solution to #math.equation(block: false, alt: "open parenthesis x prime close parenthesis squared plus x squared equals 4")[$attach(( x^(′) ), t: 2) + x^(2) = 4$] using your knowledge of derivatives of functions that you know from basic calculus. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve: + #math.equation(block: false, alt: "the fraction d A over d t equals minus 10 A , A open parenthesis 0 close parenthesis equals 5")[$frac(d A, d t) = − 10 A , #h(1em) A ( 0 ) = 5$] + #math.equation(block: false, alt: "the fraction d H over d x equals 3 H , H open parenthesis 0 close parenthesis equals 1")[$frac(d H, d x) = 3 H , #h(1em) H ( 0 ) = 1$] + #math.equation(block: false, alt: "the fraction d squared y over d x squared equals 4 y , y open parenthesis 0 close parenthesis equals 0 , y prime open parenthesis 0 close parenthesis equals 1")[$frac(d^(2) y, d x^(2)) = 4 y , #h(1em) y ( 0 ) = 0 , #h(1em) y^(′) ( 0 ) = 1$] + #math.equation(block: false, alt: "the fraction d squared x over d y squared equals minus 9 x , x open parenthesis 0 close parenthesis equals 1 , x prime open parenthesis 0 close parenthesis equals 0")[$frac(d^(2) x, d y^(2)) = − 9 x , #h(1em) x ( 0 ) = 1 , #h(1em) x^(′) ( 0 ) = 0$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Is there a solution to #math.equation(block: false, alt: "y prime equals y")[$y^(′) = y$], such that #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals y open parenthesis 1 close parenthesis")[$y ( 0 ) = y ( 1 )$]? ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ The population of city #math.equation(block: false, alt: "X")[$X$] was #math.equation(block: false, alt: "100")[$100$] thousand #math.equation(block: false, alt: "20")[$20$] years ago, and the population of city #math.equation(block: false, alt: "X")[$X$] was #math.equation(block: false, alt: "120")[$120$] thousand #math.equation(block: false, alt: "10")[$10$] years ago. Assuming constant growth, you can use the exponential population model (like for the bacteria). What do you estimate the population is now? ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Suppose that a football coach gets a salary of one million dollars now, and a raise of #math.equation(block: false, alt: "10")[$10$] every year (so exponential model, like population of bacteria). Let #math.equation(block: false, alt: "s")[$s$] be the salary in millions of dollars, and #math.equation(block: false, alt: "t")[$t$] is time in years. + What is #math.equation(block: false, alt: "s open parenthesis 0 close parenthesis")[$s ( 0 )$] and #math.equation(block: false, alt: "s open parenthesis 1 close parenthesis")[$s ( 1 )$]. + Approximately how many years will it take for the salary to be #math.equation(block: false, alt: "10")[$10$] million. + Approximately how many years will it take for the salary to be #math.equation(block: false, alt: "20")[$20$] million. + Approximately how many years will it take for the salary to be #math.equation(block: false, alt: "30")[$30$] million. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Show that #math.equation(block: false, alt: "x equals e to the power minus 2 t")[$x = e^(− 2 t)$] is a solution to #math.equation(block: false, alt: "x double prime plus 4 x prime plus 4 x equals 0")[$x^(″) + 4 x^(′) + 4 x = 0$]. Answer Compute #math.equation(block: false, alt: "x prime equals minus 2 e to the power minus 2 t")[$x^(′) = − 2 e^(− 2 t)$] and #math.equation(block: false, alt: "x double prime equals 4 e to the power minus 2 t")[$x^(″) = 4 e^(− 2 t)$]. Then #math.equation(block: false, alt: "open parenthesis 4 e to the power minus 2 t close parenthesis plus 4 open parenthesis minus 2 e to the power minus 2 t close parenthesis plus 4 open parenthesis e to the power minus 2 t close parenthesis equals 0")[$( 4 e^(− 2 t) ) + 4 ( − 2 e^(− 2 t) ) + 4 ( e^(− 2 t) ) = 0$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Is #math.equation(block: false, alt: "y equals x squared")[$y = x^(2)$] a solution to #math.equation(block: false, alt: "x squared y double prime minus 2 y equals 0")[$x^(2) y^(″) − 2 y = 0$]? Justify. Answer Yes. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Let #math.equation(block: false, alt: "x y double prime minus y prime equals 0")[$x y^(″) − y^(′) = 0$]. Try a solution of the form #math.equation(block: false, alt: "y equals x to the power r")[$y = x^(r)$]. Is this a solution for some #math.equation(block: false, alt: "r")[$r$]? If so, find all such #math.equation(block: false, alt: "r")[$r$]. Answer #math.equation(block: false, alt: "y equals x to the power r")[$y = x^(r)$] is a solution for #math.equation(block: false, alt: "r equals 0")[$r = 0$] and #math.equation(block: false, alt: "r equals 2")[$r = 2$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Verify that #math.equation(block: false, alt: "x equals C sub 1 e to the power t plus C sub 2")[$x = C_(1) e^(t) + C_(2)$] is a solution to #math.equation(block: false, alt: "x double prime minus x prime equals 0")[$x^(″) − x^(′) = 0$]. Find #math.equation(block: false, alt: "C sub 1")[$C_(1)$] and #math.equation(block: false, alt: "C sub 2")[$C_(2)$] so that #math.equation(block: false, alt: "x")[$x$] satisfies #math.equation(block: false, alt: "x open parenthesis 0 close parenthesis equals 10")[$x ( 0 ) = 10$] and #math.equation(block: false, alt: "x prime open parenthesis 0 close parenthesis equals 100")[$x^(′) ( 0 ) = 100$]. Answer #math.equation(block: true, alt: "C sub 1 equals 100")[$C_(1) = 100$], #math.equation(block: true, alt: "C sub 2 equals minus 90")[$C_(2) = − 90$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "the fraction d φ over d s equals 8 φ")[$frac(d φ, d s) = 8 φ$] and #math.equation(block: false, alt: "φ open parenthesis 0 close parenthesis equals minus 9")[$φ ( 0 ) = − 9$]. Answer #math.equation(block: true, alt: "φ equals minus 9 e to the power 8 s")[$φ = − 9 e^(8 s)$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve: + #math.equation(block: false, alt: "the fraction d x over d t equals minus 4 x , x open parenthesis 0 close parenthesis equals 9")[$frac(d x, d t) = − 4 x , #h(1em) x ( 0 ) = 9$] + #math.equation(block: false, alt: "the fraction d squared x over d t squared equals minus 4 x , x open parenthesis 0 close parenthesis equals 1 , x prime open parenthesis 0 close parenthesis equals 2")[$frac(d^(2) x, d t^(2)) = − 4 x , #h(1em) x ( 0 ) = 1 , #h(1em) x^(′) ( 0 ) = 2$] + #math.equation(block: false, alt: "the fraction d p over d q equals 3 p , p open parenthesis 0 close parenthesis equals 4")[$frac(d p, d q) = 3 p , #h(1em) p ( 0 ) = 4$] + #math.equation(block: false, alt: "the fraction d squared T over d x squared equals 4 T , T open parenthesis 0 close parenthesis equals 0 , T prime open parenthesis 0 close parenthesis equals 6")[$frac(d^(2) T, d x^(2)) = 4 T , #h(1em) T ( 0 ) = 0 , #h(1em) T^(′) ( 0 ) = 6$] Answer + #math.equation(block: false, alt: "x equals 9 e to the power minus 4 t")[$x = 9 e^(− 4 t)$] + #math.equation(block: false, alt: "x equals cos open parenthesis 2 t close parenthesis plus sin open parenthesis 2 t close parenthesis")[$x = cos ( 2 t ) + sin ( 2 t )$] + #math.equation(block: false, alt: "p equals 4 e to the power 3 q")[$p = 4 e^(3 q)$] + #math.equation(block: false, alt: "T equals 3 sinh open parenthesis 2 x close parenthesis")[$T = 3 sinh ( 2 x )$] ] === Classification of Differential Equations #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Classify the following equations. Are they ODE or PDE? Is it an equation or a system? What is the order? Is it linear or nonlinear, and if it is linear, is it homogeneous, constant coefficient? If it is an ODE, is it autonomous? + #math.equation(block: false, alt: "sin open parenthesis t close parenthesis the fraction d squared x over d t squared plus cos open parenthesis t close parenthesis x equals t squared")[$sin ( t ) frac(d^(2) x, d t^(2)) + cos ( t ) x = t^(2)$] + #math.equation(block: false, alt: "the fraction ∂ u over ∂ x plus 3 the fraction ∂ u over ∂ y equals x y")[$frac(∂ u, ∂ x) + 3 frac(∂ u, ∂ y) = x y$] + #math.equation(block: false, alt: "y double prime plus 3 y plus 5 x equals 0 , x double prime plus x minus y equals 0")[$y^(″) + 3 y + 5 x = 0 , #h(1em) x^(″) + x − y = 0$] + #math.equation(block: false, alt: "the fraction ∂ squared u over ∂ t squared plus u the fraction ∂ squared u over ∂ s squared equals 0")[$frac(∂^(2) u, ∂ t^(2)) + u frac(∂^(2) u, ∂ s^(2)) = 0$] + #math.equation(block: false, alt: "x double prime plus t x squared equals t")[$x^(″) + t x^(2) = t$] + #math.equation(block: false, alt: "the fraction d to the power 4 x over d t to the power 4 equals 0")[$frac(d^(4) x, d t^(4)) = 0$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ If #math.equation(block: false, alt: "u → equals open parenthesis u sub 1 , u sub 2 , u sub 3 close parenthesis")[$arrow(u) = ( u_(1) , u_(2) , u_(3) )$] is a vector, we have the divergence #math.equation(block: false, alt: "∇ times u → equals the fraction ∂ u sub 1 over ∂ x plus the fraction ∂ u sub 2 over ∂ y plus the fraction ∂ u sub 3 over ∂ z")[$∇ · arrow(u) = frac(∂ u_(1), ∂ x) + frac(∂ u_(2), ∂ y) + frac(∂ u_(3), ∂ z)$] and curl #math.equation(block: false, alt: "∇ times u → equals open parenthesis the fraction ∂ u sub 3 over ∂ y minus the fraction ∂ u sub 2 over ∂ z , the fraction ∂ u sub 1 over ∂ z minus the fraction ∂ u sub 3 over ∂ x , the fraction ∂ u sub 2 over ∂ x minus the fraction ∂ u sub 1 over ∂ y close parenthesis")[$∇ × arrow(u) = ( frac(∂ u_(3), ∂ y) − frac(∂ u_(2), ∂ z) , " " frac(∂ u_(1), ∂ z) − frac(∂ u_(3), ∂ x) , " " frac(∂ u_(2), ∂ x) − frac(∂ u_(1), ∂ y) )$]. Notice that curl of a vector is still a vector. Write out Maxwell’s equations in terms of partial derivatives and classify the system. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Suppose #math.equation(block: false, alt: "F")[$F$] is a linear function, that is, #math.equation(block: false, alt: "F open parenthesis x , y close parenthesis equals a x plus b y")[$F ( x , y ) = a x + b y$] for constants #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "b")[$b$]. What is the classification of equations of the form #math.equation(block: false, alt: "F open parenthesis y prime , y close parenthesis equals 0")[$F ( y^(′) , y ) = 0$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Write down an explicit example of a third order, linear, nonconstant coefficient, nonautonomous, nonhomogeneous system of two ODE such that every derivative that could appear, does appear. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Classify the following equations. Are they ODE or PDE? Is it an equation or a system? What is the order? Is it linear or nonlinear, and if it is linear, is it homogeneous, constant coefficient? If it is an ODE, is it autonomous? + #math.equation(block: false, alt: "the fraction ∂ squared v over ∂ x squared plus 3 the fraction ∂ squared v over ∂ y squared equals sin open parenthesis x close parenthesis")[$frac(∂^(2) v, ∂ x^(2)) + 3 frac(∂^(2) v, ∂ y^(2)) = sin ( x )$] + #math.equation(block: false, alt: "the fraction d x over d t plus cos open parenthesis t close parenthesis x equals t squared plus t plus 1")[$frac(d x, d t) + cos ( t ) x = t^(2) + t + 1$] + #math.equation(block: false, alt: "the fraction d to the power 7 F over d x to the power 7 equals 3 F open parenthesis x close parenthesis")[$frac(d^(7) F, d x^(7)) = 3 F ( x )$] + #math.equation(block: false, alt: "y double prime plus 8 y prime equals 1")[$y^(″) + 8 y^(′) = 1$] + #math.equation(block: false, alt: "x double prime plus t y x prime equals 0 , y double prime plus t x y equals 0")[$x^(″) + t y x^(′) = 0 , #h(1em) y^(″) + t x y = 0$] + #math.equation(block: false, alt: "the fraction ∂ u over ∂ t equals the fraction ∂ squared u over ∂ s squared plus u squared")[$frac(∂ u, ∂ t) = frac(∂^(2) u, ∂ s^(2)) + u^(2)$] Answer + PDE, equation, second order, linear, nonhomogeneous, constant coefficient. + ODE, equation, first order, linear, nonhomogeneous, not constant coefficient, not autonomous. + ODE, equation, seventh order, linear, homogeneous, constant coefficient, autonomous. + ODE, equation, second order, linear, nonhomogeneous, constant coefficient, autonomous. + ODE, system, second order, nonlinear. + PDE, equation, second order, nonlinear. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Write down the general #emph[zero]th order linear ordinary differential equation. Write down the general solution. Answer equation: #math.equation(block: false, alt: "a open parenthesis x close parenthesis y equals b open parenthesis x close parenthesis")[$a ( x ) y = b ( x )$], solution: #math.equation(block: false, alt: "y equals the fraction b open parenthesis x close parenthesis over a open parenthesis x close parenthesis")[$y = frac(b ( x ), a ( x ))$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ For which #math.equation(block: false, alt: "k")[$k$] is #math.equation(block: false, alt: "the fraction d x over d t plus x to the power k equals t to the power k plus 2")[$frac(d x, d t) + x^(k) = t^(k + 2)$] linear. Hint: there are two answers. Answer #math.equation(block: false, alt: "k equals 0")[$k = 0$] or #math.equation(block: false, alt: "k equals 1")[$k = 1$] ]