To be used for in-course assessment (Problem solving exercises 1) in MAS3801
", "licence": "None specified"}, "duration": 0, "percentPass": 0, "showQuestionGroupNames": false, "shuffleQuestionGroups": false, "showstudentname": true, "question_groups": [{"name": "Group", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questionNames": ["", "", "", ""], "variable_overrides": [[], [], [], []], "questions": [{"name": "General solution", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Toby Wood", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/400/"}], "tags": [], "metadata": {"description": "", "licence": "All rights reserved"}, "statement": "This question concerns the form of the general solution of an Ordinary Differential Equation (ODE).
\nSuppose that $p(x)$ and $q(x)$ are given functions. For each of the ODEs below, suppose that $y_1(x)$ and $y_2(x)$ are two different (i.e. independent) solutions of the ODE, and $A$ and $B$ are arbitrary constants. Identify the correct form for the general solution of the ODE.
\n[Hint: If you're unsure, consider the case where $p = 0$ and $q = 1$.]
", "advice": "(a) Because this is a second-order ODE, its general solution must involve precisely two arbitrary constants. And because this ODE is linear and homogeneous, the general solution can be written as a superposition of independent solutions, which in this case are $y_1$ and $y_2$.
\n(b) Because this is a first-order ODE, its general solution must involve precisely one arbitrary constant. And because this ODE is linear and inhomogeneous, the general solution can be written as a particular integral plus a complementary function. Since we know that $y_1$ and $y_2$ are independent solutions, we can take the particular integral to be $y_1$ and the complementary function to be $A(y_2-y_1)$.
", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": [], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "1_n_2", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "For the second-order, homogeneous ODE
\n\\[\\dfrac{\\mathrm{d}^2y}{\\mathrm{d}x^2} + p\\dfrac{\\mathrm{d}y}{\\mathrm{d}x} + qy = 0\\]
\nthe general solution can be written as:
", "minMarks": 0, "maxMarks": 0, "shuffleChoices": true, "displayType": "radiogroup", "displayColumns": "1", "showCellAnswerState": true, "choices": ["$y = Ay_1 + By_2$", "$y = (1-A)y_1 + Ay_2$", "$y = A + By_1y_2$", "$y = Ay_1 + Ay_2$"], "matrix": ["1", 0, 0, 0], "distractors": ["", "", "", ""]}, {"type": "1_n_2", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "For the first-order, inhomogeneous ODE
\n\\[\\dfrac{\\mathrm{d}y}{\\mathrm{d}x} + py + q = 0\\]
\nthe general solution can be written as:
", "minMarks": 0, "maxMarks": 0, "shuffleChoices": true, "displayType": "radiogroup", "displayColumns": "1", "showCellAnswerState": true, "choices": ["$y = Ay_1 + By_2$", "$y = (1-A)y_1 + Ay_2$", "$y = A + By_1y_2$", "$y = Ay_1 + Ay_2$"], "matrix": ["0", "1", 0, 0], "distractors": ["", "", "", ""]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always", "type": "question"}, {"name": "Integration by substitution", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Toby Wood", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/400/"}], "tags": [], "metadata": {"description": "", "licence": "All rights reserved"}, "statement": "Consider the definite integral
\n\\[ \\int_{x=\\var{c}}^{\\var{d}} \\dfrac{x}{\\sqrt{\\var{a^2}-x^2}} \\, \\mathrm{d}x \\]
", "advice": "(a) With this change of variable, we have $\\mathrm{d}x = \\dfrac{\\mathrm{d}x}{\\mathrm{d}s}\\,\\mathrm{d}s = \\var{a}\\cos(s)\\,\\mathrm{d}s$. Thus the integral becomes
\n\\[ \\int_{s=\\arcsin(\\var{c/a})}^{\\arcsin(\\var{d/a})} \\var{a}\\sin(s) \\, \\mathrm{d}s \\]
\n(b) The value of the integral is
\n\\[ \\int_{s=\\arcsin(\\var{c/a})}^{\\arcsin(\\var{d/a})} \\var{a}\\sin(s) \\, \\mathrm{d}s = - \\var{a} \\left[\\cos(s)\\right]_{s=\\arcsin(\\var{c/a})}^{\\arcsin(\\var{d/a})} \\]
\n\\[ = \\simplify{sqrt({a^2-c^2})-sqrt({a^2-d^2})} \\]
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\nwhere the constants $A,B,C$ have the values
\n$A =\\ $[[0]]
\n$B = \\arcsin(\\ $[[1]]$)$
\n$C = \\arcsin(\\ $[[2]]$)$
\nNote: Each answer can be written as a fraction or an integer.
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", "answer": "sqrt({a^2-c^2})-sqrt({a^2-d^2})", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": "0.01", "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": []}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always", "type": "question"}, {"name": "Change of variable", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Toby Wood", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/400/"}], "tags": [], "metadata": {"description": "Solve a 2nd-order ODE by making two changes of variable
", "licence": "None specified"}, "statement": "Consider the following linear Ordinary Differential Equation (ODE) for $y(x)$:
\n\\[\\simplify{x^2y'' + {1-2a-c}x}y' + (\\simplify{{a}{a+c}+{b}^2{c}^2x^{2c}})y = 0\\,.\\]
", "advice": "a)
\nWith the change of variable $t = x^\\var{c}$, we have $\\dfrac{\\mathrm{d}t}{\\mathrm{d}x} = \\simplify{{c}x^{c-1}}$, and so
\n\\[\\dfrac{\\mathrm{d}}{\\mathrm{d}x} = \\simplify{{c}x^{c-1}}\\dfrac{\\mathrm{d}}{\\mathrm{d}t} = \\var{c}t^\\simplify{{c-1}/{c}}\\dfrac{\\mathrm{d}}{\\mathrm{d}t}\\,.\\]
\nTherefore the original ODE becomes
\n\\[\\simplify{t^{2/{c}}}\\left(\\var{c}t^\\simplify{{c-1}/{c}}\\dfrac{\\mathrm{d}}{\\mathrm{d}t}\\right)^2y + (\\simplify{{1-2a-c}t^{1/{c}}})\\left(\\var{c}t^\\simplify{{c-1}/{c}}\\dfrac{\\mathrm{d}}{\\mathrm{d}t}\\right)y + (\\simplify{{a}{a+c}+{b}^2{c}^2t^2})y = 0\\,.\\]
\nAfter expanding out the derivative terms and simplifying the result, this reduces to the answer given above, i.e.
\n\\[t^2\\dfrac{\\mathrm{d}^2y}{\\mathrm{d}t^2} + (\\simplify{-2{a}/{c}t})\\dfrac{\\mathrm{d}y}{\\mathrm{d}t} + (\\simplify{{a}{a+c}/{c^2} + {b^2}t^2})y = 0\\,.\\]
\nb)
\nWith the change of variable $y = \\simplify{t^{{a}/{c}}}w(t)$, the ODE above becomes
\n\\[t^2\\dfrac{\\mathrm{d}^2}{\\mathrm{d}t^2}(\\simplify{t^{{a}/{c}}}w(t)) + (\\simplify{-2{a}/{c}t})\\dfrac{\\mathrm{d}}{\\mathrm{d}t}(\\simplify{t^{{a}/{c}}}w(t)) + (\\simplify{{a}{a+c}/{c^2} + {b^2}t^2})(\\simplify{t^{{a}/{c}}}w(t)) = 0\\,.\\]
\nExpanding out the derivative terms and simplifying, we eventually obtain
\n\\[\\dfrac{\\mathrm{d}^2w}{\\mathrm{d}t^2} + \\var{b^2}w = 0\\,.\\]
\nc)
\nThe general solution of the above ODE is
\n\\[w = \\alpha\\cos(\\var{b}t) + \\beta\\sin(\\var{b}t)\\,.\\]
\nRewriting this in terms of the original variables, $x$ and $y$, it becomes
\n\\[y = \\simplify{x^{a}}\\left[\\alpha\\cos(\\var{b}x^\\var{c}) + \\beta\\sin(\\var{b}x^\\var{c})\\right]\\,.\\]
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\n\\[t^2\\dfrac{\\mathrm{d}^2y}{\\mathrm{d}t^2} + At\\dfrac{\\mathrm{d}y}{\\mathrm{d}t} + (B + Ct^2)y = 0\\]
\nwhere
\nUnder the further change of variable $y = \\simplify{t^{{a}/{c}}}w(t)$, the ODE becomes
\n\\[\\dfrac{\\mathrm{d}^2w}{\\mathrm{d}t^2} + Dw = 0\\]
\nwhere $D = $[[0]]
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\n\\[y = x^E\\left[\\alpha\\cos(Fx^G) + \\beta\\sin(Fx^G)\\right]\\]
\nwhere $\\alpha$ and $\\beta$ are arbitrary constants and
\nIdentify a regular singularity in an ODE and find the Frobenius series solution
", "licence": "None specified"}, "statement": "The following linear ODE for $y(x)$,
\n\\[y'' + \\dfrac{2y'}{\\simplify{x-{b}}} + \\dfrac{\\simplify{{a}{1-alpha}y'}}{(\\simplify{x-{a+b}})(\\simplify{x-{b}})} + \\dfrac{\\simplify{{c}y}}{(\\simplify{x-{b}})^4} = 0\\,,\\]
\nhas two singular points: a regular singularity and an irregular (or essential) singularity.
", "advice": "a)
\nRecall the definition of singular points:
\nAn ODE of the form
\n\\[y'' + Py' + Qy = 0\\]
\nhas singular points wherever the functions $P$ and $Q$ are undefined. If the ODE has a singular point at $x=x_0$, and if the functions $(x-x_0)P(x)$ and $(x-x_0)^2Q(x)$ are well-behaved in the limit $x \\to x_0$, then we say that $x_0$ is a regular singular point. Otherwise it is an irregular singular point (also known as an essential singularity).
\nBy this definition, it's clear that our ODE has singular points at $x=\\var{a+b}$ and $x=\\var{b}$. The point $x=\\var{a+b}$ is a regular singularity (because $(\\simplify{x-{a+b}})\\times\\dfrac{\\simplify{{a}{1-alpha}}}{(\\simplify{x-{a+b}})(\\simplify{x-{b}})}$ is well-behaved as $x\\to\\var{a+b}$), whereas the point $x = \\var{b}$ is an essential singularity (because $(\\simplify{x-{b}})^2\\times\\dfrac{\\var{c}}{(\\simplify{x-{b}})^4}$ is undefined in the limit $x\\to\\var{b}$).
\nb)
\nThe change of variable from $x$ to $t$ relocates the regular singularity to $t=0$. We will now use the \"method of Frobenius\" to express the solution in the form of a power series. If we suppose that the leading-order term in the series is $y = t^\\alpha + \\ldots$, for some constant $\\alpha$, then plugging this into the ODE we obtain
\n\\[\\alpha(\\alpha-1)(\\simplify{t+{a}})^4t^{\\alpha-1} + \\ldots + (\\simplify{2t+{a}{1-alpha}})(\\simplify{t+{a}})^3\\alpha t^{\\alpha-1} + \\ldots + \\simplify{{c}t}^{\\alpha+1} + \\ldots = 0\\]
\nThe \"biggest\" terms in the limit $t\\to0$ here are
\n\\[\\var{a^4}\\alpha(\\alpha-1)t^{\\alpha-1} + (\\simplify{{a^4}{1-alpha}})\\alpha t^{\\alpha-1}\\]
\nand in order for these terms to cancel $\\alpha$ must satisfy the indicial equation
\n\\[\\simplify{alpha^2 - {alpha}alpha} = 0\\,,\\]
\nwhich has two roots: $\\alpha=0$ and $\\alpha=\\simplify{{alpha}}$. Since these two roots do not differ by an integer, we know that the general solution will be of the form
\n\\[y = A[1 + a_1t + a_2t^2 + \\ldots] + B\\simplify{t^{alpha}}[1 + b_1t+ b_2t^2 + \\ldots]\\]
\n(If the two roots of the indicial equation differ by an integer, then the solution may include logarithmic terms.)
\nTo determine the coefficients in the power series, we simply have to insert the power series into the ODE, expand all of the terms in powers of $t$, and then equate the coefficient of each power of $t$ to zero. First, let's consider the solution
\n\\[y = 1 + a_1t + a_2t^2 + \\ldots\\]
\nInserting this into the ODE, we obtain
\n\\[(\\simplify{{a}+t})^4(2a_2t + \\ldots) + (\\simplify{{a}{1-alpha}+2t})(\\simplify{{a}+t})^3(a_1 + 2a_2t + \\ldots) + (\\var{c})(t + a_1t^2 + a_2t^3 + \\ldots) = 0\\,.\\]
\nWhen we expand everything here as a power series in $t$, we find that the leading-order term is $\\simplify{{1-alpha}{a^4}}a_1$. Since this term must vanish, we conclude that $a_1=0$. (The next-order terms then turn out to be $\\simplify{2{2-alpha}{a^4}a_2 t + {c}t}$, and to make this vanish we require $a_2 = \\simplify{-{c}/(2{2-alpha}{a}^4)}$.)
\nNow, let's consider the solution
\n\\[y = \\simplify{t^{alpha}} + b_1\\simplify{t^{alpha+1}} + b_2\\simplify{t^{alpha+2}} + \\ldots\\]
\nInserting this into the ODE, we obtain
\n\\[t(\\simplify{{a}+t})^4(\\simplify{{alpha}{alpha-1}t^{alpha-2}} + \\simplify{{alpha+1}{alpha}b_1 t^{alpha-1}} + \\simplify{{alpha+2}{alpha+1}b_2 t^{alpha}} + \\ldots)\\]\\[ + (\\simplify{{a}+t})^3(\\simplify{{a}{1-alpha}+2t})(\\simplify{{alpha}t^{alpha-1}} + \\simplify{{alpha+1}b_1 t^{alpha}} + \\simplify{{alpha+2}b_2 t^{alpha+1}} + \\ldots)\\]\\[ + (\\var{c})t(\\simplify{t^{alpha}} + b_1\\simplify{t^{alpha+1}} + b_2\\simplify{t^{alpha+2}} + \\ldots) = 0\\,.\\]
\nIf we now expand everything here as a power series in $t$, retaining terms up to order $\\simplify{t^{alpha}}$, we eventually obtain
\n\\[\\simplify{{a^4}{alpha}{alpha-1}t^{alpha-1}} + (\\simplify{4{alpha-1}{alpha}{a^3} + {alpha+1}{alpha}{a^4}b_1}) t^\\var{alpha} - \\simplify{{alpha-1}{alpha}{a^4}t^{alpha-1}} + (\\simplify{3{1-alpha}{alpha}{a^3} + 2{alpha}{a^3} + {alpha+1}{1-alpha}{a^4}b_1}) t^\\var{alpha} + \\ldots = 0\\,.\\]
\nThe terms that are of order $\\simplify{t^{alpha-1}}$ cancel out (as we should expect) and we are left with simply
\n\\[\\left(\\simplify{{alpha+1}{a^4}b_1 + {alpha+1}{alpha}{a^3}}\\right)\\simplify{t^{alpha}} + \\ldots = 0\\,.\\]
\nTherefore we must have $b_1 = \\simplify{{-alpha/a}}$.
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\n\\[(\\simplify{t+{a}})^4ty'' + (\\simplify{2t+{a}{1-alpha}})(\\simplify{t+{a}})^3y' + \\simplify{{c}t y} = 0\\,,\\]
\nwhere $'$ now indicates a derivative with respect to $t$. (You should check this result!)
\n\nBy formulating the indicial equation, it can be shown that the general solution has the form
\n\\[y = A[1 + a_1t + a_2t^2 + \\ldots] + Bt^\\alpha[1 + b_1t+ b_2t^2 + \\ldots]\\]
\nwhere $A$ and $B$ are arbitrary constants. What are the values of the constants $\\alpha$ and $a_1$ in this solution?
\nFor an extra challenge, try to find the correct values for $a_2$ and $b_1$ as well. Doing so will be give you more practice at working with power series, which will be good preparation for the exam!
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