Find a second solution to an ODE by first finding the Wronskian.
", "licence": "All rights reserved"}, "statement": "Consider the homogeneous, 2nd-order ODE
\n\\begin{equation}x^2\\dfrac{\\mathrm{d}^2y}{\\mathrm{d}x^2} - \\var{2r-1}x\\dfrac{\\mathrm{d}y}{\\mathrm{d}x} + \\var{r^2}y = 0\\,. \\qquad\\qquad(1)\\end{equation}
", "advice": "a) If we substitute $y = x^r$ into the ODE, we find that $r(r-1)x^r - \\var{2r-1}x^r + \\var{r^2}x^r = 0$. This is satisfied (for all values of $x$) if and only if the constant $r$ satisfies the quadratic equation $(r-\\var{r})^2 = 0$, i.e. if $r = \\var{r}$.
\nb) For any two particular solutions of this ODE, say $y = y_1(x)$ and $y = y_2(x)$, the \"Wronskian\" $W(x) \\equiv y_1y_2^{\\prime} - y_1^{\\prime}y_2$ must satisfy \\[\\dfrac{\\mathrm{d}W}{\\mathrm{d}x} = a(x)\\,W\\,,\\] where here $a(x) = \\dfrac{\\var{2r-1}}{x}$. Solving this equation for $W$, we find that $W(x) = Ax^{\\var{2r-1}}$, where $A$ is an arbitrary constant. We know that one solution of our ODE is $y_1 = x^\\var{r}$, and so if $y_2$ is any other solution then it must satisfy \\[y_1y_2^{\\prime} - y_1^{\\prime}y_2 = W\\] and so \\[x^\\var{r}y_2 - \\var{r}x^\\var{r-1}y_2 = Ax^\\var{2r-1}\\,.\\] We can solve this first-order equation for $y_2$ using an integrating factor, which leads to \\begin{align}\\dfrac{\\mathrm{d}}{\\mathrm{d}x}[x^\\var{-r}y_2] &= \\dfrac{A}{x} \\\\ \\Rightarrow x^\\var{-r}y_2 &= A\\ln(x) + B \\\\ \\Rightarrow y_2 &= Ax^\\var{r}\\ln(x) + Bx^\\var{r}\\,,\\end{align} where $B$ is another arbitrary constant. This is, in fact, the general solution of the original ODE.
\nThe only solution that satisfies the given boundary conditions has $A=B=1$, and so $y = x^\\var{r}\\ln(x)$.
\nc) The adjoint of equation (1) is \\[\\dfrac{\\mathrm{d}^2}{\\mathrm{d}x^2}(x^2z) + \\var{2r-1}\\dfrac{\\mathrm{d}}{\\mathrm{d}x}(xz) + \\var{r^2}z = 0\\,.\\] Expanding out the derivative terms (using the product rule repeatedly) and simplifying, we end up with \\[x^2\\dfrac{\\mathrm{d}^2z}{\\mathrm{d}x^2} + \\var{2r+3}x\\dfrac{\\mathrm{d}z}{\\mathrm{d}x} + \\var{(r+1)^2}z = 0\\,.\\] So $c = \\var{2r+3}$ and $d = \\var{(r+1)^2}$.
", "rulesets": {}, "extensions": [], "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"r": {"name": "r", "group": "Ungrouped variables", "definition": "random(2,3,4)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["r"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "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": "This equation has a solution $y=x^r$, where $r$ is a constant. Enter the correct value for this constant below.
\n$r = $ [[0]]
", "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "{r}", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": []}], "sortAnswers": false}, {"type": "gapfill", "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": "Rewrite the ODE in the form
\n\\[\\dfrac{\\mathrm{d}^2y}{\\mathrm{d}x^2} = a(x)\\dfrac{\\mathrm{d}y}{\\mathrm{d}x} + b(x)y\\]
\nand then use the \"Wronskian Method\" (see Problems Sheet 1) to obtain the general solution.
\nEnter into the box the particular solution that satisfies the boundary conditions $y(1)=0$ and $y'(1) = 1$.
\n$y(x) = $ [[0]]
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\n$c = $ [[0]] and $d = $ [[1]]
\n[Hint: One solution of the adjoint equation is $z = x^\\var{-r-1}$.]
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