{
  "index": "1958-2-A-3",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "3. Under the assumption that the following set of relations has a unique solution for \\( u(t) \\), determine it.\n\\[\n\\begin{aligned}\n\\frac{d u(t)}{d t} & =u(t)+\\int_{0}^{1} u(s) d s \\\\\nu(0) & =1\n\\end{aligned}\n\\]",
  "solution": "Solution. Put\n\\[\nb=\\int_{0}^{1} u(s) d s\n\\]\n\nThen \\( b \\) is a constant and the given equation becomes the familiar linear differential equation\n\\[\nu^{\\prime}(t)=u(t)+b\n\\]\nwith the general solution\n\\[\nu(t)=-b+c e^{\\prime}\n\\]\n\nThere remains the problem of determining \\( b \\) and \\( c \\). We have\n\\[\nb=\\int_{0}^{1}\\left(-b+c e^{s}\\right) d s=-b+c(e-1)\n\\]\nand\n\\[\nu(0)=-b+c=1\n\\]\n\nThese two equations imply that\n\\[\nb=(e-1) /(3-e) \\text { and } c=2 /(3-e)\n\\]\nso\n\\[\nu(t)=\\frac{1}{3-e}\\left(2 e^{t}-e+1\\right)\n\\]\n\nIt is readily checked that this function is indeed a solution.\nThe above derivation proves the uniqueness of the solution function \\( u \\), so this assumption in the statement of the problem is unnecessary.",
  "vars": [
    "u",
    "t",
    "s"
  ],
  "params": [
    "b",
    "c"
  ],
  "sci_consts": [
    "e"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "u": "unknownu",
        "t": "variablet",
        "s": "variables",
        "b": "constb",
        "c": "constc"
      },
      "question": "3. Under the assumption that the following set of relations has a unique solution for \\( unknownu(variablet) \\), determine it.\n\\[\n\\begin{aligned}\n\\frac{d\\, unknownu(variablet)}{d\\, variablet} & = unknownu(variablet)+\\int_{0}^{1} unknownu(variables)\\, d variables \\\\\nunknownu(0) & =1\n\\end{aligned}\n\\]",
      "solution": "Solution. Put\n\\[\nconstb = \\int_{0}^{1} unknownu(variables)\\, d variables\n\\]\nThen \\( constb \\) is a constant and the given equation becomes the familiar linear differential equation\n\\[\nunknownu^{\\prime}(variablet)=unknownu(variablet)+constb\n\\]\nwith the general solution\n\\[\nunknownu(variablet) = - constb + constc e^{\\prime}\n\\]\nThere remains the problem of determining \\( constb \\) and \\( constc \\). We have\n\\[\nconstb = \\int_{0}^{1}\\left(- constb + constc e^{variables}\\right) d variables = - constb + constc (e-1)\n\\]\nand\n\\[\nunknownu(0) = - constb + constc = 1\n\\]\nThese two equations imply that\n\\[\nconstb = (e-1)/(3-e) \\text{ and } constc = 2/(3-e)\n\\]\nso\n\\[\nunknownu(variablet) = \\frac{1}{3-e}\\left(2 e^{variablet} - e + 1\\right)\n\\]\nIt is readily checked that this function is indeed a solution.\nThe above derivation proves the uniqueness of the solution function \\( unknownu \\), so this assumption in the statement of the problem is unnecessary."
    },
    "descriptive_long_confusing": {
      "map": {
        "u": "sandcastle",
        "t": "blueberry",
        "s": "stalemate",
        "b": "horizonline",
        "c": "paintbrush"
      },
      "question": "3. Under the assumption that the following set of relations has a unique solution for \\( sandcastle(blueberry) \\), determine it.\n\\[\n\\begin{aligned}\n\\frac{d sandcastle(blueberry)}{d blueberry} & =sandcastle(blueberry)+\\int_{0}^{1} sandcastle(stalemate) d stalemate \\\\\nsandcastle(0) & =1\n\\end{aligned}\n\\]",
      "solution": "Solution. Put\n\\[\nhorizonline=\\int_{0}^{1} sandcastle(stalemate) d stalemate\n\\]\n\nThen \\( horizonline \\) is a constant and the given equation becomes the familiar linear differential equation\n\\[\nsandcastle^{\\prime}(blueberry)=sandcastle(blueberry)+horizonline\n\\]\nwith the general solution\n\\[\nsandcastle(blueberry)=-horizonline+paintbrush e^{\\prime}\n\\]\n\nThere remains the problem of determining \\( horizonline \\) and \\( paintbrush \\). We have\n\\[\nhorizonline=\\int_{0}^{1}\\left(-horizonline+paintbrush e^{stalemate}\\right) d stalemate=-horizonline+paintbrush(e-1)\n\\]\nand\n\\[\nsandcastle(0)=-horizonline+paintbrush=1\n\\]\n\nThese two equations imply that\n\\[\nhorizonline=(e-1) /(3-e) \\text { and } paintbrush=2 /(3-e)\n\\]\nso\n\\[\nsandcastle(blueberry)=\\frac{1}{3-e}\\left(2 e^{blueberry}-e+1\\right)\n\\]\n\nIt is readily checked that this function is indeed a solution.\nThe above derivation proves the uniqueness of the solution function \\( sandcastle \\), so this assumption in the statement of the problem is unnecessary."
    },
    "descriptive_long_misleading": {
      "map": {
        "u": "knownval",
        "t": "distance",
        "s": "constant",
        "b": "variable",
        "c": "fluctuant"
      },
      "question": "3. Under the assumption that the following set of relations has a unique solution for \\( knownval(distance) \\), determine it.\n\\[\n\\begin{aligned}\n\\frac{d knownval(distance)}{d distance} & =knownval(distance)+\\int_{0}^{1} knownval(constant) d constant \\\\\nknownval(0) & =1\n\\end{aligned}\n\\]",
      "solution": "Solution. Put\n\\[\nvariable=\\int_{0}^{1} knownval(constant) d constant\n\\]\n\nThen \\( variable \\) is a constant and the given equation becomes the familiar linear differential equation\n\\[\nknownval^{\\prime}(distance)=knownval(distance)+variable\n\\]\nwith the general solution\n\\[\nknownval(distance)=-variable+fluctuant e^{distance}\n\\]\n\nThere remains the problem of determining \\( variable \\) and \\( fluctuant \\). We have\n\\[\nvariable=\\int_{0}^{1}\\left(-variable+fluctuant e^{constant}\\right) d constant=-variable+fluctuant(e-1)\n\\]\nand\n\\[\nknownval(0)=-variable+fluctuant=1\n\\]\n\nThese two equations imply that\n\\[\nvariable=(e-1) /(3-e) \\text { and } fluctuant=2 /(3-e)\n\\]\nso\n\\[\nknownval(distance)=\\frac{1}{3-e}\\left(2 e^{distance}-e+1\\right)\n\\]\n\nIt is readily checked that this function is indeed a solution.\nThe above derivation proves the uniqueness of the solution function \\( knownval \\), so this assumption in the statement of the problem is unnecessary."
    },
    "garbled_string": {
      "map": {
        "u": "qzxwvtnp",
        "t": "hjgrksla",
        "s": "vbcmdrqe",
        "b": "lksjdofm",
        "c": "mnbvcxzq"
      },
      "question": "3. Under the assumption that the following set of relations has a unique solution for \\( qzxwvtnp(hjgrksla) \\), determine it.\n\\[\n\\begin{aligned}\n\\frac{d qzxwvtnp(hjgrksla)}{d hjgrksla} & =qzxwvtnp(hjgrksla)+\\int_{0}^{1} qzxwvtnp(vbcmdrqe) d vbcmdrqe \\\\\nqzxwvtnp(0) & =1\n\\end{aligned}\n\\]",
      "solution": "Solution. Put\n\\[\nlksjdofm=\\int_{0}^{1} qzxwvtnp(vbcmdrqe) d vbcmdrqe\n\\]\n\nThen \\( lksjdofm \\) is a constant and the given equation becomes the familiar linear differential equation\n\\[\nqzxwvtnp^{\\prime}(hjgrksla)=qzxwvtnp(hjgrksla)+lksjdofm\n\\]\nwith the general solution\n\\[\nqzxwvtnp(hjgrksla)=-lksjdofm+mnbvcxzq e^{\\prime}\n\\]\n\nThere remains the problem of determining \\( lksjdofm \\) and \\( mnbvcxzq \\). We have\n\\[\nlksjdofm=\\int_{0}^{1}\\left(-lksjdofm+mnbvcxzq e^{vbcmdrqe}\\right) d vbcmdrqe=-lksjdofm+mnbvcxzq(e-1)\n\\]\nand\n\\[\nqzxwvtnp(0)=-lksjdofm+mnbvcxzq=1\n\\]\n\nThese two equations imply that\n\\[\nlksjdofm=(e-1) /(3-e) \\text { and } mnbvcxzq=2 /(3-e)\n\\]\nso\n\\[\nqzxwvtnp(hjgrksla)=\\frac{1}{3-e}\\left(2 e^{hjgrksla}-e+1\\right)\n\\]\n\nIt is readily checked that this function is indeed a solution.\nThe above derivation proves the uniqueness of the solution function \\( qzxwvtnp \\), so this assumption in the statement of the problem is unnecessary."
    },
    "kernel_variant": {
      "question": "Let  \n\\[\nu\\colon[-1,1]\\longrightarrow\\mathbb R,\\qquad u\\in C^{2}[-1,1],\n\\]\nbe twice continuously differentiable and assume that the {\\em non–local} second–order relation  \n\\[\nu''(t)\\;=\\;2\\,u'(t)\\;+\\;\\int_{-1}^{1}\\bigl(t-s\\bigr)\\,u(s)\\,ds\\;+\\;3\\int_{-1}^{1}u(s)\\,ds,\n\\qquad -1\\le t\\le1, \\tag{$\\ast$}\n\\]\nholds together with the two side conditions  \n\\[\nu(0)=1,\n\\qquad\\qquad\n\\int_{-1}^{1}t\\,u(t)\\,dt=0. \\tag{BC}\n\\]\n(i)  Determine the function $u(t)$ explicitly.\n\n(ii)  Prove that the function you found is the {\\em unique} real–valued $C^{2}$–solution of $(\\ast)$ subject to {\\rm(BC)}.\n\n\\vspace{6pt}",
      "solution": "{\\bf 0. Two integral moments.}  \nWrite\n\\[\nA_{0}:=\\int_{-1}^{1}u(s)\\,ds,\n\\qquad\nA_{1}:=\\int_{-1}^{1}s\\,u(s)\\,ds. \\tag{0.1}\n\\]\n\n\\medskip\n{\\bf 1. Reduction of $(\\ast)$ to an ordinary differential equation.}  \nSince  \n\\[\n\\int_{-1}^{1}(t-s)\\,u(s)\\,ds=tA_{0}-A_{1},\n\\]\nrelation $(\\ast)$ is equivalent to  \n\\[\nu''(t)-2u'(t)=A_{0}t+\\bigl(3A_{0}-A_{1}\\bigr),\\qquad -1\\le t\\le1. \\tag{1.1}\n\\]\n\n\\medskip\n{\\bf 2. General $C^{2}$–solution of (1.1).}  \nThe homogeneous equation $u''-2u'=0$ has fundamental set $\\{1,e^{2t}\\}$.  \nBecause the right–hand side of (1.1) is affine in $t$, we make the ansatz  \n\\[\nu(t)=C_{1}+C_{2}e^{2t}+\\alpha\\,t^{2}+\\beta\\,t, \\tag{2.1}\n\\]\nobtaining  \n\\[\nu''(t)-2u'(t)=2\\alpha-4\\alpha t-2\\beta. \\tag{2.2}\n\\]\nComparing (2.2) with the right–hand side of (1.1) gives  \n\\[\n-4\\alpha=A_{0},\\qquad 2\\alpha-2\\beta=3A_{0}-A_{1}. \\tag{2.3}\n\\]\nHence  \n\\[\n\\alpha=-\\frac{A_{0}}{4},\\qquad\n\\beta=-\\frac{7A_{0}}{4}+\\frac{A_{1}}{2}. \\tag{2.4}\n\\]\n\n\\medskip\n{\\bf 3. Expressing $A_{0}$ and $A_{1}$ through $C_{1},C_{2}$.}  \nDenote  \n\\[\nE:=\\int_{-1}^{1}e^{2s}\\,ds=\\frac{e^{2}-e^{-2}}{2},\\qquad\nF:=\\int_{-1}^{1}s\\,e^{2s}\\,ds=\\frac{e^{2}+3e^{-2}}{4}. \\tag{3.1}\n\\]\nWith (2.1),\n\\[\n\\begin{aligned}\nA_{0}&=\\int_{-1}^{1}\\bigl[C_{1}+C_{2}e^{2s}+\\alpha s^{2}+\\beta s\\bigr]ds\n      =2C_{1}+C_{2}E+\\frac{2\\alpha}{3},\\\\[4pt]\nA_{1}&=\\int_{-1}^{1}\\bigl[C_{1}s+C_{2}s e^{2s}+\\alpha s^{3}+\\beta s^{2}\\bigr]ds\n      =C_{2}F+\\frac{2\\beta}{3}. \n\\end{aligned}\\tag{3.2}\n\\]\n\n\\medskip\n{\\bf 4. Inserting the boundary conditions $A_{1}=0$ and $u(0)=1$.}\n\n(i)  Setting $A_{1}=0$ in (3.2) and using (2.4) yields  \n\\[\n0=C_{2}F-\\frac{7}{6}A_{0}\\quad\\Longrightarrow\\quad\nA_{0}=\\frac{6}{7}\\,C_{2}F. \\tag{4.1}\n\\]\nConsequently\n\\[\n\\alpha=-\\frac{3}{14}C_{2}F,\\qquad\n\\beta=-\\frac{3}{2}C_{2}F. \\tag{4.2}\n\\]\n\n(ii)  The point condition $u(0)=1$ gives $C_{1}+C_{2}=1$, hence  \n\\[\nC_{1}=1-C_{2}. \\tag{4.3}\n\\]\n\n\\medskip\n{\\bf 5. Determining $C_{2}$.}  \nInsert $C_{1}=1-C_{2}$, (4.1) and $\\alpha=-A_{0}/4$ into the first identity of (3.2):\n\\[\nA_{0}=2(1-C_{2})+C_{2}E-\\frac{A_{0}}{6}\\quad\\Longrightarrow\\quad\nC_{2}\\bigl(F-E+2\\bigr)=2,\n\\]\nso that  \n\\[\n\\boxed{\\; C_{2}= \\dfrac{2}{\\,F-E+2\\,} \\;}. \\tag{5.1}\n\\]\nNote that  \n\\[\nF-E+2=\\frac{-e^{2}+5e^{-2}+8}{4}\\approx0.3219>0,\n\\]\nso $C_{2}$ is well–defined.\n\n\\medskip\n{\\bf 6. Collecting all parameters.}  \nUsing (4.1)–(4.3) and (5.1),\n\\[\n\\begin{aligned}\nC_{2}&=\\dfrac{2}{\\,F-E+2\\,},\\qquad\nC_{1}=1-C_{2},\\\\[6pt]\nA_{0}&=\\dfrac{6}{7}C_{2}F,\\qquad\n\\alpha=-\\dfrac{3}{14}C_{2}F,\\qquad\n\\beta=-\\dfrac{3}{2}C_{2}F.\n\\end{aligned}\\tag{6.1}\n\\]\n\n\\medskip\n{\\bf 7. Explicit solution.}  \nSubstituting the constants from (6.1) into (2.1) gives the desired function\n\\[\n\\boxed{\\;\nu(t)=1-C_{2}+C_{2}e^{2t}-\\frac{3}{14}C_{2}F\\,t^{2}-\\frac{3}{2}C_{2}F\\,t,\n\\qquad\nC_{2}=\\frac{2}{\\,F-E+2\\,},\n\\;\nF=\\frac{e^{2}+3e^{-2}}{4},\n\\;\nE=\\frac{e^{2}-e^{-2}}{2}\n\\;}. \\tag{7.1}\n\\]\nDirect substitution shows that $u$ satisfies $(\\ast)$ and {\\rm(BC)}.\n\n\\medskip\n{\\bf 8. Uniqueness.}  \nAssume $u_{1},u_{2}\\in C^{2}[-1,1]$ both satisfy {\\rm(BC)} and $(\\ast)$.  \nPut $w:=u_{1}-u_{2}$.  Then\n\\[\nw''(t)-2w'(t)=B_{0}t+3B_{0},\\qquad w(0)=0,\\qquad\n\\int_{-1}^{1}t\\,w(t)\\,dt=0, \\tag{8.1}\n\\]\nwhere  \n\\[\nB_{0}:=\\int_{-1}^{1}w(s)\\,ds. \\tag{8.2}\n\\]\n\n\\underline{Step 1: General form of $w$.}  \nEquation (8.1) has the same structure as (1.1); therefore  \n\\[\nw(t)=D_{1}+D_{2}e^{2t}-\\frac{B_{0}}{4}t^{2}-\\frac{7B_{0}}{4}t. \\tag{8.3}\n\\]\n\n\\underline{Step 2: Inserting $w(0)=0$.}  \nFrom (8.3) we obtain $D_{1}+D_{2}=0$, that is  \n\\[\nD_{1}=-D_{2}. \\tag{8.4}\n\\]\n\n\\underline{Step 3: Expressing $B_{0}$ by $D_{2}$.}  \nUsing (3.2) with $A_{0}\\rightsquigarrow B_{0}$, $C_{1}\\rightsquigarrow D_{1}$, $C_{2}\\rightsquigarrow D_{2}$ and $\\alpha=-B_{0}/4$ gives  \n\\[\nB_{0}=2D_{1}+D_{2}E-\\frac{B_{0}}{6}. \\tag{8.5}\n\\]\nSubstituting (8.4) into (8.5) yields  \n\\[\n\\frac{7}{6}B_{0}=D_{2}(E-2)\\quad\\Longrightarrow\\quad\nD_{2}=\\frac{7B_{0}}{6\\,(E-2)}. \\tag{8.6}\n\\]\n\n\\underline{Step 4: Using $\\displaystyle\\int_{-1}^{1}t\\,w(t)\\,dt=0$.}  \nWith (3.2) (now $A_{1}=0$) and $\\beta=-7B_{0}/4$ we get  \n\\[\n0=D_{2}F-\\frac{7B_{0}}{6}. \\tag{8.7}\n\\]\nInsert $D_{2}$ from (8.6):  \n\\[\n\\frac{7B_{0}}{6}\\,\\frac{F}{E-2}-\\frac{7B_{0}}{6}=0\n\\quad\\Longrightarrow\\quad\nB_{0}\\Bigl(\\frac{F}{E-2}-1\\Bigr)=0. \\tag{8.8}\n\\]\n\n\\underline{Step 5: Eliminating the alternative $F/(E-2)=1$.}  \nA short calculation gives\n\\[\nF-E+2=\\frac{-e^{2}+5e^{-2}+8}{4}>0\\quad\\Longrightarrow\\quad\n\\frac{F}{E-2}\\neq1. \\tag{8.9}\n\\]\nHence (8.8) forces $B_{0}=0$.  By (8.6) this implies $D_{2}=0$, and (8.4) gives $D_{1}=0$.  Finally, (8.3) yields $w\\equiv0$.\n\nTherefore $u_{1}\\equiv u_{2}$; the function in (7.1) is the {\\em unique} $C^{2}$–solution of $(\\ast)$ satisfying {\\rm(BC)}.\n\n\\medskip\n{\\bf 9.  (Optional) numerical form.}  \nNumerically\n\\[\nE\\approx3.6270,\\qquad F\\approx1.9488,\\qquad C_{2}\\approx6.2113,\n\\]\nso  \n\\[\nu(t)\\approx-5.2113+6.2113\\,e^{2t}-2.5921\\,t^{2}-18.1514\\,t.\n\\]\n\n\\bigskip\nCHANGES\\_MADE:\n1. The problem statement was kept, only minor typesetting refinements added.\n\n2. Steps 1–7 (construction of $u$) are unchanged, only cosmetic LaTeX adjustments made.\n\n3. The {\\em flawed} uniqueness argument of the original solution was {\\bf completely replaced}.  \n   • We consider the difference $w=u_{1}-u_{2}$ of two potential solutions.  \n   • By repeating the moment method we express $w$ in the same four–parameter family and\n     use the conditions $w(0)=0$ and $\\int_{-1}^{1}t\\,w(t)\\,dt=0$.  \n   • This yields an algebraic relation that forces the moment $B_{0}=\\int_{-1}^{1}w(s)\\,ds$ to vanish because $F/(E-2)\\neq1$.  \n   • Consequently all coefficients of $w$ vanish and $w\\equiv0$, proving uniqueness.\n\n4. All matrices containing a wrong zero row were removed; the determinant trap is avoided.\n\n5. Every mathematical expression is written in {\\em strict} LaTeX syntax, avoiding Unicode math symbols.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.493651",
        "was_fixed": false,
        "difficulty_analysis": "• Extra Structure – The integral kernel (t−s) couples the differential equation to the first two global moments A₀, A₁, making the right-hand side depend linearly on t with unknown coefficients.  \n• Higher Interaction – One must simultaneously solve a differential equation and a nonlinear algebraic system for the moments; they interact through equations (3)–(9).  \n• Larger Algebraic System – Eliminating the moments produces several coupled equations that ultimately reduce to a single but intricate rational equation (10) involving exponentials; careless algebra quickly leads to dead ends.  \n• Non-trivial Particular Solution – Because r=0 is a root of the homogeneous equation, the usual polynomial ansatz fails and must be modified, a detail that is easy to overlook.  \n• Exact Closed Form – The final constants contain combinations of e^{±2} and rational numbers; obtaining the compact form (13) requires careful symbolic manipulation.  \n\nThese layers of technical and conceptual complexity go well beyond the original problem’s single linear differential equation with one constant integral and thus make the enhanced variant substantially harder."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\\[\nu\\colon[-1,1]\\longrightarrow\\mathbb R,\\qquad u\\in C^{2}[-1,1],\n\\]\nbe twice continuously differentiable and assume that the {\\em non–local} second–order relation  \n\\[\nu''(t)\\;=\\;2\\,u'(t)\\;+\\;\\int_{-1}^{1}\\bigl(t-s\\bigr)\\,u(s)\\,ds\\;+\\;3\\int_{-1}^{1}u(s)\\,ds,\n\\qquad -1\\le t\\le1, \\tag{$\\ast$}\n\\]\nholds together with the two side conditions  \n\\[\nu(0)=1,\n\\qquad\\qquad\n\\int_{-1}^{1}t\\,u(t)\\,dt=0. \\tag{BC}\n\\]\n(i)  Determine the function $u(t)$ explicitly.\n\n(ii)  Prove that the function you found is the {\\em unique} real–valued $C^{2}$–solution of $(\\ast)$ subject to {\\rm(BC)}.\n\n\\vspace{6pt}",
      "solution": "{\\bf 0. Two integral moments.}  \nWrite\n\\[\nA_{0}:=\\int_{-1}^{1}u(s)\\,ds,\n\\qquad\nA_{1}:=\\int_{-1}^{1}s\\,u(s)\\,ds. \\tag{0.1}\n\\]\n\n\\medskip\n{\\bf 1. Reduction of $(\\ast)$ to an ordinary differential equation.}  \nSince  \n\\[\n\\int_{-1}^{1}(t-s)\\,u(s)\\,ds=tA_{0}-A_{1},\n\\]\nrelation $(\\ast)$ is equivalent to  \n\\[\nu''(t)-2u'(t)=A_{0}t+\\bigl(3A_{0}-A_{1}\\bigr),\\qquad -1\\le t\\le1. \\tag{1.1}\n\\]\n\n\\medskip\n{\\bf 2. General $C^{2}$–solution of (1.1).}  \nThe homogeneous equation $u''-2u'=0$ has fundamental set $\\{1,e^{2t}\\}$.  \nBecause the right–hand side of (1.1) is affine in $t$, we make the ansatz  \n\\[\nu(t)=C_{1}+C_{2}e^{2t}+\\alpha\\,t^{2}+\\beta\\,t, \\tag{2.1}\n\\]\nobtaining  \n\\[\nu''(t)-2u'(t)=2\\alpha-4\\alpha t-2\\beta. \\tag{2.2}\n\\]\nComparing (2.2) with the right–hand side of (1.1) gives  \n\\[\n-4\\alpha=A_{0},\\qquad 2\\alpha-2\\beta=3A_{0}-A_{1}. \\tag{2.3}\n\\]\nHence  \n\\[\n\\alpha=-\\frac{A_{0}}{4},\\qquad\n\\beta=-\\frac{7A_{0}}{4}+\\frac{A_{1}}{2}. \\tag{2.4}\n\\]\n\n\\medskip\n{\\bf 3. Expressing $A_{0}$ and $A_{1}$ through $C_{1},C_{2}$.}  \nDenote  \n\\[\nE:=\\int_{-1}^{1}e^{2s}\\,ds=\\frac{e^{2}-e^{-2}}{2},\\qquad\nF:=\\int_{-1}^{1}s\\,e^{2s}\\,ds=\\frac{e^{2}+3e^{-2}}{4}. \\tag{3.1}\n\\]\nWith (2.1),\n\\[\n\\begin{aligned}\nA_{0}&=\\int_{-1}^{1}\\bigl[C_{1}+C_{2}e^{2s}+\\alpha s^{2}+\\beta s\\bigr]ds\n      =2C_{1}+C_{2}E+\\frac{2\\alpha}{3},\\\\[4pt]\nA_{1}&=\\int_{-1}^{1}\\bigl[C_{1}s+C_{2}s e^{2s}+\\alpha s^{3}+\\beta s^{2}\\bigr]ds\n      =C_{2}F+\\frac{2\\beta}{3}. \n\\end{aligned}\\tag{3.2}\n\\]\n\n\\medskip\n{\\bf 4. Inserting the boundary conditions $A_{1}=0$ and $u(0)=1$.}\n\n(i)  Setting $A_{1}=0$ in (3.2) and using (2.4) yields  \n\\[\n0=C_{2}F-\\frac{7}{6}A_{0}\\quad\\Longrightarrow\\quad\nA_{0}=\\frac{6}{7}\\,C_{2}F. \\tag{4.1}\n\\]\nConsequently\n\\[\n\\alpha=-\\frac{3}{14}C_{2}F,\\qquad\n\\beta=-\\frac{3}{2}C_{2}F. \\tag{4.2}\n\\]\n\n(ii)  The point condition $u(0)=1$ gives $C_{1}+C_{2}=1$, hence  \n\\[\nC_{1}=1-C_{2}. \\tag{4.3}\n\\]\n\n\\medskip\n{\\bf 5. Determining $C_{2}$.}  \nInsert $C_{1}=1-C_{2}$, (4.1) and $\\alpha=-A_{0}/4$ into the first identity of (3.2):\n\\[\nA_{0}=2(1-C_{2})+C_{2}E-\\frac{A_{0}}{6}\\quad\\Longrightarrow\\quad\nC_{2}\\bigl(F-E+2\\bigr)=2,\n\\]\nso that  \n\\[\n\\boxed{\\; C_{2}= \\dfrac{2}{\\,F-E+2\\,} \\;}. \\tag{5.1}\n\\]\nNote that  \n\\[\nF-E+2=\\frac{-e^{2}+5e^{-2}+8}{4}\\approx0.3219>0,\n\\]\nso $C_{2}$ is well–defined.\n\n\\medskip\n{\\bf 6. Collecting all parameters.}  \nUsing (4.1)–(4.3) and (5.1),\n\\[\n\\begin{aligned}\nC_{2}&=\\dfrac{2}{\\,F-E+2\\,},\\qquad\nC_{1}=1-C_{2},\\\\[6pt]\nA_{0}&=\\dfrac{6}{7}C_{2}F,\\qquad\n\\alpha=-\\dfrac{3}{14}C_{2}F,\\qquad\n\\beta=-\\dfrac{3}{2}C_{2}F.\n\\end{aligned}\\tag{6.1}\n\\]\n\n\\medskip\n{\\bf 7. Explicit solution.}  \nSubstituting the constants from (6.1) into (2.1) gives the desired function\n\\[\n\\boxed{\\;\nu(t)=1-C_{2}+C_{2}e^{2t}-\\frac{3}{14}C_{2}F\\,t^{2}-\\frac{3}{2}C_{2}F\\,t,\n\\qquad\nC_{2}=\\frac{2}{\\,F-E+2\\,},\n\\;\nF=\\frac{e^{2}+3e^{-2}}{4},\n\\;\nE=\\frac{e^{2}-e^{-2}}{2}\n\\;}. \\tag{7.1}\n\\]\nDirect substitution shows that $u$ satisfies $(\\ast)$ and {\\rm(BC)}.\n\n\\medskip\n{\\bf 8. Uniqueness.}  \nAssume $u_{1},u_{2}\\in C^{2}[-1,1]$ both satisfy {\\rm(BC)} and $(\\ast)$.  \nPut $w:=u_{1}-u_{2}$.  Then\n\\[\nw''(t)-2w'(t)=B_{0}t+3B_{0},\\qquad w(0)=0,\\qquad\n\\int_{-1}^{1}t\\,w(t)\\,dt=0, \\tag{8.1}\n\\]\nwhere  \n\\[\nB_{0}:=\\int_{-1}^{1}w(s)\\,ds. \\tag{8.2}\n\\]\n\n\\underline{Step 1: General form of $w$.}  \nEquation (8.1) has the same structure as (1.1); therefore  \n\\[\nw(t)=D_{1}+D_{2}e^{2t}-\\frac{B_{0}}{4}t^{2}-\\frac{7B_{0}}{4}t. \\tag{8.3}\n\\]\n\n\\underline{Step 2: Inserting $w(0)=0$.}  \nFrom (8.3) we obtain $D_{1}+D_{2}=0$, that is  \n\\[\nD_{1}=-D_{2}. \\tag{8.4}\n\\]\n\n\\underline{Step 3: Expressing $B_{0}$ by $D_{2}$.}  \nUsing (3.2) with $A_{0}\\rightsquigarrow B_{0}$, $C_{1}\\rightsquigarrow D_{1}$, $C_{2}\\rightsquigarrow D_{2}$ and $\\alpha=-B_{0}/4$ gives  \n\\[\nB_{0}=2D_{1}+D_{2}E-\\frac{B_{0}}{6}. \\tag{8.5}\n\\]\nSubstituting (8.4) into (8.5) yields  \n\\[\n\\frac{7}{6}B_{0}=D_{2}(E-2)\\quad\\Longrightarrow\\quad\nD_{2}=\\frac{7B_{0}}{6\\,(E-2)}. \\tag{8.6}\n\\]\n\n\\underline{Step 4: Using $\\displaystyle\\int_{-1}^{1}t\\,w(t)\\,dt=0$.}  \nWith (3.2) (now $A_{1}=0$) and $\\beta=-7B_{0}/4$ we get  \n\\[\n0=D_{2}F-\\frac{7B_{0}}{6}. \\tag{8.7}\n\\]\nInsert $D_{2}$ from (8.6):  \n\\[\n\\frac{7B_{0}}{6}\\,\\frac{F}{E-2}-\\frac{7B_{0}}{6}=0\n\\quad\\Longrightarrow\\quad\nB_{0}\\Bigl(\\frac{F}{E-2}-1\\Bigr)=0. \\tag{8.8}\n\\]\n\n\\underline{Step 5: Eliminating the alternative $F/(E-2)=1$.}  \nA short calculation gives\n\\[\nF-E+2=\\frac{-e^{2}+5e^{-2}+8}{4}>0\\quad\\Longrightarrow\\quad\n\\frac{F}{E-2}\\neq1. \\tag{8.9}\n\\]\nHence (8.8) forces $B_{0}=0$.  By (8.6) this implies $D_{2}=0$, and (8.4) gives $D_{1}=0$.  Finally, (8.3) yields $w\\equiv0$.\n\nTherefore $u_{1}\\equiv u_{2}$; the function in (7.1) is the {\\em unique} $C^{2}$–solution of $(\\ast)$ satisfying {\\rm(BC)}.\n\n\\medskip\n{\\bf 9.  (Optional) numerical form.}  \nNumerically\n\\[\nE\\approx3.6270,\\qquad F\\approx1.9488,\\qquad C_{2}\\approx6.2113,\n\\]\nso  \n\\[\nu(t)\\approx-5.2113+6.2113\\,e^{2t}-2.5921\\,t^{2}-18.1514\\,t.\n\\]\n\n\\bigskip\nCHANGES\\_MADE:\n1. The problem statement was kept, only minor typesetting refinements added.\n\n2. Steps 1–7 (construction of $u$) are unchanged, only cosmetic LaTeX adjustments made.\n\n3. The {\\em flawed} uniqueness argument of the original solution was {\\bf completely replaced}.  \n   • We consider the difference $w=u_{1}-u_{2}$ of two potential solutions.  \n   • By repeating the moment method we express $w$ in the same four–parameter family and\n     use the conditions $w(0)=0$ and $\\int_{-1}^{1}t\\,w(t)\\,dt=0$.  \n   • This yields an algebraic relation that forces the moment $B_{0}=\\int_{-1}^{1}w(s)\\,ds$ to vanish because $F/(E-2)\\neq1$.  \n   • Consequently all coefficients of $w$ vanish and $w\\equiv0$, proving uniqueness.\n\n4. All matrices containing a wrong zero row were removed; the determinant trap is avoided.\n\n5. Every mathematical expression is written in {\\em strict} LaTeX syntax, avoiding Unicode math symbols.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.413151",
        "was_fixed": false,
        "difficulty_analysis": "• Extra Structure – The integral kernel (t−s) couples the differential equation to the first two global moments A₀, A₁, making the right-hand side depend linearly on t with unknown coefficients.  \n• Higher Interaction – One must simultaneously solve a differential equation and a nonlinear algebraic system for the moments; they interact through equations (3)–(9).  \n• Larger Algebraic System – Eliminating the moments produces several coupled equations that ultimately reduce to a single but intricate rational equation (10) involving exponentials; careless algebra quickly leads to dead ends.  \n• Non-trivial Particular Solution – Because r=0 is a root of the homogeneous equation, the usual polynomial ansatz fails and must be modified, a detail that is easy to overlook.  \n• Exact Closed Form – The final constants contain combinations of e^{±2} and rational numbers; obtaining the compact form (13) requires careful symbolic manipulation.  \n\nThese layers of technical and conceptual complexity go well beyond the original problem’s single linear differential equation with one constant integral and thus make the enhanced variant substantially harder."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}