{
  "index": "1975-A-5",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "A-5. On some interval \\( I \\) of the real line, let \\( y_{1}(x) \\) and \\( y_{2}(x) \\) be linearly independent solutions of the differential equation\n\\[\ny^{\\prime \\prime}=f(x) y\n\\]\nwhere \\( f(x) \\) is a continuous real-valued function. Suppose that \\( y_{1}(x)>0 \\) and \\( y_{2}(x)>0 \\) on \\( I \\). Show that there exists a positive constant \\( c \\) such that, on \\( I \\), the function\n\\[\nz(x)=c \\sqrt{y_{1}(x) y_{2}(x)}\n\\]\nsatisfies the equation\n\\[\nz^{\\prime \\prime}+\\frac{1}{z^{3}}=f(x) z\n\\]\n\nState clearly the manner in which \\( c \\) depends on \\( y_{1}(x) \\) and \\( y_{2}(x) \\).",
  "solution": "A-5.\nThe answer for \\( c \\) is \\( \\sqrt{2 / w} \\), where \\( w \\) is the wronskian \\( y_{1} y_{2}^{\\prime}-y_{2} y_{1}^{\\prime} \\) (and will be seen below to be constant).\n\nLet \\( c^{2}=2 k \\). Then \\( z^{2} / 2=k y_{1} y_{2} \\). Differentiating twice, one has\n\\[\nz z^{\\prime}=k\\left(y_{1} y_{2}^{\\prime}+y_{2} y_{1}^{\\prime}\\right), z z^{\\prime \\prime}+\\left(z^{\\prime}\\right)^{2}=k\\left(y_{1} y_{2}^{\\prime \\prime}+y_{2} y_{1}^{\\prime \\prime}+2 y_{1}^{\\prime} y_{2}^{\\prime}\\right) .\n\\]\n\nSince \\( y_{1}^{\\prime \\prime}=f y_{1} \\) and \\( y_{2}^{\\prime \\prime}=f y_{2} \\), this implies\n\\[\nz z^{\\prime \\prime}+\\left(z^{\\prime}\\right)^{2}=2 k\\left(f y_{1} y_{2}+y_{1}^{\\prime} y_{2}{ }_{2}^{\\prime}\\right)=f\\left(2 k y_{1} y_{2}\\right)+2 k y_{1}^{\\prime} y_{2}^{\\prime}=f z^{2}+2 k y_{1}^{\\prime} y_{2}^{\\prime} .\n\\]\n\nNow\n\\[\n\\begin{array}{l}\nz^{3} z^{\\prime \\prime}+\\left(z z^{\\prime}\\right)^{2}=f z^{4}+2 k z^{2} y_{1}^{\\prime} y_{2}^{\\prime}, \\\\\nz^{3} z^{\\prime \\prime}+k^{2}\\left(y_{1} y_{2}^{\\prime}+y_{2} y_{1}^{\\prime}\\right)^{2}=f z^{4}+4 k^{2}\\left(y_{1} y_{2} y_{1}^{\\prime} y_{2}^{\\prime}\\right) \\\\\nz^{3} z^{\\prime \\prime}+k^{2}\\left(y_{1} y_{2}^{\\prime}-y_{2} y_{1}^{\\prime}\\right)^{2}=f z^{4}, \\\\\nz^{3} z^{\\prime \\prime}-f z^{4}=-k^{2}\\left(y_{1} y_{2}^{\\prime}-y_{2} y_{1}^{\\prime}\\right)^{2}=-k^{2} w^{2}=-c^{4} w^{2} / 4\n\\end{array}\n\\]\n\nSince \\( w^{\\prime}=\\left(y_{1} y_{2}^{\\prime}-y_{2} y_{1}^{\\prime}\\right)^{\\prime}=y_{1} y_{2}^{\\prime \\prime}-y_{2} y_{1}^{\\prime \\prime}=y_{1}\\left(f y_{2}\\right)-y_{2}\\left(f y_{1}\\right)=0, w \\) is a constant. Solving \\( c^{4} w^{2} / 4=1 \\) for \\( c \\) gives \\( c=\\sqrt{2 / w} \\); for this \\( c \\), (1) implies \\( \\mid z^{\\prime \\prime}-f z=-z^{-3} \\) or \\( z^{\\prime \\prime}+z^{-3}=f z \\).",
  "vars": [
    "x",
    "y",
    "y_1",
    "y_2",
    "z"
  ],
  "params": [
    "I",
    "c",
    "k",
    "w",
    "f"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "realvarx",
        "y": "depvar",
        "y_1": "firstsol",
        "y_2": "secondsol",
        "z": "combined",
        "I": "interval",
        "c": "scalec",
        "k": "auxkappa",
        "w": "wronsk",
        "f": "coefff"
      },
      "question": "A-5. On some interval \\( interval \\) of the real line, let \\( firstsol(realvarx) \\) and \\( secondsol(realvarx) \\) be linearly independent solutions of the differential equation\n\\[\ndepvar^{\\prime \\prime}=coefff(realvarx) \\, depvar\n\\]\nwhere \\( coefff(realvarx) \\) is a continuous real-valued function. Suppose that \\( firstsol(realvarx)>0 \\) and \\( secondsol(realvarx)>0 \\) on \\( interval \\). Show that there exists a positive constant \\( scalec \\) such that, on \\( interval \\), the function\n\\[\ncombined(realvarx)=scalec \\sqrt{firstsol(realvarx) \\, secondsol(realvarx)}\n\\]\nsatisfies the equation\n\\[\ncombined^{\\prime \\prime}+\\frac{1}{combined^{3}}=coefff(realvarx) \\, combined\n\\]\n\nState clearly the manner in which \\( scalec \\) depends on \\( firstsol(realvarx) \\) and \\( secondsol(realvarx) \\).",
      "solution": "A-5.\nThe answer for \\( scalec \\) is \\( \\sqrt{2 / wronsk} \\), where \\( wronsk \\) is the wronskian \\( firstsol \\, secondsol^{\\prime}-secondsol \\, firstsol^{\\prime} \\) (and will be seen below to be constant).\n\nLet \\( scalec^{2}=2 auxkappa \\). Then \\( combined^{2} / 2 = auxkappa \\, firstsol \\, secondsol \\). Differentiating twice, one has\n\\[\ncombined \\, combined^{\\prime}=auxkappa\\left(firstsol \\, secondsol^{\\prime}+secondsol \\, firstsol^{\\prime}\\right), \\quad\ncombined \\, combined^{\\prime \\prime}+\\left(combined^{\\prime}\\right)^{2}=auxkappa\\left(firstsol \\, secondsol^{\\prime \\prime}+secondsol \\, firstsol^{\\prime \\prime}+2 firstsol^{\\prime} \\, secondsol^{\\prime}\\right).\n\\]\n\nSince \\( firstsol^{\\prime \\prime}=coefff \\, firstsol \\) and \\( secondsol^{\\prime \\prime}=coefff \\, secondsol \\), this implies\n\\[\ncombined \\, combined^{\\prime \\prime}+\\left(combined^{\\prime}\\right)^{2}=2 auxkappa\\left(coefff \\, firstsol \\, secondsol + firstsol^{\\prime} \\, secondsol^{\\prime}\\right)\n=coefff\\left(2 auxkappa \\, firstsol \\, secondsol\\right)+2 auxkappa \\, firstsol^{\\prime} \\, secondsol^{\\prime}\n=coefff \\, combined^{2}+2 auxkappa \\, firstsol^{\\prime} \\, secondsol^{\\prime}.\n\\]\n\nNow\n\\[\n\\begin{array}{l}\ncombined^{3} \\, combined^{\\prime \\prime}+\\left(combined \\, combined^{\\prime}\\right)^{2}=coefff \\, combined^{4}+2 auxkappa \\, combined^{2} \\, firstsol^{\\prime} \\, secondsol^{\\prime}, \\\ncombined^{3} \\, combined^{\\prime \\prime}+auxkappa^{2}\\left(firstsol \\, secondsol^{\\prime}+secondsol \\, firstsol^{\\prime}\\right)^{2}\n=coefff \\, combined^{4}+4 auxkappa^{2}\\left(firstsol \\, secondsol \\, firstsol^{\\prime} \\, secondsol^{\\prime}\\right), \\\\\ncombined^{3} \\, combined^{\\prime \\prime}+auxkappa^{2}\\left(firstsol \\, secondsol^{\\prime}-secondsol \\, firstsol^{\\prime}\\right)^{2}=coefff \\, combined^{4}, \\\\\ncombined^{3} \\, combined^{\\prime \\prime}-coefff \\, combined^{4}=-auxkappa^{2}\\left(firstsol \\, secondsol^{\\prime}-secondsol \\, firstsol^{\\prime}\\right)^{2}=-auxkappa^{2} \\, wronsk^{2} = -scalec^{4} \\, wronsk^{2} / 4 .\n\\end{array}\n\\]\n\nSince \\( wronsk^{\\prime}=\\left(firstsol \\, secondsol^{\\prime}-secondsol \\, firstsol^{\\prime}\\right)^{\\prime}\n=firstsol \\, secondsol^{\\prime \\prime}-secondsol \\, firstsol^{\\prime \\prime}\n=firstsol\\left(coefff \\, secondsol\\right)-secondsol\\left(coefff \\, firstsol\\right)=0\\), \\( wronsk \\) is a constant. Solving \\( scalec^{4} \\, wronsk^{2} / 4 = 1 \\) for \\( scalec \\) gives \\( scalec=\\sqrt{2 / wronsk} \\); for this \\( scalec \\), the preceding relation implies \\( \\, combined^{\\prime \\prime}-coefff \\, combined=-combined^{-3} \\) or equivalently \\( combined^{\\prime \\prime}+combined^{-3}=coefff \\, combined \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "chandelier",
        "y": "bungalow",
        "y_1": "seashell",
        "y_2": "pineapple",
        "z": "harmonica",
        "I": "corridor",
        "c": "gardenias",
        "k": "motorboat",
        "w": "rainstorm",
        "f": "salamander"
      },
      "question": "A-5. On some interval \\( corridor \\) of the real line, let \\( seashell(chandelier) \\) and \\( pineapple(chandelier) \\) be linearly independent solutions of the differential equation\n\\[\nbungalow^{\\prime \\prime}=salamander(chandelier) bungalow\n\\]\nwhere \\( salamander(chandelier) \\) is a continuous real-valued function. Suppose that \\( seashell(chandelier)>0 \\) and \\( pineapple(chandelier)>0 \\) on \\( corridor \\). Show that there exists a positive constant \\( gardenias \\) such that, on \\( corridor \\), the function\n\\[\nharmonica(chandelier)=gardenias \\sqrt{seashell(chandelier) pineapple(chandelier)}\n\\]\nsatisfies the equation\n\\[\nharmonica^{\\prime \\prime}+\\frac{1}{harmonica^{3}}=salamander(chandelier) harmonica\n\\]\n\nState clearly the manner in which \\( gardenias \\) depends on \\( seashell(chandelier) \\) and \\( pineapple(chandelier) \\).",
      "solution": "A-5.\nThe answer for \\( gardenias \\) is \\( \\sqrt{2 / rainstorm} \\), where \\( rainstorm \\) is the wronskian \\( seashell pineapple^{\\prime}-pineapple seashell^{\\prime} \\) (and will be seen below to be constant).\n\nLet \\( gardenias^{2}=2 motorboat \\). Then \\( harmonica^{2} / 2=motorboat seashell pineapple \\). Differentiating twice, one has\n\\[\nharmonica harmonica^{\\prime}=motorboat\\left(seashell pineapple^{\\prime}+pineapple seashell^{\\prime}\\right),\\quad harmonica harmonica^{\\prime \\prime}+\\left(harmonica^{\\prime}\\right)^{2}=motorboat\\left(seashell pineapple^{\\prime \\prime}+pineapple seashell^{\\prime \\prime}+2 seashell^{\\prime} pineapple^{\\prime}\\right) .\n\\]\n\nSince \\( seashell^{\\prime \\prime}=salamander seashell \\) and \\( pineapple^{\\prime \\prime}=salamander pineapple \\), this implies\n\\[\nharmonica harmonica^{\\prime \\prime}+\\left(harmonica^{\\prime}\\right)^{2}=2 motorboat\\left(salamander seashell pineapple+seashell^{\\prime} pineapple^{\\prime}\\right)=salamander\\left(2 motorboat seashell pineapple\\right)+2 motorboat seashell^{\\prime} pineapple^{\\prime}=salamander harmonica^{2}+2 motorboat seashell^{\\prime} pineapple^{\\prime} .\n\\]\n\nNow\n\\[\n\\begin{array}{l}\nharmonica^{3} harmonica^{\\prime \\prime}+\\left(harmonica harmonica^{\\prime}\\right)^{2}=salamander harmonica^{4}+2 motorboat harmonica^{2} seashell^{\\prime} pineapple^{\\prime}, \\\\\nharmonica^{3} harmonica^{\\prime \\prime}+motorboat^{2}\\left(seashell pineapple^{\\prime}+pineapple seashell^{\\prime}\\right)^{2}=salamander harmonica^{4}+4 motorboat^{2}\\left(seashell pineapple seashell^{\\prime} pineapple^{\\prime}\\right), \\\\\nharmonica^{3} harmonica^{\\prime \\prime}+motorboat^{2}\\left(seashell pineapple^{\\prime}-pineapple seashell^{\\prime}\\right)^{2}=salamander harmonica^{4}, \\\\\nharmonica^{3} harmonica^{\\prime \\prime}-salamander harmonica^{4}=-motorboat^{2}\\left(seashell pineapple^{\\prime}-pineapple seashell^{\\prime}\\right)^{2}=-motorboat^{2} rainstorm^{2}=-gardenias^{4} rainstorm^{2} / 4\n\\end{array}\n\\]\n\nSince \\( rainstorm^{\\prime}=\\left(seashell pineapple^{\\prime}-pineapple seashell^{\\prime}\\right)^{\\prime}=seashell pineapple^{\\prime \\prime}-pineapple seashell^{\\prime \\prime}=seashell\\left(salamander pineapple\\right)-pineapple\\left(salamander seashell\\right)=0, rainstorm \\) is a constant. Solving \\( gardenias^{4} rainstorm^{2} / 4=1 \\) for \\( gardenias \\) gives \\( gardenias=\\sqrt{2 / rainstorm} \\); for this \\( gardenias \\), (1) implies \\( \\mid harmonica^{\\prime \\prime}-salamander harmonica=-harmonica^{-3} \\) or \\( harmonica^{\\prime \\prime}+harmonica^{-3}=salamander harmonica \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "fixedvalue",
        "y": "independent",
        "y_1": "difficultyone",
        "y_2": "difficultytwo",
        "z": "singularity",
        "I": "discreteset",
        "c": "variable",
        "k": "unstable",
        "w": "randomness",
        "f": "constantvalue"
      },
      "question": "A-5. On some interval \\( discreteset \\) of the real line, let \\( difficultyone(fixedvalue) \\) and \\( difficultytwo(fixedvalue) \\) be linearly independent solutions of the differential equation\n\\[\nindependent^{\\prime \\prime}=constantvalue(fixedvalue) independent\n\\]\nwhere \\( constantvalue(fixedvalue) \\) is a continuous real-valued function. Suppose that \\( difficultyone(fixedvalue)>0 \\) and \\( difficultytwo(fixedvalue)>0 \\) on \\( discreteset \\). Show that there exists a positive constant \\( variable \\) such that, on \\( discreteset \\), the function\n\\[\nsingularity(fixedvalue)=variable \\sqrt{difficultyone(fixedvalue)\\, difficultytwo(fixedvalue)}\n\\]\nsatisfies the equation\n\\[\nsingularity^{\\prime \\prime}+\\frac{1}{singularity^{3}}=constantvalue(fixedvalue)\\, singularity\n\\]\n\nState clearly the manner in which \\( variable \\) depends on \\( difficultyone(fixedvalue) \\) and \\( difficultytwo(fixedvalue) \\).",
      "solution": "A-5.\nThe answer for \\( variable \\) is \\( \\sqrt{2 / randomness} \\), where \\( randomness \\) is the wronskian \\( difficultyone\\, difficultytwo^{\\prime}-difficultytwo\\, difficultyone^{\\prime} \\) (and will be seen below to be constant).\n\nLet \\( variable^{2}=2\\, unstable \\). Then \\( singularity^{2} / 2=unstable\\, difficultyone\\, difficultytwo \\). Differentiating twice, one has\n\\[\nsingularity\\, singularity^{\\prime}=unstable\\left(difficultyone\\, difficultytwo^{\\prime}+difficultytwo\\, difficultyone^{\\prime}\\right),\\quad singularity\\, singularity^{\\prime \\prime}+\\left(singularity^{\\prime}\\right)^{2}=unstable\\left(difficultyone\\, difficultytwo^{\\prime \\prime}+difficultytwo\\, difficultyone^{\\prime \\prime}+2\\, difficultyone^{\\prime}\\, difficultytwo^{\\prime}\\right) .\n\\]\n\nSince \\( difficultyone^{\\prime \\prime}=constantvalue\\, difficultyone \\) and \\( difficultytwo^{\\prime \\prime}=constantvalue\\, difficultytwo \\), this implies\n\\[\nsingularity\\, singularity^{\\prime \\prime}+\\left(singularity^{\\prime}\\right)^{2}=2\\, unstable\\left(constantvalue\\, difficultyone\\, difficultytwo+difficultyone^{\\prime}\\, difficultytwo{ }_{2}^{\\prime}\\right)=constantvalue\\left(2\\, unstable\\, difficultyone\\, difficultytwo\\right)+2\\, unstable\\, difficultyone^{\\prime}\\, difficultytwo^{\\prime}=constantvalue\\, singularity^{2}+2\\, unstable\\, difficultyone^{\\prime}\\, difficultytwo^{\\prime} .\n\\]\n\nNow\n\\[\n\\begin{array}{l}\nsingularity^{3}\\, singularity^{\\prime \\prime}+\\left(singularity\\, singularity^{\\prime}\\right)^{2}=constantvalue\\, singularity^{4}+2\\, unstable\\, singularity^{2}\\, difficultyone^{\\prime}\\, difficultytwo^{\\prime}, \\\\\nsingularity^{3}\\, singularity^{\\prime \\prime}+unstable^{2}\\left(difficultyone\\, difficultytwo^{\\prime}+difficultytwo\\, difficultyone^{\\prime}\\right)^{2}=constantvalue\\, singularity^{4}+4\\, unstable^{2}\\left(difficultyone\\, difficultytwo\\, difficultyone^{\\prime}\\, difficultytwo^{\\prime}\\right) \\\\\nsingularity^{3}\\, singularity^{\\prime \\prime}+unstable^{2}\\left(difficultyone\\, difficultytwo^{\\prime}-difficultytwo\\, difficultyone^{\\prime}\\right)^{2}=constantvalue\\, singularity^{4}, \\\\\nsingularity^{3}\\, singularity^{\\prime \\prime}-constantvalue\\, singularity^{4}=-unstable^{2}\\left(difficultyone\\, difficultytwo^{\\prime}-difficultytwo\\, difficultyone^{\\prime}\\right)^{2}=-unstable^{2}\\, randomness^{2}=-variable^{4}\\, randomness^{2} / 4\n\\end{array}\n\\]\n\nSince \\( randomness^{\\prime}=\\left(difficultyone\\, difficultytwo^{\\prime}-difficultytwo\\, difficultyone^{\\prime}\\right)^{\\prime}=difficultyone\\, difficultytwo^{\\prime \\prime}-difficultytwo\\, difficultyone^{\\prime \\prime}=difficultyone\\left(constantvalue\\, difficultytwo\\right)-difficultytwo\\left(constantvalue\\, difficultyone\\right)=0, randomness \\) is a constant. Solving \\( variable^{4}\\, randomness^{2} / 4=1 \\) for \\( variable \\) gives \\( variable=\\sqrt{2 / randomness} \\); for this \\( variable \\), (1) implies \\( \\mid singularity^{\\prime \\prime}-constantvalue\\, singularity=-singularity^{-3} \\) or \\( singularity^{\\prime \\prime}+singularity^{-3}=constantvalue\\, singularity \\)."
    },
    "garbled_string": {
      "map": {
        "x": "mldkrpqa",
        "y": "glitsznu",
        "y_1": "qzxwvtnp",
        "y_2": "hjgrksla",
        "z": "rpkfgqnv",
        "I": "uavczmle",
        "c": "dfmbtose",
        "k": "xnqbrfwa",
        "w": "ptzhyclo",
        "f": "vwsejkmn"
      },
      "question": "A-5. On some interval \\( uavczmle \\) of the real line, let \\( qzxwvtnp(mldkrpqa) \\) and \\( hjgrksla(mldkrpqa) \\) be linearly independent solutions of the differential equation\n\\[\nglitsznu^{\\prime \\prime}=vwsejkmn(mldkrpqa) glitsznu\n\\]\nwhere \\( vwsejkmn(mldkrpqa) \\) is a continuous real-valued function. Suppose that \\( qzxwvtnp(mldkrpqa)>0 \\) and \\( hjgrksla(mldkrpqa)>0 \\) on \\( uavczmle \\). Show that there exists a positive constant \\( dfmbtose \\) such that, on \\( uavczmle \\), the function\n\\[\nrpkfgqnv(mldkrpqa)=dfmbtose \\sqrt{qzxwvtnp(mldkrpqa) hjgrksla(mldkrpqa)}\n\\]\nsatisfies the equation\n\\[\nrpkfgqnv^{\\prime \\prime}+\\frac{1}{rpkfgqnv^{3}}=vwsejkmn(mldkrpqa) rpkfgqnv\n\\]\n\nState clearly the manner in which \\( dfmbtose \\) depends on \\( qzxwvtnp(mldkrpqa) \\) and \\( hjgrksla(mldkrpqa) \\).",
      "solution": "A-5.\nThe answer for \\( dfmbtose \\) is \\( \\sqrt{2 / ptzhyclo} \\), where \\( ptzhyclo \\) is the wronskian \\( qzxwvtnp hjgrksla^{\\prime}-hjgrksla qzxwvtnp^{\\prime} \\) (and will be seen below to be constant).\n\nLet \\( dfmbtose^{2}=2 xnqbrfwa \\). Then \\( rpkfgqnv^{2} / 2=xnqbrfwa qzxwvtnp hjgrksla \\). Differentiating twice, one has\n\\[\nrpkfgqnv rpkfgqnv^{\\prime}=xnqbrfwa\\left(qzxwvtnp hjgrksla^{\\prime}+hjgrksla qzxwvtnp^{\\prime}\\right), \\quad rpkfgqnv rpkfgqnv^{\\prime \\prime}+\\left(rpkfgqnv^{\\prime}\\right)^{2}=xnqbrfwa\\left(qzxwvtnp hjgrksla^{\\prime \\prime}+hjgrksla qzxwvtnp^{\\prime \\prime}+2 qzxwvtnp^{\\prime} hjgrksla^{\\prime}\\right) .\n\\]\n\nSince \\( qzxwvtnp^{\\prime \\prime}=vwsejkmn qzxwvtnp \\) and \\( hjgrksla^{\\prime \\prime}=vwsejkmn hjgrksla \\), this implies\n\\[\nrpkfgqnv rpkfgqnv^{\\prime \\prime}+\\left(rpkfgqnv^{\\prime}\\right)^{2}=2 xnqbrfwa\\left(vwsejkmn qzxwvtnp hjgrksla+qzxwvtnp^{\\prime} hjgrksla^{\\prime}\\right)=vwsejkmn\\left(2 xnqbrfwa qzxwvtnp hjgrksla\\right)+2 xnqbrfwa qzxwvtnp^{\\prime} hjgrksla^{\\prime}=vwsejkmn rpkfgqnv^{2}+2 xnqbrfwa qzxwvtnp^{\\prime} hjgrksla^{\\prime} .\n\\]\n\nNow\n\\[\n\\begin{array}{l}\nrpkfgqnv^{3} rpkfgqnv^{\\prime \\prime}+\\left(rpkfgqnv rpkfgqnv^{\\prime}\\right)^{2}=vwsejkmn rpkfgqnv^{4}+2 xnqbrfwa rpkfgqnv^{2} qzxwvtnp^{\\prime} hjgrksla^{\\prime}, \\\\\nrpkfgqnv^{3} rpkfgqnv^{\\prime \\prime}+xnqbrfwa^{2}\\left(qzxwvtnp hjgrksla^{\\prime}+hjgrksla qzxwvtnp^{\\prime}\\right)^{2}=vwsejkmn rpkfgqnv^{4}+4 xnqbrfwa^{2}\\left(qzxwvtnp hjgrksla qzxwvtnp^{\\prime} hjgrksla^{\\prime}\\right), \\\\\nrpkfgqnv^{3} rpkfgqnv^{\\prime \\prime}+xnqbrfwa^{2}\\left(qzxwvtnp hjgrksla^{\\prime}-hjgrksla qzxwvtnp^{\\prime}\\right)^{2}=vwsejkmn rpkfgqnv^{4}, \\\\\nrpkfgqnv^{3} rpkfgqnv^{\\prime \\prime}-vwsejkmn rpkfgqnv^{4}=-xnqbrfwa^{2}\\left(qzxwvtnp hjgrksla^{\\prime}-hjgrksla qzxwvtnp^{\\prime}\\right)^{2}=-xnqbrfwa^{2} ptzhyclo^{2}=-dfmbtose^{4} ptzhyclo^{2} / 4 .\n\\end{array}\n\\]\n\nSince \\( ptzhyclo^{\\prime}=\\left(qzxwvtnp hjgrksla^{\\prime}-hjgrksla qzxwvtnp^{\\prime}\\right)^{\\prime}=qzxwvtnp hjgrksla^{\\prime \\prime}-hjgrksla qzxwvtnp^{\\prime \\prime}=qzxwvtnp\\left(vwsejkmn hjgrksla\\right)-hjgrksla\\left(vwsejkmn qzxwvtnp\\right)=0, ptzhyclo \\) is a constant. Solving \\( dfmbtose^{4} ptzhyclo^{2} / 4=1 \\) for \\( dfmbtose \\) gives \\( dfmbtose=\\sqrt{2 / ptzhyclo} \\); for this \\( dfmbtose \\), (1) implies \\( \\mid rpkfgqnv^{\\prime \\prime}-vwsejkmn rpkfgqnv=-rpkfgqnv^{-3} \\) or \\( rpkfgqnv^{\\prime \\prime}+rpkfgqnv^{-3}=vwsejkmn rpkfgqnv \\)."
    },
    "kernel_variant": {
      "question": "Let $I\\subset\\mathbb R$ be a non–empty open interval and assume  \n\n• $r\\in C^{1}(I)$ with $r(x)>0$ for every $x\\in I$,  \n• $s\\in C(I)$.\n\nConsider the self–adjoint second–order linear differential equation  \n\\[\n\\tag{1}\\bigl(r(x)\\,y'(x)\\bigr)'+s(x)\\,y(x)=0 ,\\qquad x\\in I .\n\\]\n\nSuppose that $y_{1},y_{2}\\in C^{2}(I)$ are linearly independent **positive** solutions of (1).  \nDefine the (weighted) Wronskian  \n\\[\n\\mathcal W(x):=r(x)\\bigl(y_{1}(x)\\,y_{2}'(x)-y_{2}(x)\\,y_{1}'(x)\\bigr),\n\\qquad x\\in I .\n\\]\n\n(A)  Prove that $\\mathcal W(x)\\equiv\\mathcal W_{0}\\;(\\neq 0)$ is constant on $I$.\n\n(B)  Show that there exists a **unique** constant $c>0$, depending only on $y_{1},y_{2}$ (equivalently only on $\\mathcal W_{0}$), such that  \n\\[\n\\boxed{\\,z(x)=c\\sqrt{y_{1}(x)\\,y_{2}(x)}\\,},\\qquad x\\in I,\n\\]\nsolves the nonlinear *Sturm–Liouville–Ermakov–Pinney equation*  \n\\[\n\\tag{2}\\bigl(r(x)\\,z'(x)\\bigr)'+s(x)\\,z(x)+\\frac{1}{r(x)\\,z^{3}(x)}=0 ,\\qquad x\\in I .\n\\]\n\n(C)  Determine $c$ explicitly in terms of the constant $\\mathcal W_{0}$.",
      "solution": "\\[\n\\text{(All functions are understood to be restricted to }I.)\n\\]\n\n--------------------------------------------------------------------\n\\textbf{Part (A) – Constancy of the weighted Wronskian}\n\nBecause $y_{1},y_{2}$ satisfy (1),\n\\[\n(r y_{j}')'=-s\\,y_{j},\\qquad j=1,2 .\n\\]\nDifferentiate $\\mathcal W$:\n\\[\n\\begin{aligned}\n\\mathcal W'\n     &=(r y_{1}y_{2}'-r y_{2}y_{1}')'                 \\\\\n     &=r'y_{1}y_{2}'+r\\,y_{1}'y_{2}'+r\\,y_{1}y_{2}''   \\\\\n     &\\phantom{=} -r'y_{2}y_{1}'-r\\,y_{2}'y_{1}'-r\\,y_{2}y_{1}''.\n\\end{aligned}\n\\]\nInsert $r y_{j}''=-(r' y_{j}'+s y_{j})$ ($j=1,2$); every term cancels and $\\mathcal W'=0$.  \nSince $y_{1},y_{2}$ are independent, $\\mathcal W_{0}:=\\mathcal W(x)\\neq 0$.\n\n--------------------------------------------------------------------\n\\textbf{Part (B) – Verification of the nonlinear equation}\n\nPut\n\\[\nP:=y_{1}y_{2}>0,\\qquad z^{2}=c^{2}P.\n\\]\n\n\\emph{Step 1 – First derivative of $z$.}\n\\[\nz'=\\frac{c^{2}}{2z}P'=\\frac{z}{2}\n      \\Bigl(\\frac{y_{1}'}{y_{1}}+\\frac{y_{2}'}{y_{2}}\\Bigr)\n     =\\frac{z}{2}\\,S,\\quad\nS:=\\frac{y_{1}'}{y_{1}}+\\frac{y_{2}'}{y_{2}}.\n\\]\n\n\\emph{Step 2 – Second derivative and the left–hand side of (2).}\n\\[\nz''=\\frac{z}{2}S'+\\frac{z}{4}S^{2},\n\\qquad\n(rz')'=(r\\tfrac{zS}{2})'\n       =\\frac{r'zS}{2}+\\frac{rzS^{2}}{4}+\\frac{rzS'}{2}.\n\\]\nHence\n\\[\n(rz')'+s z\n  =z\\Bigl[\n        \\frac{rS^{2}}{4}\n       +\\frac{r' S}{2}\n       +\\frac{r S'}{2}\n       +s\n     \\Bigr].\n\\tag{3}\n\\]\n\n\\emph{Step 3 – Computing $S'$.}  \nWrite $p:=r'/r,\\; q:=s/r$.  Then (1) becomes $y''+p\\,y'+q\\,y=0$.  For $P_{j}:=y_{j}'/y_{j}$,\n\\[\nP_{j}'=-pP_{j}-q-P_{j}^{2},\\qquad j=1,2 .\n\\]\nThus\n\\[\nS'=-pS-2q-\\bigl(P_{1}^{2}+P_{2}^{2}\\bigr).\n\\]\nSet $\\Delta:=P_{2}-P_{1}$.  A short computation gives\n\\[\nP_{1}^{2}+P_{2}^{2}=\\frac{S^{2}+\\Delta^{2}}{2}.\n\\tag{4}\n\\]\n\n\\emph{Step 4 – Insert $S'$ into (3).}  Using $r'=rp$ and (4),\n\\[\n(rz')'+s z\n        =z\\Bigl[-\\frac{r\\Delta^{2}}{4}\\Bigr].\n\\tag{5}\n\\]\n\n\\emph{Step 5 – Express $\\Delta$ via the constant $\\mathcal W_{0}$.}\n\\[\n\\Delta\n  =\\frac{y_{2}'}{y_{2}}-\\frac{y_{1}'}{y_{1}}\n  =\\frac{y_{1}y_{2}'-y_{2}y_{1}'}{y_{1}y_{2}}\n  =\\frac{\\mathcal W_{0}}{rP},\n\\qquad\n\\Delta^{2}=\\frac{\\mathcal W_{0}^{2}}{r^{2}P^{2}}.\n\\]\nInsert this into (5):\n\\[\n(rz')'+s z\n    =-\\,z\\,\\frac{\\mathcal W_{0}^{2}}{4r P^{2}}\n    =-\\frac{\\mathcal W_{0}^{2}\\,c}{4r}\\,P^{-3/2}.\n\\]\nSince $z=cP^{1/2}$,\n\\[\nP^{-3/2}=c^{3}\\,z^{-3},\n\\]\nand therefore\n\\[\n(rz')'+s z\n   =-\\frac{\\mathcal W_{0}^{2}c^{4}}{4\\,r}\\,z^{-3}.\n\\tag{6}\n\\]\n\n\\emph{Step 6 – Choice (and uniqueness) of the constant $c$.}  \nEquation (2) is equivalent to\n\\[\n(rz')'+s z=-\\frac{1}{r\\,z^{3}}.\n\\]\nComparing with (6) we need the coefficient of $z^{-3}$ to be $-1/r$, i.e.\n\\[\n\\frac{\\mathcal W_{0}^{2}c^{4}}{4}=1.\n\\]\nAs $\\mathcal W_{0}\\neq 0$, this determines the unique positive constant\n\\[\n\\boxed{\\,c=\\sqrt{\\frac{2}{|\\mathcal W_{0}|}}\\,}.\n\\]\nWith this choice of $c$, identity (6) becomes exactly equation (2), so the required $z$ indeed solves (2).\n\n--------------------------------------------------------------------\n\\textbf{Part (C) – Explicit formula for $c$}\n\nCollecting the result of Step 6,\n\\[\n\\boxed{\\displaystyle \nc=\\sqrt{\\frac{2}{\\bigl|\\mathcal W_{0}\\bigr|}}},\\qquad\n\\mathcal W_{0}=r(x)\\bigl(y_{1}(x)\\,y_{2}'(x)-y_{2}(x)\\,y_{1}'(x)\\bigr).\n\\]\nBecause $\\mathcal W_{0}$ is independent of $x$, $c$ depends only on the pair $(y_{1},y_{2})$.\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.619423",
        "was_fixed": false,
        "difficulty_analysis": "1. Extra Structure.  \n   • The original problem had the *constant-coefficient* operator \\(y''\\).  \n   • The enhanced variant works with the **self–adjoint Sturm–Liouville operator** \\((ry')'+sy\\), forcing the solver to manipulate a weight \\(r(x)\\) and a potential \\(s(x)\\).\n\n2. Variable Wronskian.  \n   • One must discover and use the *weighted* Wronskian \\(\\mathcal W=r(y_{1}y_{2}'-y_{2}y_{1}')\\).  \n   • Showing its constancy requires integration-factor ideas absent in the original problem.\n\n3. Algebraic Complexity.  \n   • The derivation of \\((rz')'+sz\\) involves several layers of substitutions (\\(S,\\;P_{j},\\;\\Delta\\)) and non-trivial identities such as  \n     \\(P_{1}^{2}+P_{2}^{2}=\\tfrac{S^{2}+\\Delta^{2}}{2}\\).\n\n4. Functional Coefficient.  \n   • The right-hand side of (2) now contains the *variable* factor \\(1/r(x)\\); isolating a **constant** \\(c\\) that makes the coefficient unity demands precise bookkeeping of how \\(r\\) enters the calculation.\n\n5. Uniqueness.  \n   • The enhanced variant asks not only for existence but also for **uniqueness** of the constant \\(c\\), adding an extra logical step.\n\n6. Intertwined Concepts.  \n   • Solving the problem requires the interplay of Sturm–Liouville theory, Wronskian invariants, nonlinear ODE manipulations, and careful algebra—far beyond the straightforward differentiation in the original setting.\n\nConsequently, the enhanced kernel variant is substantially more intricate, demands deeper theoretical insight, and entails a markedly longer chain of arguments than either the original problem or the preceding kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let $I\\subset\\mathbb R$ be a non–empty open interval and assume  \n\n• $r\\in C^{1}(I)$ with $r(x)>0$ for every $x\\in I$,  \n• $s\\in C(I)$.\n\nConsider the self–adjoint second–order linear differential equation  \n\\[\n\\tag{1}\\bigl(r(x)\\,y'(x)\\bigr)'+s(x)\\,y(x)=0 ,\\qquad x\\in I .\n\\]\n\nSuppose that $y_{1},y_{2}\\in C^{2}(I)$ are linearly independent **positive** solutions of (1).  \nDefine the (weighted) Wronskian  \n\\[\n\\mathcal W(x):=r(x)\\bigl(y_{1}(x)\\,y_{2}'(x)-y_{2}(x)\\,y_{1}'(x)\\bigr),\n\\qquad x\\in I .\n\\]\n\n(A)  Prove that $\\mathcal W(x)\\equiv\\mathcal W_{0}\\;(\\neq 0)$ is constant on $I$.\n\n(B)  Show that there exists a **unique** constant $c>0$, depending only on $y_{1},y_{2}$ (equivalently only on $\\mathcal W_{0}$), such that  \n\\[\n\\boxed{\\,z(x)=c\\sqrt{y_{1}(x)\\,y_{2}(x)}\\,},\\qquad x\\in I,\n\\]\nsolves the nonlinear *Sturm–Liouville–Ermakov–Pinney equation*  \n\\[\n\\tag{2}\\bigl(r(x)\\,z'(x)\\bigr)'+s(x)\\,z(x)+\\frac{1}{r(x)\\,z^{3}(x)}=0 ,\\qquad x\\in I .\n\\]\n\n(C)  Determine $c$ explicitly in terms of the constant $\\mathcal W_{0}$.",
      "solution": "\\[\n\\text{(All functions are understood to be restricted to }I.)\n\\]\n\n--------------------------------------------------------------------\n\\textbf{Part (A) – Constancy of the weighted Wronskian}\n\nBecause $y_{1},y_{2}$ satisfy (1),\n\\[\n(r y_{j}')'=-s\\,y_{j},\\qquad j=1,2 .\n\\]\nDifferentiate $\\mathcal W$:\n\\[\n\\begin{aligned}\n\\mathcal W'\n     &=(r y_{1}y_{2}'-r y_{2}y_{1}')'                 \\\\\n     &=r'y_{1}y_{2}'+r\\,y_{1}'y_{2}'+r\\,y_{1}y_{2}''   \\\\\n     &\\phantom{=} -r'y_{2}y_{1}'-r\\,y_{2}'y_{1}'-r\\,y_{2}y_{1}''.\n\\end{aligned}\n\\]\nInsert $r y_{j}''=-(r' y_{j}'+s y_{j})$ ($j=1,2$); every term cancels and $\\mathcal W'=0$.  \nSince $y_{1},y_{2}$ are independent, $\\mathcal W_{0}:=\\mathcal W(x)\\neq 0$.\n\n--------------------------------------------------------------------\n\\textbf{Part (B) – Verification of the nonlinear equation}\n\nPut\n\\[\nP:=y_{1}y_{2}>0,\\qquad z^{2}=c^{2}P.\n\\]\n\n\\emph{Step 1 – First derivative of $z$.}\n\\[\nz'=\\frac{c^{2}}{2z}P'=\\frac{z}{2}\n      \\Bigl(\\frac{y_{1}'}{y_{1}}+\\frac{y_{2}'}{y_{2}}\\Bigr)\n     =\\frac{z}{2}\\,S,\\quad\nS:=\\frac{y_{1}'}{y_{1}}+\\frac{y_{2}'}{y_{2}}.\n\\]\n\n\\emph{Step 2 – Second derivative and the left–hand side of (2).}\n\\[\nz''=\\frac{z}{2}S'+\\frac{z}{4}S^{2},\n\\qquad\n(rz')'=(r\\tfrac{zS}{2})'\n       =\\frac{r'zS}{2}+\\frac{rzS^{2}}{4}+\\frac{rzS'}{2}.\n\\]\nHence\n\\[\n(rz')'+s z\n  =z\\Bigl[\n        \\frac{rS^{2}}{4}\n       +\\frac{r' S}{2}\n       +\\frac{r S'}{2}\n       +s\n     \\Bigr].\n\\tag{3}\n\\]\n\n\\emph{Step 3 – Computing $S'$.}  \nWrite $p:=r'/r,\\; q:=s/r$.  Then (1) becomes $y''+p\\,y'+q\\,y=0$.  For $P_{j}:=y_{j}'/y_{j}$,\n\\[\nP_{j}'=-pP_{j}-q-P_{j}^{2},\\qquad j=1,2 .\n\\]\nThus\n\\[\nS'=-pS-2q-\\bigl(P_{1}^{2}+P_{2}^{2}\\bigr).\n\\]\nSet $\\Delta:=P_{2}-P_{1}$.  A short computation gives\n\\[\nP_{1}^{2}+P_{2}^{2}=\\frac{S^{2}+\\Delta^{2}}{2}.\n\\tag{4}\n\\]\n\n\\emph{Step 4 – Insert $S'$ into (3).}  Using $r'=rp$ and (4),\n\\[\n(rz')'+s z\n        =z\\Bigl[-\\frac{r\\Delta^{2}}{4}\\Bigr].\n\\tag{5}\n\\]\n\n\\emph{Step 5 – Express $\\Delta$ via the constant $\\mathcal W_{0}$.}\n\\[\n\\Delta\n  =\\frac{y_{2}'}{y_{2}}-\\frac{y_{1}'}{y_{1}}\n  =\\frac{y_{1}y_{2}'-y_{2}y_{1}'}{y_{1}y_{2}}\n  =\\frac{\\mathcal W_{0}}{rP},\n\\qquad\n\\Delta^{2}=\\frac{\\mathcal W_{0}^{2}}{r^{2}P^{2}}.\n\\]\nInsert this into (5):\n\\[\n(rz')'+s z\n    =-\\,z\\,\\frac{\\mathcal W_{0}^{2}}{4r P^{2}}\n    =-\\frac{\\mathcal W_{0}^{2}\\,c}{4r}\\,P^{-3/2}.\n\\]\nSince $z=cP^{1/2}$,\n\\[\nP^{-3/2}=c^{3}\\,z^{-3},\n\\]\nand therefore\n\\[\n(rz')'+s z\n   =-\\frac{\\mathcal W_{0}^{2}c^{4}}{4\\,r}\\,z^{-3}.\n\\tag{6}\n\\]\n\n\\emph{Step 6 – Choice (and uniqueness) of the constant $c$.}  \nEquation (2) is equivalent to\n\\[\n(rz')'+s z=-\\frac{1}{r\\,z^{3}}.\n\\]\nComparing with (6) we need the coefficient of $z^{-3}$ to be $-1/r$, i.e.\n\\[\n\\frac{\\mathcal W_{0}^{2}c^{4}}{4}=1.\n\\]\nAs $\\mathcal W_{0}\\neq 0$, this determines the unique positive constant\n\\[\n\\boxed{\\,c=\\sqrt{\\frac{2}{|\\mathcal W_{0}|}}\\,}.\n\\]\nWith this choice of $c$, identity (6) becomes exactly equation (2), so the required $z$ indeed solves (2).\n\n--------------------------------------------------------------------\n\\textbf{Part (C) – Explicit formula for $c$}\n\nCollecting the result of Step 6,\n\\[\n\\boxed{\\displaystyle \nc=\\sqrt{\\frac{2}{\\bigl|\\mathcal W_{0}\\bigr|}}},\\qquad\n\\mathcal W_{0}=r(x)\\bigl(y_{1}(x)\\,y_{2}'(x)-y_{2}(x)\\,y_{1}'(x)\\bigr).\n\\]\nBecause $\\mathcal W_{0}$ is independent of $x$, $c$ depends only on the pair $(y_{1},y_{2})$.\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.495063",
        "was_fixed": false,
        "difficulty_analysis": "1. Extra Structure.  \n   • The original problem had the *constant-coefficient* operator \\(y''\\).  \n   • The enhanced variant works with the **self–adjoint Sturm–Liouville operator** \\((ry')'+sy\\), forcing the solver to manipulate a weight \\(r(x)\\) and a potential \\(s(x)\\).\n\n2. Variable Wronskian.  \n   • One must discover and use the *weighted* Wronskian \\(\\mathcal W=r(y_{1}y_{2}'-y_{2}y_{1}')\\).  \n   • Showing its constancy requires integration-factor ideas absent in the original problem.\n\n3. Algebraic Complexity.  \n   • The derivation of \\((rz')'+sz\\) involves several layers of substitutions (\\(S,\\;P_{j},\\;\\Delta\\)) and non-trivial identities such as  \n     \\(P_{1}^{2}+P_{2}^{2}=\\tfrac{S^{2}+\\Delta^{2}}{2}\\).\n\n4. Functional Coefficient.  \n   • The right-hand side of (2) now contains the *variable* factor \\(1/r(x)\\); isolating a **constant** \\(c\\) that makes the coefficient unity demands precise bookkeeping of how \\(r\\) enters the calculation.\n\n5. Uniqueness.  \n   • The enhanced variant asks not only for existence but also for **uniqueness** of the constant \\(c\\), adding an extra logical step.\n\n6. Intertwined Concepts.  \n   • Solving the problem requires the interplay of Sturm–Liouville theory, Wronskian invariants, nonlinear ODE manipulations, and careful algebra—far beyond the straightforward differentiation in the original setting.\n\nConsequently, the enhanced kernel variant is substantially more intricate, demands deeper theoretical insight, and entails a markedly longer chain of arguments than either the original problem or the preceding kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}