{ "schema_version": "ocme-derivation-reader-v1", "guide_id": "derivation-mko-ordinary-differential-equation-01", "mko_id": "mko-ordinary-differential-equation", "title": { "zh-Hant": "常微分方程：判定階數並核對候選解 y=x² 與 y=x²+x", "en": "Ordinary differential equation: finding the order and checking the candidates y=x² and y=x²+x" }, "contract": { "schema_version": "ocme-derivation-guide-v0.1", "guide_id": "derivation-mko-ordinary-differential-equation-01", "source": { "mko_id": "mko-ordinary-differential-equation", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "sha256": "78a220ac6f5f781865fa9d3cdc17fd9d5f931ff83cfdd5890bfaae2b91641a13", "english_sha256": "435d009ee6ad859f9ac02c4a0cb32c615d035b5925c196f2f27b8e4f2ced3f71" }, "mode": "worked_example", "audience_prerequisites": [ "func_limit_def", "distributive", "eq_replace", "forall_intro", "quantifier_negation" ], "goal": { "zh-Hant": "在區間 I=ℝ 上考慮常微分方程 y″−2y′+y=x²−4x+2。(1) 判定它的階數；(2) 核對 y₁(x)=x² 在整個 ℝ 上是解（殘差恆為零）；(3) 證明 y₂(x)=x²+x 的殘差是 x−2，只在 x=2 為零，所以它不是 ℝ 上的解；(4) 核對初值 y(0)=0、y′(0)=0，並引用二階線性方程的存在唯一定理（含其假設）判定 y₁ 是此初值問題的唯一解；(5) 說明沒有初值時解不唯一。", "en": "On the interval I=ℝ consider the ordinary differential equation y″−2y′+y=x²−4x+2. (1) Determine its order; (2) check that y₁(x)=x² is a solution on all of ℝ (its residual is identically zero); (3) prove that y₂(x)=x²+x has residual x−2, which vanishes only at x=2, so it is not a solution on ℝ; (4) check the initial values y(0)=0, y′(0)=0 and, citing the existence and uniqueness theorem for second-order linear equations with its hypotheses, conclude that y₁ is the unique solution of this initial value problem; (5) show that without initial values the solution is not unique." }, "givens": [ { "id": "G1", "statement": { "zh-Hant": "區間 I=ℝ；未知量 y 是 ℝ 上二次可微的實值函數；方程 E：對每個 x∈I，y″(x)−2y′(x)+y(x)=x²−4x+2。", "en": "Interval I=ℝ; the unknown y is a twice differentiable real-valued function on ℝ; equation E: for every x∈I, y″(x)−2y′(x)+y(x)=x²−4x+2." } }, { "id": "G2", "statement": { "zh-Hant": "候選函數：y₁(x)=x²，y₂(x)=x²+x（整係數多項式）；另有 y₃(x)=x²+eˣ，只在 S13、S14 用來說明非唯一性。", "en": "Candidates: y₁(x)=x² and y₂(x)=x²+x (integer-coefficient polynomials); also y₃(x)=x²+eˣ, used only in S13 and S14 to show non-uniqueness." } }, { "id": "G3", "statement": { "zh-Hant": "初值條件：y(0)=0，y′(0)=0。", "en": "Initial conditions: y(0)=0 and y′(0)=0." } }, { "id": "G4", "statement": { "zh-Hant": "殘差：對候選 u，r_u(x)=u″(x)−2u′(x)+u(x)−(x²−4x+2)。u 在 I 上是 E 的解，當且僅當對每個 x∈I 有 r_u(x)=0。", "en": "Residual: for a candidate u, r_u(x)=u″(x)−2u′(x)+u(x)−(x²−4x+2). u is a solution of E on I exactly when r_u(x)=0 for every x∈I." } } ], "steps": [ { "id": "S01", "uses": [ "G1", "G4" ], "before_tex": "y''-2y'+y=x^2-4x+2\\ \\text{on }I=\\mathbb{R}", "after_tex": "F(x,y,y',y'')=y''-2y'+y-(x^2-4x+2)=0", "relation": "definition", "rule": "mko_ode_def", "conditions": [], "explanation": { "zh-Hant": "依 MKO，常微分方程是在指定區間上關聯未知單變數函數及其導數的方程。把右邊移到左邊，寫成 F(x,y,y′,y″)=0 的形式，區間明示為 I=ℝ。殘差 r_u 正是把候選 u 代入 F 所得的值（G4）。", "en": "By the MKO, an ordinary differential equation relates an unknown single-variable function and its derivatives on a specified interval. Moving the right side to the left gives the form F(x,y,y′,y″)=0, with the interval stated as I=ℝ. The residual r_u is exactly the value of F after substituting a candidate u (G4)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G1", "G4", "mko_ode_def" ], "explanation": { "zh-Hant": "此步只改寫與命名，不推出新事實；移項是等價變形。", "en": "This step only rewrites and names and derives no new fact; moving a term across is an equivalence." } }, "leaf_basis": "mko_ode_def + def_intro" }, { "id": "S02", "uses": [ "S01", "mko_ode_def" ], "before_tex": "F(x,y,y',y'')=0", "after_tex": "\\text{order}=2", "relation": "implies", "rule": "mko_ode_def", "conditions": [ "\\text{coefficient of }y''\\text{ is }1\\neq 0" ], "explanation": { "zh-Hant": "依 MKO，階數是方程中實際出現的最高導數階數。出現的導數是 y′ 與 y″，y″ 的係數是 1≠0，沒有三階或更高的導數，所以階數是 2。階數不是由左邊有三項決定，也不是由候選 x² 的次數決定。", "en": "By the MKO, the order is the highest derivative order that actually occurs. The derivatives occurring are y′ and y″, the coefficient of y″ is 1≠0, and no third or higher derivative appears, so the order is 2. It is not determined by the left side having three terms, nor by the degree of the candidate x²." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S01", "mko_ode_def" ], "explanation": { "zh-Hant": "需要確認 y″ 的係數非零；若係數為 0，y″ 並未真正出現，階數會下降。", "en": "It needs the coefficient of y″ to be nonzero; with coefficient 0, y″ would not really occur and the order would drop." } }, "leaf_basis": "mko_ode_def + def_unfold" }, { "id": "S03", "uses": [ "G2", "L_deriv_square" ], "before_tex": "y_1(x)=x^2", "after_tex": "y_1'(x)=2x", "relation": "implies", "rule": "L_deriv_square", "conditions": [], "explanation": { "zh-Hant": "依引理 L_deriv_square（在 S03.1、S03.2 由導數的極限定義證明），y₁′(x)=2x 對每個實數 x 成立。", "en": "By Lemma L_deriv_square (proved from the limit definition of the derivative in S03.1 and S03.2), y₁′(x)=2x for every real x." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "L_deriv_square" ], "explanation": { "zh-Hant": "依賴該引理；不宣稱反向（導數為 2x 的函數不只 x²，還可差一個常數）。", "en": "It depends on the lemma; no converse is claimed (functions with derivative 2x are not only x², they may differ by a constant)." } }, "leaf_basis": "func_limit_def + distributive" }, { "id": "S03.1", "uses": [ "distributive" ], "before_tex": "(x+h)^2-x^2", "after_tex": "h(2x+h)", "relation": "equivalent", "rule": "distributive", "conditions": [], "explanation": { "zh-Hant": "展開 (x+h)²=x²+2xh+h²，減去 x² 得 2xh+h²，再提出 h 得 h(2x+h)。這是對所有實數 x、h 成立的多項式恆等式（N1）。", "en": "Expand (x+h)²=x²+2xh+h², subtract x² to get 2xh+h², and factor out h to get h(2x+h). This is a polynomial identity valid for all real x and h (N1)." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "distributive", "add_inverse" ], "explanation": { "zh-Hant": "展開 h(2x+h) 得 2xh+h²，再加回 x² 即得 (x+h)²；兩式對所有 x、h 相等。", "en": "Expanding h(2x+h) gives 2xh+h², and adding back x² gives (x+h)²; the two expressions are equal for all x and h." } }, "leaf_basis": "distributive + add_assoc + add_comm + add_inverse" }, { "id": "S03.2", "uses": [ "S03.1", "func_limit_def" ], "before_tex": "\\frac{(x+h)^2-x^2}{h}=2x+h\\ (h\\neq 0)", "after_tex": "\\lim_{h\\to 0}\\frac{(x+h)^2-x^2}{h}=2x", "relation": "implies", "rule": "func_limit_def", "conditions": [ "h\\neq 0" ], "explanation": { "zh-Hant": "對 h≠0，可以除以 h，差商等於 2x+h。對任意 ε>0 取 δ=ε：當 0<|h|<δ 時，|(2x+h)−2x|=|h|<ε。依極限定義差商趨於 2x，所以 (x²)′=2x。S03.1 與 S03.2 合起來證得引理 L_deriv_square。", "en": "For h≠0 we may divide by h, and the difference quotient equals 2x+h. Given ε>0 take δ=ε: when 0<|h|<δ, |(2x+h)−2x|=|h|<ε. By the definition of limit the quotient tends to 2x, so (x²)′=2x. S03.1 and S03.2 together prove Lemma L_deriv_square." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S03.1", "func_limit_def", "mul_inverse_nonzero" ], "explanation": { "zh-Hant": "除以 h 需要 h≠0；極限定義只看 0<|h|<δ 的 h，所以 h=0 從不需要代入。", "en": "Dividing by h needs h≠0; the limit definition only looks at h with 0<|h|<δ, so h=0 is never substituted." } }, "leaf_basis": "func_limit_def + mul_inverse_nonzero + abs_rules" }, { "id": "S04", "uses": [ "S03", "deriv_rules" ], "before_tex": "y_1'(x)=2x", "after_tex": "y_1''(x)=2", "relation": "implies", "rule": "deriv_rules", "conditions": [], "explanation": { "zh-Hant": "再求一次導數：(2x)′=2·(x)′=2·1=2（常數倍規則與 (x)′=1）。", "en": "Differentiate once more: (2x)′=2·(x)′=2·1=2 (constant-multiple rule and (x)′=1)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S03", "deriv_rules" ], "explanation": { "zh-Hant": "只用到求導規則；y₁′ 在整個 ℝ 上可微，所以 y₁″ 處處有定義。", "en": "Only differentiation rules are used; y₁′ is differentiable on all of ℝ, so y₁″ is defined everywhere." } }, "leaf_basis": "deriv_rules + mul_one" }, { "id": "S05", "uses": [ "S03", "S04", "G2", "G4" ], "before_tex": "y_1''(x)-2y_1'(x)+y_1(x)", "after_tex": "2-2(2x)+x^2", "relation": "equivalent", "rule": "eq_replace", "conditions": [], "explanation": { "zh-Hant": "依 MKO 的核對方法，把重算出的 y₁、y₁′、y₁″ 代入同一個方程的左邊：y₁″=2，y₁′=2x，y₁=x²。", "en": "Following the MKO's checking method, substitute the recomputed y₁, y₁′ and y₁″ into the left side of the same equation: y₁″=2, y₁′=2x, y₁=x²." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "eq_replace", "S03", "S04" ], "explanation": { "zh-Hant": "等值代換，雙向成立：兩式在每個 x 都表示同一個數。", "en": "Replacement of equals, valid both ways: the two expressions denote the same number at every x." } }, "leaf_basis": "eq_replace + mko_ode_check" }, { "id": "S06", "uses": [ "S05" ], "before_tex": "2-2(2x)+x^2", "after_tex": "x^2-4x+2", "relation": "equivalent", "rule": "distributive", "conditions": [], "explanation": { "zh-Hant": "2(2x)=4x，重新排列各項得 x²−4x+2。這是多項式恆等式（N2），對每個實數 x 成立。", "en": "2(2x)=4x, and rearranging the terms gives x²−4x+2. This is a polynomial identity (N2), valid for every real x." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "mul_assoc", "add_comm" ], "explanation": { "zh-Hant": "只用到結合律與交換律，雙向成立。", "en": "Only associativity and commutativity are used; the step holds both ways." } }, "leaf_basis": "mul_assoc + add_comm + add_assoc" }, { "id": "S07", "uses": [ "S06", "G4" ], "before_tex": "r_{y_1}(x)=(x^2-4x+2)-(x^2-4x+2)", "after_tex": "\\forall x\\in\\mathbb{R}:\\ r_{y_1}(x)=0", "relation": "implies", "rule": "forall_intro", "conditions": [], "explanation": { "zh-Hant": "依 G4，殘差是左邊減右邊；由 S06 左邊等於 x²−4x+2，所以 r_{y₁}(x)=0。這個計算對 x 沒有任何額外假設，x 是任意實數，依全稱推廣 r_{y₁} 在整個 ℝ 上恆為零，所以 y₁ 是 E 在 I=ℝ 上的解。這是符號化簡得到的，不是抽樣點核對。", "en": "By G4 the residual is the left side minus the right side; by S06 the left side equals x²−4x+2, so r_{y₁}(x)=0. The computation makes no extra assumption on x, so x is an arbitrary real number and by generalization r_{y₁} vanishes on all of ℝ; hence y₁ solves E on I=ℝ. This comes from symbolic simplification, not from checking sample points." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S06", "G4", "forall_intro" ], "explanation": { "zh-Hant": "全稱推廣可用，因為 S03 到 S06 對任意 x 成立。結論只關於 y₁ 是一個解，不宣稱它是唯一的解（見 S12 至 S14）。", "en": "Generalization is allowed because S03 to S06 hold for arbitrary x. The conclusion is only that y₁ is a solution, not that it is the only one (see S12 to S14)." } }, "leaf_basis": "forall_intro + add_inverse" }, { "id": "S08", "uses": [ "G2", "L_deriv_square", "deriv_rules" ], "before_tex": "y_2(x)=x^2+x", "after_tex": "y_2'(x)=2x+1,\\ \\ y_2''(x)=2", "relation": "implies", "rule": "deriv_rules", "conditions": [], "explanation": { "zh-Hant": "依和的求導規則、L_deriv_square 與 (x)′=1，y₂′=2x+1；再求一次導數，y₂″=2+0=2。", "en": "By the sum rule, L_deriv_square and (x)′=1, y₂′=2x+1; differentiating again, y₂″=2+0=2." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "L_deriv_square", "deriv_rules" ], "explanation": { "zh-Hant": "每個導數都要重新計算，不能沿用 y₁ 的導數；兩者只差在 +x 一項，y₂′ 因而多了 1。", "en": "Each derivative must be recomputed rather than copied from y₁; the candidates differ by the term +x, so y₂′ gains 1." } }, "leaf_basis": "deriv_rules + L_deriv_square" }, { "id": "S09", "uses": [ "S08", "G4" ], "before_tex": "r_{y_2}(x)=2-2(2x+1)+(x^2+x)-(x^2-4x+2)", "after_tex": "r_{y_2}(x)=x-2", "relation": "equivalent", "rule": "distributive", "conditions": [], "explanation": { "zh-Hant": "代入並化簡，細節在 S09.1、S09.2：先算左邊得 x²−3x，再減去右邊得 x−2。殘差是一個非零多項式。", "en": "Substitute and simplify, with details in S09.1 and S09.2: the left side is x²−3x, and subtracting the right side gives x−2. The residual is a nonzero polynomial." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "distributive", "add_inverse" ], "explanation": { "zh-Hant": "每一步都是對所有 x 成立的多項式恆等式（N3），雙向成立。", "en": "Each move is a polynomial identity valid for all x (N3), so the step holds both ways." } }, "leaf_basis": "distributive + add_assoc + add_comm + add_inverse" }, { "id": "S09.1", "uses": [ "S08" ], "before_tex": "2-2(2x+1)+(x^2+x)", "after_tex": "x^2-3x", "relation": "equivalent", "rule": "distributive", "conditions": [], "explanation": { "zh-Hant": "2(2x+1)=4x+2，所以 2−4x−2+x²+x=x²−3x。", "en": "2(2x+1)=4x+2, so 2−4x−2+x²+x=x²−3x." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "distributive", "add_inverse" ], "explanation": { "zh-Hant": "分配律與合併同類項，對所有 x 雙向成立。", "en": "Distributivity and collecting like terms, valid both ways for all x." } }, "leaf_basis": "distributive + add_inverse + add_comm" }, { "id": "S09.2", "uses": [ "S09.1" ], "before_tex": "(x^2-3x)-(x^2-4x+2)", "after_tex": "x-2", "relation": "equivalent", "rule": "distributive", "conditions": [], "explanation": { "zh-Hant": "去括號：x²−3x−x²+4x−2，x² 相消，−3x+4x=x，得 x−2。", "en": "Remove the brackets: x²−3x−x²+4x−2; the x² terms cancel and −3x+4x=x, giving x−2." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "neg_mul", "add_inverse" ], "explanation": { "zh-Hant": "負號分配與加法反元素，對所有 x 雙向成立。", "en": "Distributing the minus sign and additive inverses, valid both ways for all x." } }, "leaf_basis": "neg_mul + distributive + add_inverse" }, { "id": "S10", "uses": [ "S09", "G4" ], "before_tex": "r_{y_2}(x)=x-2", "after_tex": "r_{y_2}(2)=0,\\ r_{y_2}(0)=-2\\neq 0\\ \\Rightarrow\\ y_2\\ \\text{not a solution on }\\mathbb{R}", "relation": "implies", "rule": "quantifier_negation", "conditions": [], "explanation": { "zh-Hant": "在 x=2 殘差為 0，單點核對會誤判 y₂ 為解；但 r_{y₂}(0)=−2≠0。「對所有 x∈ℝ 殘差為零」只要一個反例就被否定，所以 y₂ 不是 ℝ 上的解。更進一步，x−2=0 ⇔ x=2，所以在任何含有 2 以外的點的區間上，y₂ 都不是解。", "en": "At x=2 the residual is 0, so a one-point check would wrongly accept y₂; but r_{y₂}(0)=−2≠0. 'The residual vanishes for all x∈ℝ' is refuted by one counterexample, so y₂ is not a solution on ℝ. Further, x−2=0 ⇔ x=2, so y₂ is not a solution on any interval containing a point other than 2." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S09", "quantifier_negation", "numeric_evaluation" ], "explanation": { "zh-Hant": "否定全稱只需一個反例 x=0；r_{y₂}(2)=0 這個事實並不支持 y₂ 是解。", "en": "Negating the universal statement needs one counterexample, x=0; the fact r_{y₂}(2)=0 gives no support to y₂ being a solution." } }, "leaf_basis": "quantifier_negation + numeric_evaluation" }, { "id": "S11", "uses": [ "G2", "G3", "S03" ], "before_tex": "y_1(0),\\ y_1'(0)", "after_tex": "y_1(0)=0,\\ y_1'(0)=0", "relation": "implies", "rule": "numeric_evaluation", "conditions": [], "explanation": { "zh-Hant": "y₁(0)=0²=0，y₁′(0)=2·0=0，所以 y₁ 滿足 G3 的兩個初值條件。初值在這裡起篩選作用。", "en": "y₁(0)=0²=0 and y₁′(0)=2·0=0, so y₁ satisfies both initial conditions of G3. Here the initial values act as a filter." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S03", "numeric_evaluation" ], "explanation": { "zh-Hant": "只是在 x=0 求值；滿足初值本身不證明 y₁ 是唯一的解，唯一性由 S12 的定理提供。", "en": "This is evaluation at x=0; satisfying the initial values does not by itself make y₁ the only solution; uniqueness comes from the theorem in S12." } }, "leaf_basis": "numeric_evaluation + mul_zero" }, { "id": "S12", "uses": [ "S07", "S11", "G1", "G3", "thm_linear_ivp", "poly_exp_continuous" ], "before_tex": "y''+p\\,y'+q\\,y=g,\\ p=-2,\\ q=1,\\ g=x^2-4x+2,\\ y(0)=0,\\ y'(0)=0", "after_tex": "y_1\\ \\text{is the unique solution of the IVP on }\\mathbb{R}", "relation": "theorem_application", "rule": "thm_linear_ivp", "conditions": [ "p,q,g\\ \\text{continuous on }\\mathbb{R}", "0\\in\\mathbb{R},\\ \\mathbb{R}\\ \\text{open}" ], "explanation": { "zh-Hant": "把 E 寫成標準形 y″+p y′+q y=g，p(x)=−2、q(x)=1 是常數函數，g(x)=x²−4x+2 是多項式，三者在開區間 ℝ 上連續，且 x₀=0∈ℝ。引用定理 thm_linear_ivp（Y₀=0、Y₁=0）：初值問題恰有一個解，且在整個 ℝ 上存在。S07 與 S11 說明 y₁ 就是這樣一個解，所以它就是那唯一的解。唯一性來自定理與其連續性假設，而不是來自方程本身。", "en": "Write E in the standard form y″+p y′+q y=g with p(x)=−2 and q(x)=1 constant and g(x)=x²−4x+2 a polynomial; all three are continuous on the open interval ℝ, and x₀=0∈ℝ. By the cited theorem thm_linear_ivp (Y₀=0, Y₁=0), the initial value problem has exactly one solution, and it exists on all of ℝ. S07 and S11 show that y₁ is such a solution, so it is that unique solution. Uniqueness comes from the theorem and its continuity hypotheses, not from the equation itself." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "thm_linear_ivp", "poly_exp_continuous", "S07", "S11" ], "explanation": { "zh-Hant": "定理需要：方程為線性且 y″ 係數為 1、p、q、g 在含 x₀ 的開區間上連續、給定兩個初值；四項都已核對。定理本身未在本稿證明（citation_verified）。不宣稱非線性方程有同樣結論。", "en": "The theorem needs: a linear equation with leading coefficient 1, p, q, g continuous on an open interval containing x₀, and two initial values; all four are checked. The theorem itself is not proved here (citation_verified). No such conclusion is claimed for nonlinear equations." } }, "leaf_basis": "thm_linear_ivp + poly_exp_continuous" }, { "id": "S13", "uses": [ "G2", "deriv_rules", "L_deriv_square" ], "before_tex": "y_3(x)=x^2+e^x", "after_tex": "y_3'(x)=2x+e^x,\\ \\ y_3''(x)=2+e^x", "relation": "implies", "rule": "deriv_rules", "conditions": [], "explanation": { "zh-Hant": "為了說明沒有初值時解不唯一，取 y₃=x²+eˣ。依和的規則、L_deriv_square 與 (eˣ)′=eˣ：y₃′=2x+eˣ，y₃″=2+eˣ。", "en": "To show that without initial values the solution is not unique, take y₃=x²+eˣ. By the sum rule, L_deriv_square and (eˣ)′=eˣ: y₃′=2x+eˣ and y₃″=2+eˣ." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "deriv_rules", "L_deriv_square" ], "explanation": { "zh-Hant": "只用到求導規則；y₃ 不是多項式，超出 MKO 題庫的候選類，只作說明用途。", "en": "Only differentiation rules are used; y₃ is not a polynomial and lies outside the MKO exercise class, so it is used for illustration only." } }, "leaf_basis": "deriv_rules + L_deriv_square" }, { "id": "S14", "uses": [ "S13", "G4", "G3" ], "before_tex": "r_{y_3}(x)=(2+e^x)-2(2x+e^x)+(x^2+e^x)-(x^2-4x+2)", "after_tex": "r_{y_3}(x)=0\\ \\forall x,\\ \\ y_3(0)=1\\neq 0", "relation": "implies", "rule": "distributive", "conditions": [], "explanation": { "zh-Hant": "展開：eˣ 的係數 1−2+1=0，x 的係數 −4+4=0，x² 的係數 1−1=0，常數 2−2=0，所以對每個 x 殘差為 0（N7 把 eˣ 當作獨立符號 E 做恆等式核對，因此對 eˣ 的任何值都成立）。y₃ 也是 E 在 ℝ 上的解，但 y₃(0)=0+1=1≠0，不滿足初值。所以方程本身有不只一個解，初值才把 y₁ 選出來。", "en": "Expand: the coefficient of eˣ is 1−2+1=0, of x is −4+4=0, of x² is 1−1=0, and the constant is 2−2=0, so the residual is 0 for every x (N7 checks this as an identity with eˣ treated as an independent symbol E, so it holds whatever the value of eˣ). So y₃ is also a solution of E on ℝ, but y₃(0)=0+1=1≠0 fails the initial condition. Hence the equation alone has more than one solution, and the initial values are what select y₁." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S13", "G4", "distributive", "add_inverse" ], "explanation": { "zh-Hant": "需要逐項係數相消；這證明了 E 的解不唯一，與 S12 不矛盾，因為 S12 的唯一性是在加上初值之後。", "en": "It needs the coefficients to cancel term by term; this shows that solutions of E are not unique, which does not contradict S12, whose uniqueness holds only after the initial values are added." } }, "leaf_basis": "distributive + add_inverse + numeric_evaluation" }, { "id": "S15", "uses": [ "S02", "S07", "S10", "S12", "S14" ], "before_tex": "\\text{proved facts about }E", "after_tex": "\\text{order }2;\\ y_1\\ \\text{solves};\\ y_2\\ \\text{does not};\\ y_1\\ \\text{unique for the IVP};\\ E\\ \\text{alone not unique}", "relation": "implies", "rule": "and_intro_elim", "conditions": [], "explanation": { "zh-Hant": "合併結論：E 的階數為 2；y₁=x² 在 ℝ 上是解；y₂=x²+x 不是（殘差 x−2）；在初值 y(0)=0、y′(0)=0 下 y₁ 是唯一解（依引用定理）；沒有初值時 y₃=x²+eˣ 也是解。", "en": "Combine the conclusions: E has order 2; y₁=x² solves it on ℝ; y₂=x²+x does not (residual x−2); under y(0)=0, y′(0)=0, y₁ is the unique solution (by the cited theorem); without initial values y₃=x²+eˣ is also a solution." }, "reverse_audit": { "kind": "sufficient_goal_reduction", "required_ids": [ "S02", "S07", "S10", "S12", "S14", "and_intro_elim" ], "explanation": { "zh-Hant": "目標 G：本稿目標的五項判定。足夠條件 H：S02、S07、S10、S12、S14（H⇒G 由合取引入）。H 由給定條件與引用定理正向推出，沒有把 G 當作前提；唯一性依賴的定理標為 citation_verified。", "en": "Goal G: the five determinations in the goal. Sufficient condition H: S02, S07, S10, S12 and S14 (H⇒G by conjunction introduction). H is derived forwards from the givens and the cited theorem, without using G as a premise; the theorem behind uniqueness is marked citation_verified." } }, "leaf_basis": "and_intro_elim" } ], "result": { "zh-Hant": "y″−2y′+y=x²−4x+2 是二階常微分方程。y₁=x² 的殘差在整個 ℝ 上恆為零，所以是解；y₂=x²+x 的殘差是 x−2，只在 x=2 為零，所以不是 ℝ 上的解。y₁ 滿足 y(0)=0、y′(0)=0，依二階線性方程的存在唯一定理（係數連續，citation_verified）它是此初值問題的唯一解；沒有初值時 y₃=x²+eˣ 也是解，所以方程本身不決定唯一解。", "en": "y″−2y′+y=x²−4x+2 is a second-order ordinary differential equation. The residual of y₁=x² vanishes identically on ℝ, so it is a solution; the residual of y₂=x²+x is x−2, zero only at x=2, so it is not a solution on ℝ. y₁ satisfies y(0)=0 and y′(0)=0, and by the existence and uniqueness theorem for second-order linear equations (continuous coefficients, citation_verified) it is the unique solution of this initial value problem; without initial values y₃=x²+eˣ is also a solution, so the equation alone does not determine a unique solution." }, "negative_controls": [ { "claim": { "zh-Hant": "r_{y₂}(2)=0，所以 y₂=x²+x 是此方程在 ℝ 上的解", "en": "r_{y₂}(2)=0, so y₂=x²+x solves the equation on ℝ" }, "expected": "reject", "reason": { "zh-Hant": "r_{y₂}(0)=−2≠0；殘差 x−2 只在 x=2 為零（S10）。", "en": "r_{y₂}(0)=−2≠0; the residual x−2 vanishes only at x=2 (S10)." } }, { "claim": { "zh-Hant": "左邊有三項，所以這是三階方程", "en": "The left side has three terms, so the equation has order 3" }, "expected": "reject", "reason": { "zh-Hant": "階數是實際出現的最高導數階數，這裡是 y″，所以是二階（S02）。", "en": "The order is that of the highest derivative actually occurring, here y″, so the order is 2 (S02)." } }, { "claim": { "zh-Hant": "任何一階方程加上一個初值都恰有一個解", "en": "Every first-order equation with one initial value has exactly one solution" }, "expected": "reject", "reason": { "zh-Hant": "y′=3y^(2/3)、y(0)=0 有 y=0 與 y=x³ 兩個解；唯一性需要 Lipschitz 等假設。", "en": "y′=3y^(2/3) with y(0)=0 has the two solutions y=0 and y=x³; uniqueness needs hypotheses such as a Lipschitz condition." } }, { "claim": { "zh-Hant": "方程本身就決定了解是 y=x²", "en": "The equation by itself determines the solution y=x²" }, "expected": "reject", "reason": { "zh-Hant": "y₃=x²+eˣ 也滿足方程（S14）；要初值才把 y₁ 選出來。", "en": "y₃=x²+eˣ also satisfies the equation (S14); initial values are needed to single out y₁." } } ], "status": "candidate", "gaps": [], "verification": { "structure": "checked_local_script", "mathematics": "candidate_argument", "formalization": "not_formalized" }, "rule_glossary": [ { "id": "func_limit_def", "kind": "prerequisite", "statement": { "zh-Hant": "函數極限：lim_{x→a} f(x)=L 表示對每個 ε>0，存在 δ>0，使所有滿足 0<|x−a|<δ 的定義域點 x 都有 |f(x)−L|<ε（a 須為定義域的極限點）。", "en": "Limit of a function: lim_{x→a} f(x)=L means that for every ε>0 there is δ>0 such that |f(x)−L|<ε for every domain point x with 0<|x−a|<δ (a must be a limit point of the domain)." } }, { "id": "distributive", "kind": "prerequisite", "statement": { "zh-Hant": "分配律：u(v+w)=uv+uw；配合乘法交換律也有 (v+w)u=vu+wu。", "en": "Distributivity: u(v+w)=uv+uw; with commutativity also (v+w)u=vu+wu." } }, { "id": "eq_replace", "kind": "prerequisite", "statement": { "zh-Hant": "等值代換：命題中的一個式子可換成與它相等的式子，命題的真假不變。", "en": "Replacement of equals: an expression inside a statement may be replaced by an equal expression without changing whether the statement is true." } }, { "id": "forall_intro", "kind": "prerequisite", "statement": { "zh-Hant": "∀引入（全稱推廣）：若某命題對一個「任意的」元素 x 證成，而證明沒有對 x 加任何額外假設，則它對論域中所有 x 成立。", "en": "∀-introduction (generalization): if a statement is proved for an arbitrary element x, with no extra assumption on x, it holds for every x in the domain." } }, { "id": "quantifier_negation", "kind": "prerequisite", "statement": { "zh-Hant": "量詞否定：¬(∀x, P(x)) 等價於 ∃x, ¬P(x)；¬(∃x, P(x)) 等價於 ∀x, ¬P(x)。", "en": "Negating quantifiers: ¬(∀x, P(x)) is equivalent to ∃x, ¬P(x), and ¬(∃x, P(x)) is equivalent to ∀x, ¬P(x)." } }, { "id": "def_intro", "kind": "prerequisite", "statement": { "zh-Hant": "引入定義：可為一個命題、式子或物件取名；之後名稱與被定義者可互相替換。", "en": "Introducing a definition: a statement, expression or object may be given a name, after which the name and what it stands for are interchangeable." } }, { "id": "def_unfold", "kind": "prerequisite", "statement": { "zh-Hant": "展開定義：把一個已定義的詞換成它的定義內容（或反向收回），命題的真假不變。", "en": "Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true." } }, { "id": "add_assoc", "kind": "prerequisite", "statement": { "zh-Hant": "加法結合律：(u+v)+w=u+(v+w)。", "en": "Associativity of addition: (u+v)+w=u+(v+w)." } }, { "id": "add_comm", "kind": "prerequisite", "statement": { "zh-Hant": "加法交換律：u+v=v+u。", "en": "Commutativity of addition: u+v=v+u." } }, { "id": "add_inverse", "kind": "prerequisite", "statement": { "zh-Hant": "加法反元素：每個 u 都有 −u，使 u+(−u)=0；u−v 定義為 u+(−v)。", "en": "Additive inverse: every u has −u with u+(−u)=0; u−v means u+(−v)." } }, { "id": "mul_inverse_nonzero", "kind": "prerequisite", "statement": { "zh-Hant": "非零數的倒數：在 ℚ、ℝ、ℂ 中，若 u≠0，存在 1/u 使 u·(1/u)=1；對 u≠0 定義 v/u 為 v·(1/u)。u=0 時沒有倒數，v/0 不定義。", "en": "Inverse of a nonzero number: in ℚ, ℝ or ℂ, if u≠0 there is 1/u with u·(1/u)=1, and v/u means v·(1/u). 0 has no inverse, so v/0 is undefined." } }, { "id": "abs_rules", "kind": "prerequisite", "statement": { "zh-Hant": "絕對值（實數）：|u|=u（u≥0）、|u|=−u（u<0）；|u|≥0；|uv|=|u||v|；三角不等式 |u+v|≤|u|+|v|；|u|