{"schema_version":"ocme-site-results-v1","locale":"zh-Hant","page_type":"derivations","facts":{"questions":36000,"reviewed":36000,"display_candidates":0,"knowledge":302,"guides":299,"paths":59,"batches":360,"families":1441,"atlas":302,"core":80,"evidence":9,"formal_evidence":5,"formal_subjects":4,"lean_files":6,"lean_toolchain":"leanprover/lean4:v4.30.0","question_state":"candidate","english_state":"ai_candidate","question_Lean":"NOT_RUN","canonical_truth_promotion":false},"filters":{"q":"mko-ordinary-differential-equation"},"kind":"guides","total":1,"pages":1,"page":1,"entries":[{"id":"derivation-mko-ordinary-differential-equation-01","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"},"summary":{"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."},"mko_ids":["mko-ordinary-differential-equation"]}],"title":"概念推導導讀"}