--- title: "Solve xp²−2yp+ax=0" subject: "Mathematics I" chapter: 4 key: "semester_1/mathematics_i:de-xp2-minus-2yp-plus-ax" source: https://www.bctengineer.com/q/semester_1/mathematics_i/de-xp2-minus-2yp-plus-ax site: BCT Engineering --- # Solve xp²−2yp+ax=0 Mathematics I · Chapter 4: Ordinary Differential Equations and their Applications - Asked in: 2073 Shrawan; 2071 Shrawan, 5 marks; 2068 Chaitra (asked 3 times) - Video: [Solve xp^2 - 2yp + x = 0, try hard , easy solving method](https://www.youtube.com/watch?v=sifvJtfjfGI) — TK peakeducation - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/mathematics_i/de-xp2-minus-2yp-plus-ax - Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-4 ## Question Solve: xp² − 2yp + ax = 0, where p = dy/dx. ## Short answer Solve for y, differentiate w.r.t. x, and separate variables to get p = cx. The general solution is 2cy = c²x² + a, with singular solution y² = ax². ## Answer The equation xp² − 2yp + ax = 0 (p = dy/dx) is solvable for y. #### Solve for y 2yp = xp² + ax → y = x(p² + a)/(2p) = x(p/2 + a/(2p)) #### Differentiate w.r.t. x p = (p² + a)/(2p) + x(1/2 − a/(2p²)) dp/dx p − (p² + a)/(2p) = (p² − a)/(2p), so (p² − a)/(2p) = x (p² − a)/(2p²) · dp/dx Cancel (p² − a)/(2p) (the case p² = a is treated below): 1 = (x/p) dp/dx → dp/p = dx/x → p = cx #### Eliminate p y = x(c²x² + a)/(2cx) = (c²x² + a)/(2c) **General solution: 2cy = c²x² + a** Check: p = cx, then xp² − 2yp + ax = c²x³ − (c²x² + a)x + ax = 0 ✓. #### Singular solution The factor p² = a gives y = ±√a x, which satisfies the given equation. It is the envelope of the family 2cy = c²x² + a: treating it as a quadratic in c, c²x² − 2cy + a = 0, the discriminant vanishes when 4y² − 4ax² = 0, i.e. **y² = ax²** (singular solution) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which variable is the equation xp^2 - 2yp + ax = 0 solved for before differentiating? A. It is Clairaut's form B. p, then integrate directly C. y, then differentiate w.r.t. x D. x, then differentiate w.r.t. y Answer: C. It gives y = x(p^2 + a)/(2p), so y is differentiated with respect to x. (Key point: Method) 2. After cancelling (p^2 - a)/(2p), what does dp/p = dx/x give? A. p = c/x B. p = cx^2 C. p = c + x D. p = cx Answer: D. Integrating gives ln p = ln x + ln c, so p = cx. (Key point: Relation for p) 3. Eliminating p using p = cx, the general solution is: A. 2cy = c^2 x^2 + a B. y = cx + a C. cy = x^2 + a D. 2cy = c^2 x^2 - a Answer: A. y = x(c^2 x^2 + a)/(2cx) = (c^2 x^2 + a)/(2c). (Key point: General solution) 4. How is the singular solution y^2 = ax^2 found from the family? A. Differentiating y^2 = ax^2 twice B. Setting p = 0 C. Setting c = 0 D. Setting the discriminant of c^2 x^2 - 2cy + a = 0 (in c) to zero Answer: D. The envelope comes from the c-discriminant: 4y^2 - 4ax^2 = 0. (Key point: Singular solution) ## Related in Mathematics I - [Solve (1+y²)+(x−e^(tan⁻¹y))dy/dx=0](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Ade-1-plus-y2-x-minus-e-atan-y) — Mathematics I, chapter 4, asked 6× - [Solve dy/dx = y/x + tan(y/x)](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Ade-dy-dx-y-over-x-tan) — Mathematics I, chapter 4, asked 6× - [Solve dy/dx − 2y tan x = y² tan x](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Ade-bernoulli-2y-tan-x) — Mathematics I, chapter 4, asked 6× - [Solve (1+x²)dy/dx+y=e^(tan⁻¹x)](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Ade-1-plus-x2-dy-dx-etan) — Mathematics I, chapter 4, asked 6× - [Solve x dy/dx + y log y = xy eˣ](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Ade-x-dy-dx-y-log-y) — Mathematics I, chapter 4, asked 4× - [Solve x log x dy/dx + y = 2 log x](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Ade-x-log-x-dy-dx) — Mathematics I, chapter 4, asked 4× ## Related in other subjects - [RK-4: dy/dx = (1+x²)y](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Ark4-y-1-x2) — Numerical Method, chapter 6 - [Polar coordinates: ∫∫(x²+y²) dy dx](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Apolar-coordinates-x2-y2-dy) — Mathematics, chapter 2, asked 5× - [Evaluate ∫∫ e^(y/x) dy dx](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Aevaluate-e-y-x-dy) — Mathematics, chapter 2 - [Evaluate ∫∫ e^(x²) dx dy](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Aevaluate-e-x2-dx-dy) — Mathematics, chapter 2