--- title: "Solve y = yp²+2px" subject: "Mathematics I" chapter: 4 key: "semester_1/mathematics_i:de-y-yp2-plus-2px" source: https://www.bctengineer.com/q/semester_1/mathematics_i/de-y-yp2-plus-2px site: BCT Engineering --- # Solve y = yp²+2px Mathematics I · Chapter 4: Ordinary Differential Equations and their Applications - Asked in: 2074 Chaitra, 5 marks; 2071 Chaitra; 2070 Asar; 2067 Asar (asked 4 times) - Video: [y=2px+yp^2 Simple Concept Differential Equations In this Video.](https://www.youtube.com/watch?v=E_ZOM7oLODk) — Ambition Mathematics - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/mathematics_i/de-y-yp2-plus-2px - Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-4 ## Question Solve: y = yp² + 2px, where p = dy/dx. ## Short answer Solve for x, differentiate w.r.t. y, and separate variables to get yp = c. Eliminating p gives the general solution y² = 2cx + c². ## Answer The equation y = yp² + 2px, with p = dy/dx, is solvable for x, so differentiate with respect to y (treating x as a function of y). #### Solve for x 2px = y − yp² → x = y(1 − p²)/(2p) = y(1/(2p) − p/2) #### Differentiate w.r.t. y dx/dy = 1/p, so 1/p = (1/(2p) − p/2) + y(−1/(2p²) − 1/2) dp/dy 1/p − (1 − p²)/(2p) = (1 + p²)/(2p), so (1 + p²)/(2p) = −y (1 + p²)/(2p²) · dp/dy Cancel (1 + p²)/(2p) (it is non-zero for real p): 1 = −(y/p) dp/dy → dy/y + dp/p = 0 Integrating: ln y + ln p = ln c, so **yp = c**, p = c/y. #### Eliminate p Substitute p = c/y in the given equation: y = y(c²/y²) + 2(c/y)x → y² = c² + 2cx **General solution: y² = 2cx + c²** (c an arbitrary constant). Check: differentiating, 2yy' = 2c, so p = c/y; then yp² + 2px = c²/y + 2cx/y = (c² + 2cx)/y = y ✓. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is y = yp^2 + 2px best handled by solving for x and differentiating with respect to y? A. It is a Clairaut equation in x B. It is already exact C. It is solvable for x, so x = function of y and p D. It is linear in y Answer: C. The equation can be rearranged as x = y(1 - p^2)/(2p), so x is expressed in y and p. (Key point: Choice of variable) 2. After differentiating, which factor is cancelled from both sides? A. y/p B. 2p C. (1 - p^2)/(2p) D. (1 + p^2)/(2p) Answer: D. Both sides contain (1 + p^2)/(2p), which is non-zero for real p. (Key point: Cancelling the common factor) 3. What does the separated equation dy/y + dp/p = 0 give? A. y/p = c B. y p^2 = c C. yp = c D. y + p = c Answer: C. Integrating gives ln y + ln p = ln c, so yp = c. (Key point: Relation between y and p) 4. Putting p = c/y into the original equation gives which general solution? A. y^2 = 2cx - c^2 B. y^2 = 2cx + c^2 C. y^2 = c^2 x + 2c D. y = cx + c^2 Answer: B. y = c^2/y + 2cx/y gives y^2 = c^2 + 2cx. (Key point: General 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