JEE Advanced 2015Paper 1
- Exam
- JEE Advanced
- Year
- 2015
- Paper
- Paper 1
- Subject
- Mathematics
- Chapter
- Differential Equations
JEE Advanced 2015 · Paper 1 · Mathematics · Differential Equations
Consider the family of all circles whose centers lie on the...
Consider the family of all circles whose centers lie on the straight line If this family of circles is represented by the differential equation where are functions of , then which of the following statements is (are) true ?
Options
- A
- B
- CP + Q = 1 - x + y + {y}^{′} + {\left({y}^{′}\right)}^{2}
- DP - Q = x + y - {y}^{′} - {\left({y}^{′}\right)}^{2}
Correct answer
B.
Step-by-step solution
Let required circle be
Differentiation
2x + 2y{y}^{′} + 2g\left(1 + {y}^{′}\right) = 0
⇒ x + y{y}^{′} + g\left(1 + {y}^{′}\right) = 0 ...(i)
Again differentiation
1 + y{y}^{′} + {\left(y\right)}^{2} + g\left({y}^{′}\right) = 0 ...(ii)
Putting value of g in equation (ii),
1 + y{y}^{′} + {\left({y}^{′}\right)}^{2} - \left(\dfrac{x + y{y}^{′}}{1 + {y}^{′}}\right){y}^{′′} = 0
So, P = y - x, Q = 1 + {y}^{′} + {\left({y}^{′}\right)}^{2}
i.e., P + Q = 1 - x + y + {y}^{′}{\left({y}^{′}\right)}^{2}
Differentiation
2x + 2y{y}^{′} + 2g\left(1 + {y}^{′}\right) = 0
⇒ x + y{y}^{′} + g\left(1 + {y}^{′}\right) = 0 ...(i)
Again differentiation
1 + y{y}^{′} + {\left(y\right)}^{2} + g\left({y}^{′}\right) = 0 ...(ii)
Putting value of g in equation (ii),
1 + y{y}^{′} + {\left({y}^{′}\right)}^{2} - \left(\dfrac{x + y{y}^{′}}{1 + {y}^{′}}\right){y}^{′′} = 0
So, P = y - x, Q = 1 + {y}^{′} + {\left({y}^{′}\right)}^{2}
i.e., P + Q = 1 - x + y + {y}^{′}{\left({y}^{′}\right)}^{2}