Differential Equations — Class 12 Maths Solution

exemplar objective MCQ NCERT EXEMP.Q.48,Page.197
Question

${\tan ^{ - 1}}x + {\tan ^{ - 1}}y = C$ is general Solution of the differential equation

  • (a) $\frac{{dy}}{{dx}} = \frac{{1 + {y^2}}}{{1 + {x^2}}}$
  • (b) $\frac{{dy}}{{dx}} = \frac{{1 + {x^2}}}{{1 + {y^2}}}$
  • (c) $\left( {1 + {x^2}} \right)dy + \left( {1 + {y^2}} \right)dx = 0$ ✓ Correct
  • (d) $\left( {1 + {x^2}} \right)dx + \left( {1 + {y^2}} \right)dy = 0$
Step-by-step Solution
Correct answer: option (c)

Given that, ${\tan ^{ - 1}}x + {\tan ^{ - 1}}y = C$

On differentiating w.r.t. $x$,

we get
$\frac{1}{{1 + {x^2}}} + \frac{1}{{1 + {y^2}}} \cdot \frac{{dy}}{{dx}} = 0$

$\Rightarrow$ $\frac{1}{{1 + {y^2}}} \cdot \frac{{dy}}{{dx}} = - \frac{1}{{1 + {x^2}}}$

$\Rightarrow$ $\left( {1 + {x^2}} \right)dy + \left( {1 + {y^2}} \right)dx = 0$

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NCERT & Exemplar solution for CBSE Class 12 Mathematics, Differential Equations. Curated by Sachin Sharma. Free for all students.