%%

If y = y(x) is the solution of the differential equation $${{5 + {e^x}} \over {2 + y}}.{{dy} \over {dx}} + {e^x} = 0$$ satisfying y(0) = 1, then a value of y(loge13) is :

JEE · Math · previous-year question

  1. A.-1correct
  2. B.1
  3. C.0
  4. D.2

Answer

A. -1

Explanation

$${{5 + {e^x}} \over {2 + y}}.{{dy} \over {dx}} + {e^x} = 0$$ $$ \Rightarrow $$ $${{5 + {e^x}} \over {2 + y}}.{{dy} \over {dx}} = -{e^x} $$ Integrating both sides, $$ \Rightarrow $$ $$\int {{{dy} \over {2 + y}}} = \int {{{ - {e^x}} \over {{e^x} + 5}}} dx$$ $$ \Rightarrow $$ ln (y + 2) = – ln(ex + 5) + k $$ \Rightarrow $$ (y + 2) (ex + 5) = C $$ \because $$ y(0) = 1 $$ \Rightarrow $$ C = 18 $$ \therefore $$ y + 2 = $${{{18} \over {{e^x} + 5}}}$$ At x = loge13 y + 2 = $${{{18} \over {13 + 5}}}$$ = 1 $$ \Rightarrow $$ y = -1

Practice more JEE questions

Answer thousands more real JEE previous-year questions free, get graded instantly, and climb the global ranked leaderboard.

Practice JEE free →

More JEE Math questions