JEE MAIN - Mathematics (2020 - 5th September Morning Slot - No. 19)
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 :
equation $${{5 + {e^x}} \over {2 + y}}.{{dy} \over {dx}} + {e^x} = 0$$ satisfying y(0) = 1, then a value of y(loge13) is :
-1
1
0
2
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
$$ \Rightarrow $$ $${{5 + {e^x}} \over {2 + y}}.{{dy} \over {dx}} = -{e^x} $$
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
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