JEE MAIN - Mathematics (2019 - 12th April Morning Slot - No. 18)
Consider the differential equation, $${y^2}dx + \left( {x - {1 \over y}} \right)dy = 0$$, If value of y is 1 when x = 1, then the value of x
for which y = 2, is :
$${3 \over 2} - {1 \over {\sqrt e }}$$
$${1 \over 2} + {1 \over {\sqrt e }}$$
$${5 \over 2} + {1 \over {\sqrt e }}$$
$${3 \over 2} - \sqrt e $$
Explanation
$${y^2}dx + \left( {x - {1 \over y}} \right)dy = 0$$
$$ \Rightarrow {{dx} \over {dy}} + {x \over {{y^2}}} = {1 \over {{y^3}}}$$
Integrating factor (I.F) = $${e^{ - {1 \over y}}}$$
Now x.$${e^{ - {1 \over y}}}$$ = $$\int {{e^{ - {1 \over y}}}{1 \over {{y^3}}}dy} $$ by putting $${ - {1 \over y}}$$ = t
x.et = $$\int {{e^t}( - t)dt} $$
$$ \Rightarrow x{e^t} = - (t.{e^t} - {e^t}) + c$$
$$ \Rightarrow x{e^{ - {1 \over y}}} = {e^{ - {1 \over y}}}\left( {1 + {1 \over y}} \right) + c$$
$$ \Rightarrow x = 1 + {1 \over y} + c.e^{1 \over y}$$
It passes through (1, 1)
$$ \therefore $$ c = $$-{1 \over e}$$
Equation of the curve is
$$x = 1 + {1 \over y} - {e^{{1 \over y} - 1}}$$ It passes through (k, 2)
$$ \therefore $$ k = $$1 + {1 \over 2} - {e^{{1 \over 2} - 1}}$$ = $$3 \over 2$$ - $$1 \over{\sqrt e}$$
$$ \Rightarrow {{dx} \over {dy}} + {x \over {{y^2}}} = {1 \over {{y^3}}}$$
Integrating factor (I.F) = $${e^{ - {1 \over y}}}$$
Now x.$${e^{ - {1 \over y}}}$$ = $$\int {{e^{ - {1 \over y}}}{1 \over {{y^3}}}dy} $$ by putting $${ - {1 \over y}}$$ = t
x.et = $$\int {{e^t}( - t)dt} $$
$$ \Rightarrow x{e^t} = - (t.{e^t} - {e^t}) + c$$
$$ \Rightarrow x{e^{ - {1 \over y}}} = {e^{ - {1 \over y}}}\left( {1 + {1 \over y}} \right) + c$$
$$ \Rightarrow x = 1 + {1 \over y} + c.e^{1 \over y}$$
It passes through (1, 1)
$$ \therefore $$ c = $$-{1 \over e}$$
Equation of the curve is
$$x = 1 + {1 \over y} - {e^{{1 \over y} - 1}}$$ It passes through (k, 2)
$$ \therefore $$ k = $$1 + {1 \over 2} - {e^{{1 \over 2} - 1}}$$ = $$3 \over 2$$ - $$1 \over{\sqrt e}$$
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