JEE MAIN - Mathematics (2021 - 22th July Evening Shift - No. 23)
Let y = y(x) be the solution of the differential equation $$\left( {(x + 2){e^{\left( {{{y + 1} \over {x + 2}}} \right)}} + (y + 1)} \right)dx = (x + 2)dy$$, y(1) = 1. If the domain of y = y(x) is an open interval ($$\alpha$$, $$\beta$$), then | $$\alpha$$ + $$\beta$$| is equal to ______________.
Answer
4
Explanation
Let y + 1 = Y and x + 2 = X
dy = dY
dx = dX
$$\left( {X{e^{{X \over X}}} + Y} \right)dX = XdY$$
$$ \Rightarrow {{XdY - YdX} \over {{X^2}}} = {{{e^{{Y \over X}}}} \over X}dX$$
$$ \Rightarrow {e^{ - {Y \over X}}}d\left( {{Y \over X}} \right) = {{dX} \over X}$$
$$ \Rightarrow - {e^{ - {Y \over X}}} = \ln |X| + c$$
$$ \Rightarrow - {e^{ - \left( {{{y + 1} \over {x + 2}}} \right)}} = \ln |x + 2| + c$$
$$\because$$ (1, 1) satisfy this equation
So, $$c = - {e^{ - {2 \over 3}}} - \ln 3$$
Now, $$y = - 1 - (x + 2)\ln \left( {\ln \left( {\left| {{3 \over {x + 2}}} \right|} \right) + {e^{ - {2 \over 3}}}} \right)$$
Domain :
$$\ln \left| {{3 \over {x + 2}}} \right| > {e^{ - {e^{ - {2 \over 3}}}}}$$
$$ \Rightarrow {3 \over {\left| {x + 2} \right|}} > {e^{ - {e^{ - {2 \over 3}}}}}$$
$$ \Rightarrow \left| {x + 2} \right| < 3{e^{{e^{ - {2 \over 3}}}}}$$
$$ \Rightarrow - 3{e^{{e^{ - {2 \over 3}}}}} - 2 < x < 3{e^{{e^{ - {2 \over 3}}}}} - 2$$
So, $$\alpha + \beta = - 4$$
$$ \Rightarrow \left| {\alpha + \beta } \right| = 4$$
dy = dY
dx = dX
$$\left( {X{e^{{X \over X}}} + Y} \right)dX = XdY$$
$$ \Rightarrow {{XdY - YdX} \over {{X^2}}} = {{{e^{{Y \over X}}}} \over X}dX$$
$$ \Rightarrow {e^{ - {Y \over X}}}d\left( {{Y \over X}} \right) = {{dX} \over X}$$
$$ \Rightarrow - {e^{ - {Y \over X}}} = \ln |X| + c$$
$$ \Rightarrow - {e^{ - \left( {{{y + 1} \over {x + 2}}} \right)}} = \ln |x + 2| + c$$
$$\because$$ (1, 1) satisfy this equation
So, $$c = - {e^{ - {2 \over 3}}} - \ln 3$$
Now, $$y = - 1 - (x + 2)\ln \left( {\ln \left( {\left| {{3 \over {x + 2}}} \right|} \right) + {e^{ - {2 \over 3}}}} \right)$$
Domain :
$$\ln \left| {{3 \over {x + 2}}} \right| > {e^{ - {e^{ - {2 \over 3}}}}}$$
$$ \Rightarrow {3 \over {\left| {x + 2} \right|}} > {e^{ - {e^{ - {2 \over 3}}}}}$$
$$ \Rightarrow \left| {x + 2} \right| < 3{e^{{e^{ - {2 \over 3}}}}}$$
$$ \Rightarrow - 3{e^{{e^{ - {2 \over 3}}}}} - 2 < x < 3{e^{{e^{ - {2 \over 3}}}}} - 2$$
So, $$\alpha + \beta = - 4$$
$$ \Rightarrow \left| {\alpha + \beta } \right| = 4$$
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