JEE MAIN - Mathematics (2020 - 6th September Morning Slot - No. 2)
The general solution of the differential equation
$$\sqrt {1 + {x^2} + {y^2} + {x^2}{y^2}} $$ + xy$${{dy} \over {dx}}$$ = 0 is :
(where C is a constant of integration)
$$\sqrt {1 + {x^2} + {y^2} + {x^2}{y^2}} $$ + xy$${{dy} \over {dx}}$$ = 0 is :
(where C is a constant of integration)
$$\sqrt {1 + {y^2}} + \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} - 1} \over {\sqrt {1 + {x^2}} + 1}}} \right) + C$$
$$\sqrt {1 + {y^2}} - \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} - 1} \over {\sqrt {1 + {x^2}} + 1}}} \right) + C$$
$$\sqrt {1 + {y^2}} + \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} + 1} \over {\sqrt {1 + {x^2}} - 1}}} \right) + C$$
$$\sqrt {1 + {y^2}} - \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} + 1} \over {\sqrt {1 + {x^2}} - 1}}} \right) + C$$
Explanation
$$\sqrt {1 + {x^2} + {y^2} + {x^2}{y^2}} $$ + xy$${{dy} \over {dx}}$$ = 0
$$ \Rightarrow $$ $$\sqrt {\left( {1 + {x^2}} \right)\left( {1 + {y^2}} \right)} $$ + xy$${{dy} \over {dx}}$$ = 0
$$ \Rightarrow $$ $$\sqrt {\left( {1 + {x^2}} \right)} \sqrt {\left( {1 + {y^2}} \right)} $$ = -xy$${{dy} \over {dx}}$$
$$ \Rightarrow $$ $$\int {{{ydy} \over {\sqrt {1 + {y^2}} }}} = - \int {{{\sqrt {1 + {x^2}} } \over x}} dx$$ .....(1)
Now put 1 + x2 = u2 and 1 + y2 = v2
2xdx = 2udu and 2ydy = 2vdv
$$ \Rightarrow $$ xdx = udu and ydy = vdv
Substitute these values in equation (1)
$$\int {{{vdv} \over v}} = - \int {{{{u^2}du} \over {{u^2} - 1}}} $$
$$ \Rightarrow $$ $$\int {dv} = - \int {{{{u^2} - 1 + 1} \over {{u^2} - 1}}} du$$
$$ \Rightarrow $$ v = $$ - \int {\left( {1 + {1 \over {{u^2} - 1}}} \right)} du$$
$$ \Rightarrow $$ v = -u - $${1 \over 2}{\log _e}\left| {{{u - 1} \over {u + 1}}} \right|$$ + C
$$ \Rightarrow $$ $$\sqrt {1 + {y^2}} $$ = $$ - \sqrt {1 + {x^2}} $$ + $${1 \over 2}{\log _e}\left| {{{\sqrt {1 + {x^2}} + 1} \over {\sqrt {1 + {x^2}} - 1}}} \right|$$ + C
$$ \Rightarrow $$ $$\sqrt {1 + {y^2}} + \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} + 1} \over {\sqrt {1 + {x^2}} - 1}}} \right) + C$$
$$ \Rightarrow $$ $$\sqrt {\left( {1 + {x^2}} \right)\left( {1 + {y^2}} \right)} $$ + xy$${{dy} \over {dx}}$$ = 0
$$ \Rightarrow $$ $$\sqrt {\left( {1 + {x^2}} \right)} \sqrt {\left( {1 + {y^2}} \right)} $$ = -xy$${{dy} \over {dx}}$$
$$ \Rightarrow $$ $$\int {{{ydy} \over {\sqrt {1 + {y^2}} }}} = - \int {{{\sqrt {1 + {x^2}} } \over x}} dx$$ .....(1)
Now put 1 + x2 = u2 and 1 + y2 = v2
2xdx = 2udu and 2ydy = 2vdv
$$ \Rightarrow $$ xdx = udu and ydy = vdv
Substitute these values in equation (1)
$$\int {{{vdv} \over v}} = - \int {{{{u^2}du} \over {{u^2} - 1}}} $$
$$ \Rightarrow $$ $$\int {dv} = - \int {{{{u^2} - 1 + 1} \over {{u^2} - 1}}} du$$
$$ \Rightarrow $$ v = $$ - \int {\left( {1 + {1 \over {{u^2} - 1}}} \right)} du$$
$$ \Rightarrow $$ v = -u - $${1 \over 2}{\log _e}\left| {{{u - 1} \over {u + 1}}} \right|$$ + C
$$ \Rightarrow $$ $$\sqrt {1 + {y^2}} $$ = $$ - \sqrt {1 + {x^2}} $$ + $${1 \over 2}{\log _e}\left| {{{\sqrt {1 + {x^2}} + 1} \over {\sqrt {1 + {x^2}} - 1}}} \right|$$ + C
$$ \Rightarrow $$ $$\sqrt {1 + {y^2}} + \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} + 1} \over {\sqrt {1 + {x^2}} - 1}}} \right) + C$$
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