JEE MAIN - Mathematics (2019 - 11th January Evening Slot - No. 16)
The solution of the differential equation,
$${{dy} \over {dx}}$$ = (x – y)2, when y(1) = 1, is :
$${{dy} \over {dx}}$$ = (x – y)2, when y(1) = 1, is :
$$-$$ loge $$\left| {{{1 + x - y} \over {1 - x + y}}} \right|$$ = x + y $$-$$ 2
loge $$\left| {{{2 - x} \over {2 - y}}} \right|$$ = x $$-$$ y
loge $$\left| {{{2 - y} \over {2 - x}}} \right|$$ = 2(y $$-$$ 1)
$$-$$ loge $$\left| {{{1 - x + y} \over {1 + x - y}}} \right|$$ = 2(x $$-$$ 1)
Explanation
x $$-$$ y = t
$$ \Rightarrow $$ $${{dy} \over {dx}} = 1 - {{dt} \over {dx}}$$
$$ \Rightarrow $$ 1 $$-$$ $${{dt} \over {dx}}$$ = t2 $$ \Rightarrow $$ $$\int {{{dt} \over {1 - {t^2}}}} $$ = $$\int {1dx} $$
$$ \Rightarrow $$ $${1 \over 2}\ell n\left( {{{1 + t} \over {1 - t}}} \right) = x + \lambda $$
$$ \Rightarrow $$ $${1 \over 2}\ell n\left( {{{1 + x - y} \over {1 - x + y}}} \right) = x + \lambda $$ given y(1) = 1
$$ \Rightarrow $$ $${1 \over 2}\ell n(1) = 1 + \lambda \Rightarrow \lambda = - 1$$
$$ \Rightarrow $$ $$\ell n\left( {{{1 + x - y} \over {1 - x + y}}} \right)$$ = 2(x $$-$$ 1)
$$ \Rightarrow $$ $$ - \ell n\left( {{{1 - x + y} \over {1 + x - y}}} \right)$$ = 2(x $$-$$ 1)
$$ \Rightarrow $$ $${{dy} \over {dx}} = 1 - {{dt} \over {dx}}$$
$$ \Rightarrow $$ 1 $$-$$ $${{dt} \over {dx}}$$ = t2 $$ \Rightarrow $$ $$\int {{{dt} \over {1 - {t^2}}}} $$ = $$\int {1dx} $$
$$ \Rightarrow $$ $${1 \over 2}\ell n\left( {{{1 + t} \over {1 - t}}} \right) = x + \lambda $$
$$ \Rightarrow $$ $${1 \over 2}\ell n\left( {{{1 + x - y} \over {1 - x + y}}} \right) = x + \lambda $$ given y(1) = 1
$$ \Rightarrow $$ $${1 \over 2}\ell n(1) = 1 + \lambda \Rightarrow \lambda = - 1$$
$$ \Rightarrow $$ $$\ell n\left( {{{1 + x - y} \over {1 - x + y}}} \right)$$ = 2(x $$-$$ 1)
$$ \Rightarrow $$ $$ - \ell n\left( {{{1 - x + y} \over {1 + x - y}}} \right)$$ = 2(x $$-$$ 1)
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