JEE MAIN - Mathematics (2019 - 11th January Morning Slot - No. 5)
If y(x) is the solution of the differential equation $${{dy} \over {dx}} + \left( {{{2x + 1} \over x}} \right)y = {e^{ - 2x}},\,\,x > 0,\,$$ where $$y\left( 1 \right) = {1 \over 2}{e^{ - 2}},$$ then
y(loge2) = loge4
y(x) is decreasing in (0, 1)
y(loge2) = $${{{{\log }_e}2} \over 4}$$
y(x) is decreasing in $$\left( {{1 \over 2},1} \right)$$
Explanation
$${{dy} \over {dx}} + \left( {{{2x + 1} \over x}} \right)y = {e^{ - 2x}}$$
I.F. $$ = {e^{\int {\left( {{{2x + 1} \over x}} \right)dx} }} = {e^{\int {\left( {2 + {1 \over x}} \right)dx} }} = {e^{2x + \ell nx}} = {e^{2x}}.x$$
So, $$y\left( {x{e^{2x}}} \right) = \int {{e^{ - 2x}}.x{e^{2x}} + C} $$
$$ \Rightarrow xy{e^{2x}} = \int {xdx + C} $$
$$ \Rightarrow 2xy{e^{2x}} = {x^2} + 2C$$
It passes through $$\left( {1,{1 \over 2}{e^{ - 2}}} \right)$$ we get C $$=$$ 0
$$y = {{x{e^{ - 2x}}} \over 2}$$
$$ \Rightarrow {{dy} \over {dx}} = {1 \over 2}{e^{ - 2x}}\left( { - 2x + 1} \right)$$
$$ \Rightarrow f(x)$$ is decreasing in $$\left( {{1 \over 2},1} \right)$$
$$y\left( {{{\log }_e}2} \right) = {{\left( {{{\log }_e}2} \right){e^{ - 2({{\log }_e}2)}}} \over 2}$$
$$ = {1 \over 8}{\log _e}2$$
I.F. $$ = {e^{\int {\left( {{{2x + 1} \over x}} \right)dx} }} = {e^{\int {\left( {2 + {1 \over x}} \right)dx} }} = {e^{2x + \ell nx}} = {e^{2x}}.x$$
So, $$y\left( {x{e^{2x}}} \right) = \int {{e^{ - 2x}}.x{e^{2x}} + C} $$
$$ \Rightarrow xy{e^{2x}} = \int {xdx + C} $$
$$ \Rightarrow 2xy{e^{2x}} = {x^2} + 2C$$
It passes through $$\left( {1,{1 \over 2}{e^{ - 2}}} \right)$$ we get C $$=$$ 0
$$y = {{x{e^{ - 2x}}} \over 2}$$
$$ \Rightarrow {{dy} \over {dx}} = {1 \over 2}{e^{ - 2x}}\left( { - 2x + 1} \right)$$
$$ \Rightarrow f(x)$$ is decreasing in $$\left( {{1 \over 2},1} \right)$$
$$y\left( {{{\log }_e}2} \right) = {{\left( {{{\log }_e}2} \right){e^{ - 2({{\log }_e}2)}}} \over 2}$$
$$ = {1 \over 8}{\log _e}2$$
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