JEE MAIN - Mathematics (2021 - 26th August Evening Shift - No. 5)
Let y(x) be the solution of the differential equation
2x2 dy + (ey $$-$$ 2x)dx = 0, x > 0. If y(e) = 1, then y(1) is equal to :
2x2 dy + (ey $$-$$ 2x)dx = 0, x > 0. If y(e) = 1, then y(1) is equal to :
0
2
loge 2
loge (2e)
Explanation
$$2{x^2}dy + ({e^y} - 2x)dx = 0$$
$${{dy} \over {dx}} + {{{e^y} - 2x} \over {2{x^2}}} = 0 \Rightarrow {{dy} \over {dx}} + {{{e^y}} \over {2{x^2}}} - {1 \over x} = 0$$
$${e^{ - y}}{{dy} \over {dx}} - {{{e^{ - y}}} \over x} = - {1 \over {2{x^2}}} \Rightarrow $$ Put $${e^{ - y}} = z$$
$${{ - dz} \over {dx}} - {z \over x} = - {1 \over {2{x^2}}} \Rightarrow xdz + zdx = {{dx} \over {2x}}$$
$$d(xz) = {{dx} \over {2x}} \Rightarrow xz = {1 \over 2}{\log _e}x + c$$
$$x{e^{ - y}} = {1 \over 2}{\log _e}x + c$$, passes through (e, 1)
$$ \Rightarrow C = {1 \over 2}$$
$$x{e^{ - y}} = {{{{\log }_e}ex} \over 2}$$
$${e^{ - y}} = {1 \over 2} \Rightarrow y = {\log _e}2$$
$${{dy} \over {dx}} + {{{e^y} - 2x} \over {2{x^2}}} = 0 \Rightarrow {{dy} \over {dx}} + {{{e^y}} \over {2{x^2}}} - {1 \over x} = 0$$
$${e^{ - y}}{{dy} \over {dx}} - {{{e^{ - y}}} \over x} = - {1 \over {2{x^2}}} \Rightarrow $$ Put $${e^{ - y}} = z$$
$${{ - dz} \over {dx}} - {z \over x} = - {1 \over {2{x^2}}} \Rightarrow xdz + zdx = {{dx} \over {2x}}$$
$$d(xz) = {{dx} \over {2x}} \Rightarrow xz = {1 \over 2}{\log _e}x + c$$
$$x{e^{ - y}} = {1 \over 2}{\log _e}x + c$$, passes through (e, 1)
$$ \Rightarrow C = {1 \over 2}$$
$$x{e^{ - y}} = {{{{\log }_e}ex} \over 2}$$
$${e^{ - y}} = {1 \over 2} \Rightarrow y = {\log _e}2$$
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