JEE MAIN - Mathematics (2021 - 1st September Evening Shift - No. 5)
If y = y(x) is the solution curve of the differential equation $${x^2}dy + \left( {y - {1 \over x}} \right)dx = 0$$ ; x > 0 and y(1) = 1, then $$y\left( {{1 \over 2}} \right)$$ is equal to :
$${3 \over 2} - {1 \over {\sqrt e }}$$
$$3 + {1 \over {\sqrt e }}$$
3 + e
3 $$-$$ e
Explanation
$${x^2}dy + \left( {y - {1 \over x}} \right)dx = 0$$ ; x > 0, y(1) = 1
$${x^2}dy + {{(xy - 1)} \over x}dx = 0$$
$${x^2}dy = {{(xy - 1)} \over x}dx$$
$${{dy} \over {dx}} = {{1 - xy} \over {{x^3}}}$$
$${{dy} \over {dx}} = {1 \over {{x^3}}} - {y \over {{x^2}}}$$
$${{dy} \over {dx}} = {1 \over {{x^2}}}.y = {1 \over {{x^3}}}$$
If $${e^{\int {{1 \over {{x^2}}}dx} }} = {e^{ - {1 \over x}}}$$
$$y{e^{ - {1 \over x}}} = \int {{1 \over {{x^3}}}.{e^{ - {1 \over x}}}} $$
$$y{e^{ - {1 \over x}}} = {e^{ - x}}\left( {1 + {1 \over x}} \right) + C$$
$$1.\,{e^{ - 1}} = {e^{ - 1}}(2) + C$$
$$C = - {e^{ - 1}} = - {1 \over e}$$
$$y{e^{ - {1 \over x}}} = {e^{ - {1 \over x}}}\left( {1 + {1 \over x}} \right) - {1 \over e}$$
$$y\left( {{1 \over 2}} \right) = 3 - {1 \over e} \times {e^2}$$
$$y\left( {{1 \over 2}} \right) = 3 - e$$
$${x^2}dy + {{(xy - 1)} \over x}dx = 0$$
$${x^2}dy = {{(xy - 1)} \over x}dx$$
$${{dy} \over {dx}} = {{1 - xy} \over {{x^3}}}$$
$${{dy} \over {dx}} = {1 \over {{x^3}}} - {y \over {{x^2}}}$$
$${{dy} \over {dx}} = {1 \over {{x^2}}}.y = {1 \over {{x^3}}}$$
If $${e^{\int {{1 \over {{x^2}}}dx} }} = {e^{ - {1 \over x}}}$$
$$y{e^{ - {1 \over x}}} = \int {{1 \over {{x^3}}}.{e^{ - {1 \over x}}}} $$
$$y{e^{ - {1 \over x}}} = {e^{ - x}}\left( {1 + {1 \over x}} \right) + C$$
$$1.\,{e^{ - 1}} = {e^{ - 1}}(2) + C$$
$$C = - {e^{ - 1}} = - {1 \over e}$$
$$y{e^{ - {1 \over x}}} = {e^{ - {1 \over x}}}\left( {1 + {1 \over x}} \right) - {1 \over e}$$
$$y\left( {{1 \over 2}} \right) = 3 - {1 \over e} \times {e^2}$$
$$y\left( {{1 \over 2}} \right) = 3 - e$$
Comments (0)
