JEE MAIN - Mathematics (2003 - No. 22)
The solution of the differential equation
$$\left( {1 + {y^2}} \right) + \left( {x - {e^{{{\tan }^{ - 1}}y}}} \right){{dy} \over {dx}} = 0,$$ is :
$$\left( {1 + {y^2}} \right) + \left( {x - {e^{{{\tan }^{ - 1}}y}}} \right){{dy} \over {dx}} = 0,$$ is :
$$x{e^{2{{\tan }^{ - 1}}y}} = {e^{{{\tan }^{ - 1}}y}} + k$$
$$\left( {x - 2} \right) = k{e^{2{{\tan }^{ - 1}}y}}$$
$$2x{e^{{{\tan }^{ - 1}}y}} = {e^{2{{\tan }^{ - 1}}y}} + k$$
$$x{e^{{{\tan }^{ - 1}}y}} = {\tan ^{ - 1}}y + k$$
Explanation
$$\left( {1 + {y^2}} \right) + \left( {x - {e^{{{\tan }^{ - 1}}y}}} \right){{dy} \over {dx}} = 0$$
$$ \Rightarrow {{dx} \over {dy}} + {x \over {\left( {1 + {y^2}} \right)}} = {{{e^{{{\tan }^{ - 1}}y}}} \over {\left( {1 + {y^2}} \right)}}$$
$$I.F = e{}^{\int {{1 \over {\left( {1 + {y^2}} \right)}}dy} } = {e^{{{\tan }^{ - 1}}y}}$$
$$x\left( {{e^{{{\tan }^{ - 1}}y}}} \right) = \int {{{{e^{{{\tan }^{ - 1}}y}}} \over {1 + {y^2}}}} {e^{{{\tan }^{ - 1}}y}}\,dy$$
$$x\left( {{e^{{{\tan }^{ - 1}}y}}} \right) = {{{e^{2{{\tan }^{ - 1}}y}}} \over 2} + C$$
$$\therefore$$ $$\,\,\,\,\,2x{e^{{{\tan }^{ - 1}}y}} = {e^{2{{\tan }^{ - 1}}y}} + k$$
$$ \Rightarrow {{dx} \over {dy}} + {x \over {\left( {1 + {y^2}} \right)}} = {{{e^{{{\tan }^{ - 1}}y}}} \over {\left( {1 + {y^2}} \right)}}$$
$$I.F = e{}^{\int {{1 \over {\left( {1 + {y^2}} \right)}}dy} } = {e^{{{\tan }^{ - 1}}y}}$$
$$x\left( {{e^{{{\tan }^{ - 1}}y}}} \right) = \int {{{{e^{{{\tan }^{ - 1}}y}}} \over {1 + {y^2}}}} {e^{{{\tan }^{ - 1}}y}}\,dy$$
$$x\left( {{e^{{{\tan }^{ - 1}}y}}} \right) = {{{e^{2{{\tan }^{ - 1}}y}}} \over 2} + C$$
$$\therefore$$ $$\,\,\,\,\,2x{e^{{{\tan }^{ - 1}}y}} = {e^{2{{\tan }^{ - 1}}y}} + k$$
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