JEE MAIN - Mathematics (2010 - No. 7)
Solution of the differential equation
$$\cos x\,dy = y\left( {\sin x - y} \right)dx,\,\,0 < x <{\pi \over 2}$$ is :
$$\cos x\,dy = y\left( {\sin x - y} \right)dx,\,\,0 < x <{\pi \over 2}$$ is :
$$y\sec x = \tan x + c$$
$$y\tan x = \sec x + c$$
$$\tan x = \left( {\sec x + c} \right)y$$
$$\sec x = \left( {\tan x + c} \right)y$$
Explanation
$$\cos xdy = y\left( {\sin x - y} \right)dx$$
$${{dy} \over {dx}} = y\tan x - {y^2}\,\,\sec \,x$$
$${1 \over {{y^2}}}{{dy} \over {dx}} - {1 \over y}\,\tan \,x = - \sec \,x\,....\left( i \right)$$
Let $$\,\,\,\,{1 \over y} = t \Rightarrow - {1 \over {{y^2}}}{{dy} \over {dx}} = {{dt} \over {dx}}$$
From equation $$(i)$$
$$ - {{dt} \over {dx}} - \tan x = - \sec x$$
$$ \Rightarrow {{dt} \over {dx}} + \left( {\tan x} \right)t = \sec x$$
$${\rm I}.F. = {e^{\int {\tan xdx} }} = {\left( e \right)^{\log \left| {\sec x} \right|}}\sec x$$
Solution $$:$$ $$\,\,t\left( {{\rm I}.F} \right) = \int {\left( {{\rm I}.F} \right)\sec xdx} $$
$$ \Rightarrow {1 \over y}\sec x = \tan x + c$$
$${{dy} \over {dx}} = y\tan x - {y^2}\,\,\sec \,x$$
$${1 \over {{y^2}}}{{dy} \over {dx}} - {1 \over y}\,\tan \,x = - \sec \,x\,....\left( i \right)$$
Let $$\,\,\,\,{1 \over y} = t \Rightarrow - {1 \over {{y^2}}}{{dy} \over {dx}} = {{dt} \over {dx}}$$
From equation $$(i)$$
$$ - {{dt} \over {dx}} - \tan x = - \sec x$$
$$ \Rightarrow {{dt} \over {dx}} + \left( {\tan x} \right)t = \sec x$$
$${\rm I}.F. = {e^{\int {\tan xdx} }} = {\left( e \right)^{\log \left| {\sec x} \right|}}\sec x$$
Solution $$:$$ $$\,\,t\left( {{\rm I}.F} \right) = \int {\left( {{\rm I}.F} \right)\sec xdx} $$
$$ \Rightarrow {1 \over y}\sec x = \tan x + c$$
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