Solution of the differential equation [ x,dy = y( x - y )dx,,,0

Question

Solution of the differential equation

$$\cos x\,dy = y\left( {\sin x - y} \right)dx,\,\,0 < x <{\pi  \over 2}$$ is :

Accepted Answer

$$\cos xdy = y\left( {\sin x - y} \right)dx$$

$${{dy} \over {dx}} = y\tan x - {y^2}\,\,\sec \,x$$

$${1 \over {{y^2}}}{{dy} \over {dx}} - {1 \over y}\,\tan \,x =  - \sec \,x\,....\left( i \right)$$

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 $:$ $$\,\,t\left( {{\rm I}.F} \right) = \int {\left( {{\rm I}.F} \right)\sec xdx} $$

$$ \Rightarrow {1 \over y}\sec x = \tan x + c$$

Solutions · 1