If [y = ( 2 x - 1 ) cosec,x] is the solution of the differential equation, [dy dx + p( x )…

Question

If $$y = \left( {{2 \over \pi }x - 1} \right) cosec\,x$$  is the solution of the
differential equation,

$${{dy} \over {dx}} + p\left( x \right)y = {2 \over \pi } cosec\,x$$,

$$0 < x < {\pi  \over 2}$$,  then the function p(x) is equal to :

Accepted Answer

$$y = \left( {{2 \over \pi }x - 1} \right) cosec\,x$$

Differentiate w.r.t x

$${{dy} \over {dx}} = {2 \over \pi }$$cosec x - $$\left( {{2 \over \pi }x - 1} \right)$$cosec x.cot x

$$ \Rightarrow $$ $${{dy} \over {dx}}$$ + $$\left( {{2 \over \pi }x - 1} \right)$$cosec x.cot x = $${2 \over \pi }$$cosec x

$$ \Rightarrow $$ $${{dy} \over {dx}}$$ + ycot x = $${2 \over \pi }$$cosec x

Compare this differential equation with given differential equation, we get

p(x) = cot x

Solutions · 1