IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

tejasvi2389

p+ Cospx Siny = SinpxCosy p=dy/dx then what is/are the soln of this diff. equation

rsharma.ranchi_000

dear,

first consider dy/dx = sinpx.cospy

i.e. secpy dy = sinpx dx

integrating both sides

we get,

log(secpy+tanpy)/p = -cospx/p

or log(secpy+tanpy)=-cospx

now you take dy/dx=p+cospxsiny=p-(-cospx)(siny)

putting the value of (-cos x) in the above equation,

dy/dx=p-log(secpy+tanpy)(siny)

i.e. dy/(p-log(secy+tany)(siny))=dx

now integrate both sides

and get your answer.

if you will not be able to integrate i will solveit for you.

amar.gupta

dear tejasvi2389

please check it out and ask if doubt any.

tejasvi2389

understood