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diff equations
Posted:
Tue Mar 11, 2008 10:39 pmby pkciit
sin-1dy/dx =x+y
it's sin inverse don't get confused
plz give sol to the above
Re: diff equations
Posted:
Wed Mar 12, 2008 1:02 amby kauser
sin-1dy/dx=x+y
dy/dx=sin(x+y)
dy=(sinxcosy+sinycosx)dx
integrating both sides
y= /sinxcosydx +/sinycosxdx
y= -cosycosx +sinysinx + C
/ is the integral sign
Re: diff equations
Posted:
Wed Mar 12, 2008 4:10 pmby kauser
the previous method is applicable if the question had been sin-1 (dy/dx)=x+a
(replacing all ys on the RHS by a)
Re: diff equations
Posted:
Wed Mar 12, 2008 4:14 pmby kauser
sin-1 (dy/dx)= x+y
dy/dx= sin(x+y).........1
z = x+y
dz/dx = 1+ dy/dx
dy/dx = dz/dx- 1
from 1
sinz = dz/dx-1
sinz+1= dz/dx
dz/(sinz+1) = dx
integrating both sides
..............
tanz - sec z = x+c
tan(x+y) - sec(x+y) = x+ c is the rqrd diff eqn
Re: diff equations
Posted:
Wed Mar 12, 2008 9:19 pmby pkciit
i had solved all the que's of diff equ but i was really not able to solve this
so thanks a lot buddy !!!!!!!!!!!!!!
Re: diff equations
Posted:
Wed Mar 12, 2008 11:21 pmby kauser
can any body pls give me the solutions.....atleast the starting for miscellaneous diff eqns Q.11, 12, 13, 14