sin-1dy/dx =x+y
it's sin inverse don't get confused
plz give sol to the above
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diff equations
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diff equationssin-1dy/dx =x+y
it's sin inverse don't get confused plz give sol to the above
Re: diff equationssin-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 equationsthe 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 equationssin-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 equationsi 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 equationscan any body pls give me the solutions.....atleast the starting for miscellaneous diff eqns Q.11, 12, 13, 14
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