1∫√(cosθ) dθ = ????????
If y is a fn. of x such that y(x-y)2=x , then
Find ∫dx/(x-3y)
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22 Answers
1look 4 some substitutn.......
that seems 2 be only way out....... as the functn is defined implicitly........
one mite not b able 2 make y the subj of eqtn
11Another way of proceeding......This can be done only if RHS is known
(x-y)2 = x/y
2 log (x-y) = log x - log y
Differentiating,
2{1/(x-y)}.{1-y'} = [(1/x) - y'/(y)]
2/(x-y) - 2y'/(x-y) = 1/x - y'/y
y'[1/y - 2/(x-y)] = [1/x - 2/(x-y)]
y' = - (y/x) . {(x+y)/(x-3y)}
RHS = (1/2) log [(x-y)2 - 1]
= (1/2) log [(x/y) - 1] since (x-y)2=x/y
= (1/2) log [(x-y)/y]
= (1/2) log (x-y) - (1/2) log y
d(RHS)/dx = (1/2) [ {(1-y')/(x-y)} - y'/y ]
= (1/2) [ {1/(x-y)} - y'. { x/(y(x-y)) } ]
= (1/2) [ {1/(x-y)} + (y/x) . {(x+y)/(x-3y)} . {x/(y(x-y))} ]
= (1/2) [ {1/(x-y)} + (x+y)/{(x-3y)(x-y)}]
= (1/2) [ {(x-3y) + (x+y)}/{(x-3y)(x-y)} ]
= 1/(x-3y)
Therefore,
∫ dx/(x-3y) = RHS = (1/2) log [(x-y)2 - 1] [1][1][1]
21boy o boy!!!!
sir, itnaaaa bada solution.. [5][5]
can this type b ther in JEE (i know ur ans wud be no.... but 4 confirming)
62differentiate..
y(x-y)2=x
(x-y)2 dy/dx + y2(x-y)(1-dy/dx) = 1
thus dy/dx(x2+y2-4xy+2y2) = 1+2y2-2xy
thus,
dy/dx(x2+y2-2xy - 2xy +2y2) = 1+2y2-2xy
dy/dx(x - y)2 +2y(y-x)dy/dx = 1+2y(y-x)
thus, dy/dx = (-2xy+2y2+1)/(x2-4xy+3y2)
thus, 1- dy/dx =1-(-2xy+2y2+1)/(x2-4xy+3y2)
thus, 1- dy/dx = {(x-y)2-1}/{(x-3y)(x-y)}
thus, (1- dy/dx)(x-y) = 1/{(x-y)2-1} . {(x-3y)
thus, (1- dy/dx)(x-y)/{(x-y)2-1} = 1/(x-3y)
LHS is derivative of log{(x-y)2-1}
now integrate.. wrt x
we get the final answer
(I did tihs by back calculation :P)
I dont see how a sane person could have done it without that :P
21cud ne1 manage it now,
I havnt given it a whole hearted try yet!!!
21rk is the questn this Prove : ∫ dx/(x-3y) = (1/2) [log (x-y)2 - 1]
or
Prove : ∫ dx/(x-3y) = (1/2) [log {(x-y)2 - 1}]???
11yeah..even i feel integrating the given fn is a bit difficult.
well...the actual ques is not this!!!
It is actually :-
If y(x-y)2 = x , then
Prove : ∫ dx/(x-3y) = (1/2) log [(x-y)2 - 1]
which is simpler than integrating...!!
Try this...and i'll post the soln. tomorrow.
21Looks like road block ahead........
And I m not gettiin how wud partial derivative help [7]
1hey got something... dun ask y i did it ...:P
partial diff of y kping x constant:
y'(x-y)(x-3y) =0
partial diff of x kping y const :
x'[2y(x-y) - 1] = 0
can we do somehtning from here ?? [12]
processing......[12] [12] [12] [12] [12] [12]
21not gettin u exactly unique!!!
i got ∂y/∂x = (1 + y^2 -2xy ) / (x^2 - 4xy)
wat 2 do next?
1tapan, have the equation's derivative n find the integration of the equation on the other side of dx/(x-3y)
i.e; ∫(x-y)dy
but constants can vary ..but i dont think it matters !
21answer to v can know only once rkrish logs in and posts da ams..........
can u tell how did u simplify the integrand??
1its a good one !!
is it xy-y2/2 + c ?
if not , then simplifying it may get u there
11a really long method
expanding the equation and solving the quadratic in x
we get x in terms of y then it becomes easy but not an elegant expression we get
1can we try using some parametric eqns? x/y = t may work I think .asking I dunno .whether right or wrong.
