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Vikas TU
14149 Points
4 years ago
we realize that on the correct side of the condition we have a which is a consistent subsequently its subordinate is zero. Taking first subordinate we have 1/2√(y+x) + 1/2√(y-x)=0. Presently again taking subordinate we utilize remainder run and get 1/(2√(y+x)(y+x))+ 1/(2√(y-x)(y-x))=0.
somi teez
105 Points
4 years ago
Hi vikas,
i too solved in that way....but
you have to bring it in terms of a...answer is 2/a^2...and when you differentiate  1/2√(y+x) + 1/2√(y-x)=0 it is
1/4(y+x)^3/2 +1/4(y-x)^3/2=0
manogna
16 Points
4 years ago
Q: √y+x + √y-x=a
(Differentiate)
Step:1 1/2√y+x[dy/dx+1] + 1/2√y-x[dy/dx-1]=0
Step:2  dy/dx[1/√y+x +1/√y-x] + [1/√y+x – 1/√y-x]=0
Step:3 dy/dx [√y-x + √y-x /√y+x.√y-x]=1/√y-x -1/√y+x
(then by substituting ‘a’ from the question we get)
Step:4 dy/dx(a) =√y+x -√y-x
TO FIND THE REQUIRED SOLUTION
(Mutiply ‘a’ on bothsides in step:4)
A2.dy/dx =[√y+x - √y-x].[√y+x +√y-x]
A2dy/dx = 2x
Therefore,
A2d2y/dx2 =2
I think this would be helpful (manogna chandrika)