Question

find d^2y/dx^2,if √x+y +√y-x=c if any one think that he is much intelligent then solve it
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Answer 1

sqrt(x+y) +sqrt(y-x) = c

squaring on both sides,  

x+y + y-x + 2* sqrt((x+y)*(y-x)) = c2

x+y + y-x + 2* sqrt(y2  – x2) =  c2

2*sqrt(y2  – x2) =  c2 – 2y

squaring on both sides,  

4(y2  – x2) = c4 + 4y2 – 4c2y

-4x2 = c4  – 4c2y

y = (c2/4)+ (x2)/c2

dy/dx = (2x)/c2

d2y/dx2 = 2/c2

Answer 2

Therefore the value of d²y/dx² is { ( 1/4 ) [   $x^-^3^/^2$ + dy/dx.( $y^-^3^/^2$) ] } / { [ 1 + ( 1/2 ) dy/dx ] }.

Given:

The equation: √x + y + √y - x = c

To Find:

The value of d²y / d²x.

Solution:

The given question can be easily solved as shown below.

Given equation: √x + y + √y - x = c

Differentiating with respect to 'x' on both sides,

⇒ d ( √x + y + √y - x ) / dx = dc / dx

⇒ d√x / dx + dy/dx + d√y/dx - dx/dx= dc/dx

[  d√x / dx = ( 1/2 ) $x^-^1^/^2$;

d√y/dx = ( 1/2 ) $y^-^1^/^2$ dy/dx;

dx/dx = 1;

dc/dx =0 ]

⇒ ( 1/2 ) $x^-^1^/^2$ + dy/dx + ( 1/2 ) $y^-^1^/^2$ dy/dx - 1 = 0

⇒ ( 1/2 ) $x^-^1^/^2$ + dy/dx + ( 1/2 ) $y^-^1^/^2$ dy/dx = 1

Now again differentiating with respect to 'x' in both sides,

⇒ d [( 1/2 ) $x^-^1^/^2$ + dy/dx + ( 1/2 ) $y^-^1^/^2$ dy/dx ] / dx = d1/dx

⇒ d( 1/2 ) $x^-^1^/^2$ /dx + d (dy/dx) / dx + d(( 1/2 ) $y^-^1^/^2$ dy/dx)/ dx = d1/dx

⇒ ( 1/2 ) d $x^-^1^/^2$ /dx + d (dy/dx) / dx + ( 1/2 ) d( $y^-^1^/^2$ dy/dx)/ dx = d1/dx

[ d $x^-^1^/^2$ /dx = -$x^-^3^/^2$/2;  

d( $y^-^1^/^2$ dy/dx)/ dx = dy/dx ( d $y^-^1^/^2$/dx ) + $y^-^1^/^2$ ( d ( dy/dx ) / dx = dy/dx.( -$y^-^3^/^2$/2 ) + dy/dx. d²y/dx²;

d (dy/dx) / dx = d²y/dx²;

d1/dx =0 ]

⇒ ( 1/2 ) -$x^-^3^/^2$/2 + d²y/dx² + ( 1/2 ) dy/dx.( -$y^-^3^/^2$/2 ) + ( 1/2 ) dy/dx. d²y/dx² = 0

Rearranging the terms,

⇒ d²y/dx² + ( 1/2 ) dy/dx. d²y/dx² + ( 1/2 ) -$x^-^3^/^2$/2 + ( 1/2 ) dy/dx.( -$y^-^3^/^2$/2 ) = 0

⇒ d²y/dx² [ 1 + ( 1/2 ) dy/dx ] - ( 1/4 ) [   $x^-^3^/^2$ + dy/dx.( $y^-^3^/^2$) ] = 0

⇒ d²y/dx² [ 1 + ( 1/2 ) dy/dx ] = ( 1/4 ) [   $x^-^3^/^2$ + dy/dx.( $y^-^3^/^2$) ]

⇒ d²y/dx² = { ( 1/4 ) [   $x^-^3^/^2$ + dy/dx.( $y^-^3^/^2$) ] } / { [ 1 + ( 1/2 ) dy/dx ] }

Therefore the value of d²y/dx² is { ( 1/4 ) [   $x^-^3^/^2$ + dy/dx.( $y^-^3^/^2$) ] } / { [ 1 + ( 1/2 ) dy/dx ] }.

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