If , y=sqrt(x/a)-sqrt(a/x)y=xaax So, Prove that ?? 2xy(dy/dx)= x/a-a/x2xy(dydx)=xaax

If , y=sqrt(x/a)-sqrt(a/x)y=xaax So, Prove that ??
2xydy/dx= x/a-a/x2xydydx=xaax

3 Answers
Jun 17, 2018

Given: y=sqrt(x/a)-sqrt(a/x)y=xaax

Differentiating:

dy/dx = (sqrt(a/x) + sqrt(x/a))/(2 x)dydx=ax+xa2x

Multiply both sides by2xy:

2xydy/dx = 2xy(sqrt(a/x) + sqrt(x/a))/(2 x)2xydydx=2xyax+xa2x

(2x)/(2x)2x2x becomes 1:

2xydy/dx = y(sqrt(a/x) + sqrt(x/a))2xydydx=y(ax+xa)

Substitute y=sqrt(x/a)-sqrt(a/x)y=xaax on the right side only:

2xydy/dx = (sqrt(x/a)-sqrt(a/x))(sqrt(a/x) + sqrt(x/a))2xydydx=(xaax)(ax+xa)

Please observe that the right side is the pattern (a - b)(a+b) = a^2-b^2(ab)(a+b)=a2b2 where a = sqrt(x/a)a=xa and b = sqrt(a/x)b=ax

2xydy/dx = x/a-a/x" Q.E.D."2xydydx=xaax Q.E.D.

Jun 17, 2018

Kindly see a Proof in Explanation.

Explanation:

We have, y=sqrt(x/a)-sqrt(a/x)y=xaax.

Diff.ing w.r.t. x, dy/dx=1/sqrta*(1/2x^(-1/2))-sqrta*(-1/2x^(-3/2))x,dydx=1a(12x12)a(12x32).

:. dy/dx=1/(2sqrt(ax))+sqrta/(2xsqrtx).

Multiplying by 2x, 2xdy/dx=sqrt(x/a)+sqrt(a/x).

Again multiplying by y, 2xydy/dx=y(sqrt(x/a)+sqrt(a/x)),

i.e., 2xydy/dx=(sqrt(x/a)-sqrt(a/x))(sqrt(x/a)+sqrt(a/x)).

rArr 2xydy/dx=x/a-a/x, as desired!

Jun 17, 2018

We seek to show that:

2xydy/dx= x/a-a/x where y=sqrt(x/a)-sqrt(a/x)

Squaring the expression, and then expanding, we get:

y^2 = (sqrt(x/a)-sqrt(a/x))^2
\ \ \ \= (sqrt(x/a))^2 - 2(sqrt(x/a))(sqrt(a/x)) + (sqrt(a/x))^2
\ \ \ \= x/a - 2 + a/x

Differentiating Implicitly, we get

2y dy/dx = 1/a -a/x^2

Finally, multiplying by x we get the desired result:

2xy dy/dx = x/a -a/x \ \ \ QED