If, y = (√x)+(1/√x) so prove that ? 2x(dy/dx)+y = 2√x

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Answer 1 · 2018-06-17T17:26:32

$y+2x(dy/dx)= 2sqrt(x)$

$y = sqrt(x) + 1 /sqrt(x)$

$sqrt(x).y = x +1$

$(1/(2*sqrt(x))).y+sqrt(x).dy/dx = 1$

$y=2.sqrt(x)(1-sqrt(x).dy/dx)$

$y=2.sqrt(x) - 2x .dy/dx$

$y + 2x .(dy/dx) = 2.sqrt(x)$

Answer 2 · 2018-06-19T12:07:14

Please see a Proof in the Explanation.

Given that, $y=sqrtx+1/sqrtx=x^(1/2)+x^(-1/2)$ .

Recall that, $d/dx(x^n)=n*x^(n-1)$ .

$:. dy/dx=1/2*x^(1/2-1)+(-1/2)*x^(-1/2-1)$ .

$=1/2*x^(-1/2)-1/2*x^(-3/2)$ ,

$:. dy/dx=1/2{x^(-1/2)-x^(-3/2)}$ .

Multiplying this eqn. by $2x$ , we get,

$2xdy/dx=cancel(2)x*1/cancel(2){x^(-1/2)-x^(-3/2)}$ ,

$=x*x^(-1/2)-x*x^(-3/2)$ ,

$i.e., 2xdy/dx=x^(1/2)-x^(-1/2)=sqrtx-1/sqrtx$ .

Finally, adding $y=sqrtx+1/sqrtx$ , we have,

$2xdy/dx+y=(sqrtxcancel(-1/sqrtx))+(sqrtxcancel(+1/sqrtx))$ ,

$or, 2xdy/dx+y=2sqrtx$ ,

as Respected Abhishek Malviya has readily derived!