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

Question

37733 views around the world

You can reuse this answer
Creative Commons License

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!