Question

Question #60b11

Answer 1

When `y=sqrt(x/m)+sqrt(m/x)`, we observe that `2xydy/dx-x/m+m/x=0`.

Explanation:

`y=sqrt(x/m)+sqrt(m/x)`

Then I assume you're asking, what's the expression `2xydy/dx-x/m+m/x` equivalent to?

Well, we need to find `dy/dx`. First let's rewrite `y`.

`y=m^(-1/2)x^(1/2)+m^(1/2)x^(-1/2)`

So now we can differentiate with the power rule:

`dy/dx=1/2m^(-1/2)x^(-1/2)-1/2m^(1/2)x^(-3/2)=1/(2sqrt(mx))-sqrtm/(2xsqrtx)`

So then we want to know about:

`2xydy/dx-x/m+m/x`

`=2x(sqrt(x/m)+sqrt(m/x))(1/(2sqrt(mx))-sqrtm/(2xsqrtx))-x/m+m/x`

FOIL the two binomials:

`=2x(sqrtx/sqrtm1/(2sqrtmsqrtx)-sqrtx/sqrtmsqrtm/(2xsqrtx)+sqrtm/sqrtx1/(2sqrtmsqrtx)-sqrtm/sqrtxsqrtm/(2xsqrtx))-x/m+m/x`

And simplify:

`=2x(1/(2m)-1/(2x)+1/(2x)-m/(2x^2))-x/m+m/x`

Distribute:

`=(x/m-m/x)-x/m+m/x`

`=0`

Cool!