show that

Question:
If $y=\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)$, show that $2 x \cdot \frac{d y}{d x}+y=2 \sqrt{x}$

Solution:

To show:

$2 x \cdot \frac{d y}{d x}+y=2 \sqrt{x}$

Differentiating with respect to $\mathrm{x}$

$\frac{d y}{d x}=\frac{d}{d x}\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)=\frac{1}{2 \sqrt{x}}-\frac{1}{2 x^{\frac{3}{2}}}$

Now,

$\mathrm{LHS}=2 \mathrm{x} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}$

$L H S=2 x \times\left(\frac{1}{2 \sqrt{x}}-\frac{1}{2 x^{\frac{3}{2}}}\right)+\sqrt{x}+\frac{1}{\sqrt{x}}$

$L H S=\sqrt{x}-\frac{1}{\sqrt{x}}+\sqrt{x}+\frac{1}{\sqrt{x}}$

$L H S=2 \sqrt{x}$

$\therefore \mathrm{LHS}=\mathrm{RHS}$