Let's solve the problem step-by-step to show that [tex]\( 2x \frac{dy}{dx} + y = 2\sqrt{x} \)[/tex] given [tex]\( y = \sqrt{x} - \frac{1}{\sqrt{x}} \)[/tex].
1. Define [tex]\( y \)[/tex]:
[tex]\( y = \sqrt{x} - \frac{1}{\sqrt{x}} \)[/tex]
2. Find the derivative [tex]\(\frac{dy}{dx}\)[/tex]:
To find [tex]\(\frac{dy}{dx}\)[/tex], we differentiate [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex].
[tex]\[
y = x^{1/2} - x^{-1/2}
\][/tex]
Using the power rule [tex]\( \frac{d}{dx} [x^n] = nx^{n-1} \)[/tex],
[tex]\[
\frac{dy}{dx} = \frac{d}{dx} [x^{1/2}] - \frac{d}{dx} [x^{-1/2}]
= \frac{1}{2} x^{-\frac{1}{2}} + \frac{1}{2} x^{-\frac{3}{2}}
= \frac{1}{2\sqrt{x}} + \frac{1}{2x^{3/2}}
\][/tex]
3. Compute [tex]\( 2x \frac{dy}{dx} \)[/tex]:
Now, we multiply the derivative by [tex]\( 2x \)[/tex].
[tex]\[
2x \frac{dy}{dx} = 2x \left( \frac{1}{2\sqrt{x}} + \frac{1}{2x^{3/2}} \right)
\][/tex]
Simplify each term separately:
[tex]\[
2x \cdot \frac{1}{2\sqrt{x}} = x \cdot \frac{1}{\sqrt{x}} = \sqrt{x}
\][/tex]
[tex]\[
2x \cdot \frac{1}{2x^{3/2}} = x \cdot \frac{1}{x^{3/2}} = x \cdot x^{-3/2} = x^{-1/2} = \frac{1}{\sqrt{x}}
\][/tex]
Combine these simplified terms:
[tex]\[
2x \frac{dy}{dx} = \sqrt{x} + \frac{1}{\sqrt{x}}
\][/tex]
4. Add [tex]\( y \)[/tex] to [tex]\( 2x \frac{dy}{dx} \)[/tex]:
We have [tex]\( y \)[/tex] and [tex]\( 2x \frac{dy}{dx} \)[/tex]. Let's add them together:
[tex]\[
2x \frac{dy}{dx} + y = \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right) + \left( \sqrt{x} - \frac{1}{\sqrt{x}} \right)
\][/tex]
Combine the like terms:
[tex]\[
2x \frac{dy}{dx} + y = \sqrt{x} + \sqrt{x} = 2\sqrt{x}
\][/tex]
Hence, we have shown that:
[tex]\[
2x \frac{dy}{dx} + y = 2\sqrt{x}
\][/tex]
This concludes the proof.
