Question

The general solution of the differential equation \(\frac{dy}{dx} = \frac{\sqrt{y}}{\sqrt{x}}\) is

• A. \(\log\sqrt{y} = \log\sqrt{x} + C\)
• B. \(\sqrt{y} + \sqrt{x} = C\)
• C. \(\sqrt{y} - \sqrt{x} = C\)
• D. \(\log\sqrt{y} + \log\sqrt{x} = C\)

Hint:

Separate the variables to get dy/sqrt(y) = dx/sqrt(x) and integrate both sides.

Answer 1

The Correct Option is C

Step 1: Identify the type:
The right side is a function of \(y\) divided by a function of \(x\).
So the variables can be separated.

Step 2: Separate the variables:
Multiply by \(dx\) and divide by \(\sqrt{y}\).
\[ \frac{dy}{\sqrt{y}} = \frac{dx}{\sqrt{x}} \]

Step 3: Integrate both sides:
We use \(\int t^{-1/2}\,dt = 2\sqrt{t}\).
\[ 2\sqrt{y} = 2\sqrt{x} + C_{1} \]
Divide by 2 and rename the constant \(\frac{C_1}{2}\) as \(C\).
\[ \sqrt{y} - \sqrt{x} = C \]

Step 4: Check the other options:
Option (A): differentiating \(\log\sqrt{y} = \log\sqrt{x} + C\) gives \(\frac{y^{\prime}}{2y} = \frac{1}{2x}\), so \(\frac{dy}{dx} = \frac{y}{x}\). This is a different equation.
Option (B): differentiating \(\sqrt{y} + \sqrt{x} = C\) gives \(\frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}}\). The sign is wrong.
Option (D): \(\log\sqrt{xy} = C\) means \(xy\) is constant, which gives \(\frac{dy}{dx} = -\frac{y}{x}\). This is also different.

Final Answer:
The general solution is \(\sqrt{y} - \sqrt{x} = C\), which is option (C). \[ \boxed{\sqrt{y} - \sqrt{x} = C} \]