Question

The order and degree of the following differential equation:
\(y=px+\sqrt{a^2p^2+b^2}\), where \(p=\frac{dy}{dx}\), are

• A. Order = 2 : degree = 2
• B. Order = 1 : degree = 1
• C. Order = 1 : degree = 2
• D. Order = 2 : degree = 1

Hint:

Only \(\frac{dy}{dx}\) appears. Square to remove the root, then see the highest power of \(p\).

Answer 1

The Correct Option is C

Step 1: Recall the definitions.
The order of a differential equation is the order of the highest derivative in it. The degree is the power of that highest derivative after the equation is written as a polynomial in the derivatives (free of roots and fractions of derivatives).

Step 2: Find the order.
Here \(p=\frac{dy}{dx}\), so only the first derivative appears. No \(\frac{d^2y}{dx^2}\) is present. The order is 1.

Step 3: Remove the square root.
Move \(px\) to the left side and square both sides.
\[ y-px=\sqrt{a^2p^2+b^2} \] \[ (y-px)^2=a^2p^2+b^2 \]

Step 4: Write it as a polynomial in p.
Expand the left side and collect the powers of \(p\).
\[ y^2-2pxy+p^2x^2=a^2p^2+b^2 \] \[ (x^2-a^2)p^2-2xy\,p+(y^2-b^2)=0 \] The highest power of \(p=\frac{dy}{dx}\) is 2, and the coefficient \(x^2-a^2\) is not zero in general. So the degree is 2.

Step 5: Check the options.
Degree 1 (options 2 and 4) would be true only if the square root could be removed without raising a power of \(p\), which is not the case. Order 2 (options 1 and 4) is wrong because no second derivative appears. Only order 1 and degree 2 fits.

Final Answer:
The order is 1 and the degree is 2, which is option 3. \[ \boxed{\text{Order}=1,\ \text{Degree}=2} \]