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The differential equation y=xp+x2p3+4y = xp + \sqrt{x^2 p^3 + 4} where p=dydxp = \frac{dy}{dx} is :

(A) of order 1

(B) of degree 1

(C) of order 2

(D) of degree 3

Choose the correct answer from the options given below :

Solution

✅ Correct Option: 2

Highest derivative is p=dy/dxp=dy/dx, so order is 1.

To find degree, isolate radical: (y−xp)2=x2p3+4(y-xp)^2=x^2 p^3+4. Highest power of pp is 3, so degree is 3.

Hence (A) and (D) are correct.

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