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Q.The degree of the differential equation (y′′)2+(y′)3=xsin⁡(y′)(y'')^2 + (y')^3 = x \sin(y') is: (A) 11 (B) 22 (C) 33 (D) not defined

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The degree of a differential equation is defined only when the equation is a polynomial in the derivatives. Here, the term sin⁡(y′)\sin(y') is non-polynomial, so the degree is not defined. The correct option is (D).

The degree of a differential equation is a precise, formal property — it is not just the highest power of the highest derivative you see. For the degree to exist, the equation must be a polynomial in all the derivatives that appear. That means every derivative term (like y′y', y′′y'', etc.) must be raised only to a non-negative integer power, and no transcendental functions (sine, cosine, exponential, log) can wrap around any derivative.

Here, the equation is:

(y′′)2+(y′)3=xsin⁡(y′)(y'')^2 + (y')^3 = x \sin(y')

The left-hand side is fine: (y′′)2(y'')^2 and (y′)3(y')^3 are polynomial in y′′y'' and y′y'. But the right-hand side contains sin⁡(y′)\sin(y') — the sine of the first derivative. That is not a polynomial in y′y'; it is a transcendental function of y′y'. So the equation as a whole is not a polynomial in the derivatives.

Because the definition of degree requires a polynomial form, the degree simply does not exist here.

Let’s walk through the reasoning step by step.

  1. Recall the definition of degree.

    The degree of a differential equation is the power of the highest-order derivative, provided the equation is a polynomial in all the derivatives. If any derivative appears inside a non-polynomial function (like sin⁡\sin, cos⁡\cos, ee, log⁡\log, etc.), the degree is not defined.

  2. Identify the highest-order derivative.

    The highest derivative present is y′′y'' (second order). It appears as (y′′)2(y'')^2, which is polynomial. So the order is 22, but that is not what we are asked.

  3. Check the condition for degree. …

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