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Exercise 9.1 · Q11

Q.Determine the order and degree, if defined, of the differential equation: The degree of the differential equation (d2ydx2)3+(dydx)2+sin⁡(dydx)+1=0\left(\frac{d^2 y}{dx^2}\right)^3 + \left(\frac{dy}{dx}\right)^2 + \sin\left(\frac{dy}{dx}\right) + 1 = 0 is (A) 3 (B) 2 (C) 1 (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⁡(dydx)\sin\left(\frac{dy}{dx}\right) is not a polynomial in dydx\frac{dy}{dx}, so the degree is not defined. The correct option is (D).

  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 that appear. If the equation contains non-polynomial expressions like sin⁡(y′)\sin(y'), ey′e^{y'}, or log⁡(y′′)\log(y''), the degree is simply not defined.

  2. Identify the highest-order derivative.

    The given equation is:

(d2ydx2)3+(dydx)2+sin⁡(dydx)+1=0\left(\frac{d^2 y}{dx^2}\right)^3 + \left(\frac{dy}{dx}\right)^2 + \sin\left(\frac{dy}{dx}\right) + 1 = 0

The highest-order derivative present is d2ydx2\frac{d^2 y}{dx^2} (order 2). That part is fine — it appears as a cube, which is a polynomial term.

  1. Check the problematic term.

    Look at sin⁡(dydx)\sin\left(\frac{dy}{dx}\right). This is not a polynomial in dydx\frac{dy}{dx}; it is a transcendental function of the first derivative. No amount of algebraic manipulation (squaring, cubing, etc.) can turn sin⁡(y′)\sin(y') into a polynomial in y′y' — it is fundamentally non-polynomial.

  2. Apply the definition strictly. …

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