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NCERT Exemplar · Q56

Q.The degree of the differential equation (d2ydx2)2+(dydx)2=xsin⁡(dydx)\left(\frac{d^2y}{dx^2}\right)^2+\left(\frac{dy}{dx}\right)^2=x\sin\left(\frac{dy}{dx}\right) is:
(A) 1
(B) 2
(C) 3
(D) not defined

Punjab PsebMCQ· 1mImportance★★★★★
Appeared in past exams:CBSE 2025· Set 65/4/1· 1mreworded
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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) makes the equation non-polynomial, so the degree is not defined.

The degree of a differential equation is a surprisingly subtle idea. Many students rush to look at the highest power of the highest derivative — here they see (d2ydx2)2\left(\frac{d^2y}{dx^2}\right)^2 and immediately say "degree 2". But that's only half the story.

The key condition: For a differential equation to have a degree, it must be expressible as a polynomial in the derivatives (with the dependent variable and its derivatives as the only variables). No trigonometric functions, no exponentials, no logarithms of derivatives — those break the polynomial form.

Let's check our equation step by step.

  1. Write the equation clearly

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

  1. Identify the derivatives present

    • First derivative: dydx\frac{dy}{dx}
    • Second derivative: d2ydx2\frac{d^2y}{dx^2} The equation involves both.
  2. Check if it is a polynomial in derivatives

    The left-hand side is fine: (d2ydx2)2\left(\frac{d^2y}{dx^2}\right)^2 and (dydx)2\left(\frac{dy}{dx}\right)^2 are polynomial terms (powers of derivatives).

    But the right-hand side contains sin⁡(dydx)\sin\left(\frac{dy}{dx}\right). This is a trigonometric function of a derivative, not a polynomial in it. You cannot expand sin⁡(y′)\sin(y') as a finite sum of powers of y′y' — it's transcendental. …

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