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NCERT Exemplar · Q57

Q.The degree of the differential equation [1+(dydx)2]3/2=d2ydx2\left[1+\left(\frac{dy}{dx}\right)^2\right]^{3/2}=\frac{d^2y}{dx^2} is:
(A) 4
(B) 32\frac{3}{2}
(C) not defined
(D) 2

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Appeared in past exams:AP EAPCET 2026· Set eng-2026-05-14-AN· 1mrewordedGUJCET 2020· Set 07· 1mreworded
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The degree of a differential equation is defined only when the equation is a polynomial in the derivatives. Here, raising both sides to the power 2 gives a polynomial, so the degree is 2.

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 involves fractional powers, roots, or non-polynomial functions of the derivatives, the degree is not defined until we rewrite it as a polynomial — if possible.

In this problem, the highest-order derivative is d2ydx2\frac{d^2y}{dx^2}, and it appears inside a square root (via the exponent 32\frac{3}{2} on the left side). That fractional exponent is the obstacle: the equation is not yet a polynomial in the derivatives. So we must first remove the fractional power by raising both sides to a suitable integer power, then check the polynomial form.

Let’s work through it step by step.

  1. Identify the highest-order derivative. The equation is

[1+(dydx)2]3/2=d2ydx2.\left[1+\left(\frac{dy}{dx}\right)^2\right]^{3/2} = \frac{d^2y}{dx^2}.

The highest derivative present is d2ydx2\frac{d^2y}{dx^2} (order 2). The degree concerns this derivative, but only after the equation is made polynomial in all derivatives.

  1. Remove the fractional exponent. The left side has exponent 3/23/2. To clear the square root, square both sides:

([1+(dydx)2]3/2)2=(d2ydx2)2.\left( \left[1+\left(\frac{dy}{dx}\right)^2\right]^{3/2} \right)^2 = \left( \frac{d^2y}{dx^2} \right)^2.

This gives

[1+(dydx)2]3=(d2ydx2)2.\left[1+\left(\frac{dy}{dx}\right)^2\right]^{3} = \left( \frac{d^2y}{dx^2} \right)^2.

  1. Check if the equation is now a polynomial in derivatives. …

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