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NCERT Exemplar · Q34

Q.(i) The degree of the differential equation d2ydx2+edydx=0\frac{d^2y}{dx^2}+e^{\frac{dy}{dx}}=0 is ______.

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The degree of a differential equation is defined only when the equation is a polynomial in the derivatives. Since edy/dxe^{dy/dx} is not a polynomial term, the degree is not defined.

The concept of degree of a differential equation is often misunderstood. Let’s clarify it first.

What does "degree" really mean?

The order of a differential equation is the highest derivative present — that’s straightforward. The degree, however, is trickier. It is defined only when the differential equation is a polynomial in all the derivatives that appear. That means every term involving yy, dydx\frac{dy}{dx}, d2ydx2\frac{d^2y}{dx^2}, etc., must be a polynomial (i.e., raised to a whole-number power, no exponentials, no trigonometric functions of derivatives, no logarithms of derivatives).

If the equation contains something like edy/dxe^{dy/dx}, sin⁡(dy/dx)\sin(dy/dx), or log⁡(d2y/dx2)\log(d^2y/dx^2), then it is not a polynomial in the derivatives — and the degree is simply not defined.

Watch out

A common mistake is to try to "force" a degree by expanding or ignoring the exponential. But the definition is strict: if any derivative appears inside a non-polynomial function (exponential, trigonometric, logarithmic), the degree is undefined. Do not write "1" or "0" — that would be incorrect.

Now let’s apply this to the given equation.


  1. Identify the derivatives present The equation is:

d2ydx2+edydx=0\frac{d^2y}{dx^2} + e^{\frac{dy}{dx}} = 0

The derivatives are d2ydx2\frac{d^2y}{dx^2} (second order) and dydx\frac{dy}{dx} (first order). The order is clearly 2.

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

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