The order and degree of the differential equation $\frac{d}{dx}(e^y) = 0$ respectively are

Q: The order and degree of the differential equation $\frac{d}{dx}(e^y) = 0$ respectively are

**Correct option: (B) 1, 1** Expanding $\frac{d}{dx}(e^y) = e^y \left(\frac{dy}{dx}\right) = 0$. The highest order derivative present is dy/dx (order 1), and since e^y is never zero, this reduces to $\frac{dy}{dx} = 0$, a polynomial equation in the derivative with power 1, so the degree is also 1.

Differential Equations

Order and Degree of a Differential Equation

Options

(A)0, 1

(B)1, 1

(D)1, not defined

Official Answer

Correct option: (B) 1, 1

Expanding $\frac{d}{dx}(e^y) = e^y \left(\frac{dy}{dx}\right) = 0$ . The highest order derivative present is dy/dx (order 1), and since e^y is never zero, this reduces to $\frac{dy}{dx} = 0$ , a polynomial equation in the derivative with power 1, so the degree is also 1.

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Marking Scheme

11 mark: awarded only for selecting option (B) 1, 1; no partial credit.

Hint

Expand $\frac{d}{dx}(e^y)$ using the chain rule first: it equals $e^y \cdot \frac{dy}{dx}$ . Then order and degree follow directly from the resulting expression.

Quick Oral Answer

$\frac{d}{dx}(e^y)$ equals e^y times dy/dx by the chain rule; setting this to zero gives $\frac{dy}{dx}=0$ , which has order 1 and degree 1.

Analysis & Explanation

Concept: Order = highest derivative present; Degree = power of the highest derivative once the equation is a polynomial in derivatives.

By the chain rule, $\frac{d}{dx}(e^y) = e^y \left(\frac{dy}{dx}\right)$ . Setting this equal to 0 gives $e^y \left(\frac{dy}{dx}\right) = 0$ . The only derivative appearing is dy/dx, so the order is 1. The equation is already a polynomial in dy/dx (linear, since dy/dx appears to the first power, multiplied by the non-derivative factor e^y), so the degree is also 1.

Why the distractors fail:

(A) 0, 1 wrongly assigns order 0, but a derivative dy/dx does appear (order cannot be 0 whenever a derivative is present).
(C) 2, 1 wrongly assumes a second derivative is involved, perhaps by misreading the expression; only a first derivative appears here.
(D) 1, not defined would apply if the derivative appeared inside a non-polynomial (transcendental) form so that degree could not be defined — but here, after simplification, dy/dx appears in simple polynomial (linear) form, so the degree IS defined and equals 1.

Exam tip: Always fully expand/simplify a differential equation using calculus rules (chain rule, product rule) before reading off order and degree — an unsimplified expression can be misleading.

Common Mistakes

1Not expanding $\frac{d}{dx}(e^y)$ via the chain rule first, and misjudging the order/degree from the unexpanded form.
2Assuming degree is 'not defined' whenever a transcendental function like $e^y$ appears anywhere in the equation, even when the derivative itself appears in polynomial form.
3Confusing order 0 with 'no explicit dy/dx symbol visible before expansion.'

Interesting Facts

Degree is called 'not defined' only when the differential equation cannot be expressed as a polynomial in derivatives — for example, if $\frac{dy}{dx}$ itself appears inside a sine, log or exponential function.

The distinction between order and degree is one of the most repeatedly tested 1-mark concepts across CBSE Class 12 board papers, appearing in nearly every year's question paper in some form.

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Frequently Asked Questions

How do you find order and degree when the equation isn't already in derivative form?

First fully expand/simplify the expression using standard calculus rules (chain rule, product rule, etc.) to explicitly show all derivatives, then identify order as the highest derivative present and degree as its power once the equation is a polynomial in derivatives.

When is degree considered 'not defined'?

When the differential equation cannot be written as a polynomial in its derivatives — e.g., if a derivative appears inside a trigonometric, logarithmic, or exponential function, or under a radical with a non-integer power.