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CBSE Board Examination 2022 (Term 2) · Section A · 2 marks

Q1: Sum of the order and degree of (x+dydx)2=(dydx)2+1\left(x + \frac{dy}{dx}\right)^2 = \left(\frac{dy}{dx}\right)^2 + 1

Board 2022 T2Ch 9 · Differential Equations
The question

Find the sum of the order and the degree of the differential equation :

(x+dydx)2=(dydx)2+1\left(x + \frac{dy}{dx}\right)^2 = \left(\frac{dy}{dx}\right)^2 + 1
Exam answer

The given differential equation can be written as

x2+(dydx)2+2xdydx=(dydx)2+1\begin{aligned} &x^2 + \left(\frac{dy}{dx}\right)^2 + 2x\frac{dy}{dx} \\ &= \left(\frac{dy}{dx}\right)^2 + 1 \end{aligned}
Why?

The left side is the square of a sum, so expand it first:

(a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2

Here a=xa = x and b=dydxb = \frac{dy}{dx}⁠.

(x+dydx)2=(x)2+2(x)(dydx)+(dydx)2=x2+2xdydx+(dydx)2\begin{aligned} &\left(x + \frac{dy}{dx}\right)^2 \\ &= (x)^2 + 2(x)\left(\frac{dy}{dx}\right) \\ &\quad + \left(\frac{dy}{dx}\right)^2 \\ &= x^2 + 2x\frac{dy}{dx} + \left(\frac{dy}{dx}\right)^2 \end{aligned}

The right side stays as it is.

i.e. x2+2xdydx=1\text{i.e. } x^2 + 2x\frac{dy}{dx} = 1
Why?

The term (dydx)2\left(\frac{dy}{dx}\right)^2 is on both sides, so subtracting it from both sides removes it.

Left: x2+2xdydx+(dydx)2−(dydx)2=x2+2xdydxRight: (dydx)2+1−(dydx)2=1\begin{aligned} &\text{Left: } x^2 + 2x\frac{dy}{dx} \\ &\quad + \left(\frac{dy}{dx}\right)^2 - \left(\frac{dy}{dx}\right)^2 \\ &= x^2 + 2x\frac{dy}{dx} \\[4pt] &\text{Right: } \left(\frac{dy}{dx}\right)^2 + 1 \\ &\quad - \left(\frac{dy}{dx}\right)^2 = 1 \end{aligned}

Watch out: in the equation as printed, dydx\frac{dy}{dx} is squared, which tempts you to say degree 2. Simplify first. Once the squares cancel, the power is only 1.

The highest order derivative present is dydx\frac{dy}{dx}⁠, so order = 1. Its highest power is 1, so degree = 1.

Why?

The order is the order of the highest derivative in the equation. Only the first derivative dydx\frac{dy}{dx} appears, so the order is 1.

The degree is the highest power of that highest order derivative, once the equation is a polynomial in the derivatives. In x2+2xdydx=1x^2 + 2x\frac{dy}{dx} = 1 it appears to the power 1.

∴\therefore Sum of order and degree =1+1=2= 1 + 1 = 2⁠.

Where this comes from

Chapter 9, Differential Equations: Order and degree of a differential equation.

Question type: Order and degree

How to spot it: click to see

Type 1. Order and degree: the highest derivative, and its power

Spot it: "find the order and degree (if defined)". Often an MCQ.

Here: Expand the left side first: the (dydx)2\left(\frac{dy}{dx}\right)^2 terms cancel, leaving x2+2xdydx=1x^2 + 2x\frac{dy}{dx} = 1⁠. Now the highest derivative is dydx\frac{dy}{dx}⁠, to power 1, so the order and the degree are both 1.

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