Find the differential equation of all circles in the first quadrant which touches both the coordinate axes.

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Q. Find the differential equation of all circles in the first quadrant which touches both the coordinate axes.

CBSECBSE Class XII Board 2024Subjective· 3mImportance★★★★★

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Concept understanding — Order and Degree of Differential Equations

Order and Degree of Differential Equations

The Intuition: What Are We Counting?

A differential equation is an equation that involves derivatives — rates of change. When you see something like

\frac{dy}{dx} = 3x^2

\frac{d^2y}{dx^2} + 5\frac{dy}{dx} + 6y = 0

you are looking at relationships between a function and its derivatives. The order and degree are two numbers that classify such equations. Think of them as the "size" and "shape" of the derivative information.

Order tells you the highest number of times the function has been differentiated. If the highest derivative is $\frac{dy}{dx}$ (first derivative), the order is 1. If it's $\frac{d^2y}{dx^2}$ (second derivative), the order is 2. Simple.

Degree is trickier. It tells you the power to which the highest-order derivative is raised — after the equation has been cleared of radicals and fractions involving derivatives. If the highest derivative appears squared, the degree is 2. If it appears inside a square root, you must first remove that root before declaring the degree.

The Precise Definitions

Important

Order of a differential equation is the order of the highest derivative present in the equation.

Important

Degree of a differential equation is the power of the highest-order derivative, provided the equation is a polynomial equation in all the derivatives (i.e., no fractional powers, no radicals, no trigonometric functions of derivatives).

The "provided" part is critical. You cannot read the degree directly from a messy equation — you must first rewrite it so that every derivative appears with a whole-number exponent.

Examples That Build Understanding

Example 1: $\frac{dy}{dx} + y = x$

Highest derivative is $\frac{dy}{dx}$ (first derivative). Order = 1. That derivative appears to the power 1. Degree = 1.

Example 2: $\left(\frac{d^2y}{dx^2}\right)^3 + 5\frac{dy}{dx} = \sin x$

Highest derivative is $\frac{d^2y}{dx^2}$ (second derivative). Order = 2. That derivative is raised to the power 3. Degree = 3.

Example 3: $\sqrt{1 + \left(\frac{dy}{dx}\right)^2} = \frac{d^2y}{dx^2}$

Here the highest derivative is $\frac{d^2y}{dx^2}$ (order 2). But the equation is not a polynomial in derivatives — there is a square root. To find the degree, square both sides:

1 + \left(\frac{dy}{dx}\right)^2 = \left(\frac{d^2y}{dx^2}\right)^2

Now the highest derivative $\frac{d^2y}{dx^2}$ appears with power 2. Degree = 2.

Watch out

Never read the degree from an equation that still has radicals, fractional powers, or trigonometric functions of derivatives. You must first make it a polynomial in the derivatives.

Example 4: $\frac{d^2y}{dx^2} = \sin\left(\frac{dy}{dx}\right)$

Order = 2. But the equation contains $\sin$ of a derivative — it is not a polynomial in derivatives at all. Degree is not defined for such equations. Many exam questions test exactly this: if the derivative appears inside a trigonometric, logarithmic, or exponential function, the degree is simply not defined.

A Quick Reference Table

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Family: $(x-a)^2+(y-a)^2=a^2$ with one parameter $a$ ; differentiate once, eliminate $a$ , and obtain a first-order (first-degree in the family, second-degree in $y'$ ) differential equation.

(a,a)

Equation of the family (one parameter $a>0$ ): $(x-a)^2+(y-a)^2=a^2$ . ...(1)
Differentiate w.r.t. $x$ : $2(x-a)+2(y-a)\dfrac{dy}{dx}=0$ , i.e. $(x-a)+(y-a)\dfrac{dy}{dx}=0$ .
Solve for $a$ : $x+y\dfrac{dy}{dx}=a\left(1+\dfrac{dy}{dx}\right)\Rightarrow a=\dfrac{x+y\frac{dy}{dx}}{1+\frac{dy}{dx}}$ .
Then $x-a=\dfrac{\frac{dy}{dx}(x-y)}{1+\frac{dy}{dx}}$ and $y-a=\dfrac{y-x}{1+\frac{dy}{dx}}$ .
Substitute into (1): $\dfrac{\left(\frac{dy}{dx}\right)^2(x-y)^2+(x-y)^2}{\left(1+\frac{dy}{dx}\right)^2}=\dfrac{\left(x+y\frac{dy}{dx}\right)^2}{\left(1+\frac{dy}{dx}\right)^2}$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2024Set 465/RQPS/41 markMCQ
$y\,dx + x\log\left(\dfrac{y}{x}\right)dy - 2x\,dy = 0$
›Reveal solutionSolution
Only a first derivative appears, to the first power, so order $=1$ and degree $=1$ .

$Order = the highest-order derivative in the equation; Degree = the power of that highest derivative when the equation is a polynomial in the derivatives.$

Divide the equation $y\,dx+x\log(\tfrac{y}{x})\,dy-2x\,dy=0$ by $dx$ : $y+x\log(\tfrac{y}{x})\dfrac{dy}{dx}-2x\dfrac{dy}{dx}=0$ .

The highest derivative is $\dfrac{dy}{dx}$ , so the order is $1$ . …
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CBSE 2024Set 465/S/RQPS/41 markMCQ
$\left(\dfrac{d^2y}{dx^2}\right)^3 + \left(\dfrac{dy}{dx}\right)^2 + \sin\left(\dfrac{dy}{dx}\right) + 1 = 0$
›Reveal solutionSolution
Treating the degree as the highest power (3) of the highest-order derivative, both statements are true and R correctly explains A — the answer the marking scheme accepts, with a noted caveat about the $\sin$ term.

$Degree of a differential equation = the highest power of the highest-order derivative, after the equation is made a polynomial in its derivatives (it is undefined if that is impossible).$

The highest-order derivative here is $\dfrac{d^2y}{dx^2}$ (order 2), appearing with power 3 in $\left(\dfrac{d^2y}{dx^2}\right)^3$ .

Reading the degree directly as that power gives degree $= 3$ , so Assertion (A) is taken as true.

Reason (R) states the standard definition of degree correctly, and it is exactly the rule used in step 1 — so R is the correct explanation of A. …
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CBSE 2023Set 465/EF1GH/41 markMCQ
$y^2 = 4ax$
(c) and
(d) as given below.
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.
›Reveal solutionSolution
Eliminating the single parameter $a$ gives $2x y'-y=0$ , so the stated equation $x y'-2y=0$ is wrong — A false; R (n parameters $\Rightarrow$ differentiate $n$ times) true — answer (d).

$n$

Family: $y^2=4ax$ (one parameter $a$ ). Differentiate once: $2y\dfrac{dy}{dx}=4a$ , so $4a=2y\dfrac{dy}{dx}$ .

Substitute $4a$ back into $y^2=4ax$ : $y^2=\left(2y\dfrac{dy}{dx}\right)x$ . …
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Q.Find the differential equation of all circles in the first quadrant which touches both the coordinate axes.

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