Form the differential equation of all circles which pass through origin and whose centres lie on y-axis.

NCERT Exemplar · Q14

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Karnataka PUCShort· 3mImportance★★★★★

Appeared in past exams:MHT-CET 2023· Set pcm-2023-05-09-E· 2mexactMHT-CET 2021· Set pcm-2021-09-20-M· 2mexact

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Concept understanding — Eliminating Arbitrary Constant

Eliminating Arbitrary Constants – The Core Idea

Consider a family of curves — say all circles centred at the origin: $x^2 + y^2 = r^2$ . Here $r$ is an arbitrary constant: each value gives one specific circle, but the whole family shares the same shape.

Suppose you're asked: "What differential equation does every member satisfy?" You need to eliminate the arbitrary constant $r$ to get an equation that holds for all such circles, regardless of $r$ .

That is the goal: start with an equation containing arbitrary constants, differentiate enough times to remove them, and obtain a differential equation representing the whole family.

Why Differentiate?

Arbitrary constants are constants — their derivative is zero. So differentiating either makes the constant disappear or lets you eliminate it by combining the original equation with its derivatives.

Differentiate $x^2 + y^2 = r^2$ with respect to $x$ : $\;2x + 2y \frac{dy}{dx} = 0$ . Divide by 2:

x + y \frac{dy}{dx} = 0

The constant $r$ is gone. The differential equation $x + y y' = 0$ is satisfied by every circle centred at the origin.

Note

We needed one differentiation because there was one arbitrary constant. In general, $n$ arbitrary constants require $n$ differentiations.

The Precise Statement

Eliminating arbitrary constants means: given an equation involving $x$ , $y$ , and $n$ arbitrary constants, differentiate it $n$ times (with respect to the independent variable) and then algebraically eliminate the $n$ constants from the system of $n+1$ equations (the original plus the $n$ derivatives). The result is an $n$ -th order ordinary differential equation representing the entire family of curves.

A Second Example (Two Constants)

Take all curves $y = A x + B x^2$ , where $A$ and $B$ are arbitrary constants.

Differentiate once: $\;y' = A + 2B x$

Differentiate again: $\;y'' = 2B$

Now we have three equations:

$y = A x + B x^2$
$y' = A + 2B x$
$y'' = 2B$

From (3), $B = \frac{y''}{2}$ . Substitute into (2): $y' = A + x y''$ , so $A = y' - x y''$ .

Substitute $A$ and $B$ into (1):

y = (y' - x y'') x + \left(\frac{y''}{2}\right) x^2 = x y' - \frac{x^2 y''}{2}

Multiply through by 2 and rearrange:

x^2 y'' - 2x y' + 2y = 0

This second-order ODE is satisfied by every curve $y = A x + B x^2$ , no matter what $A$ and $B$ are.

Watch out

A common mistake: trying to eliminate constants by substitution before differentiating enough times. You must differentiate first, then eliminate — substituting too early loses information.

Why This Matters …

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Eliminating the single parameter $a$ from $x^2+y^2=2ay$ gives $(x^2-y^2)\frac{dy}{dx}=2xy$ .

Set up the family

A circle whose centre lies on the $y$ -axis has centre $(0,a)$ . Passing through the origin forces its radius to equal the distance from $(0,a)$ to $(0,0)$ , namely $|a|$ . Hence

x^2+(y-a)^2=a^2.

Expand and cancel $a^2$ :

x^2+y^2-2ay=0\quad\Rightarrow\quad x^2+y^2=2ay.\qquad(1)

Here $a$ is the one arbitrary constant, so a single differentiation will remove it.

Differentiate

2x+2y\frac{dy}{dx}=2a\frac{dy}{dx}\quad\Rightarrow\quad x+y\frac{dy}{dx}=a\frac{dy}{dx}.\qquad(2)

Eliminate $a$ …

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

COMEDK 2023Set 2023-M1 markMCQ
$\frac{d^2 y}{d x^2}=0$
›Reveal solutionSolution

Non-vertical lines are $y=mx+c$ ; eliminating the two constants $m,c$ needs two differentiations, giving $y''=0$ .

Every non-vertical line has finite slope, so it can be written $y=mx+c$ with two arbitrary parameters $m$ and $c$ . To form the differential equation we eliminate both constants, which requires differentiating twice.

$\frac{dy}{dx}=m,\qquad \frac{d^2y}{dx^2}=0.$ …
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KCET 2020Set A-11 markMCQ
$x^2 dy - 2xy dx = x^4 \cos x dx$
›Reveal solutionSolution

Rearrange into the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$ , find the integrating factor $\text{I.F.} = e^{\int P\,dx} = x^{-2}$ , and integrate.

Step 1 — Put the equation in linear form
$x^2\,dy - 2xy\,dx = x^4\cos x\,dx$
Divide throughout by $x^2\,dx$ (with $x \neq 0$ ):
$\frac{dy}{dx} - \frac{2y}{x} = x^2\cos x$
This is linear in $y$ : $\dfrac{dy}{dx} + P(x)\,y = Q(x)$ with
$P(x) = -\frac{2}{x}, \qquad Q(x) = x^2\cos x$
Step 2 — Integrating factor

The I.F. is chosen so that the left side becomes an exact derivative:
$\text{I.F.} = e^{\int P\,dx} = e^{\int -\frac{2}{x}dx} = e^{-2\log x} = x^{-2} = \frac{1}{x^2}$
Step 3 — Multiply through and recognise the product rule
$\frac{1}{x^2}\frac{dy}{dx} - \frac{2}{x^3}y = \cos x \;\Longleftrightarrow\; \frac{d}{dx}\!\left(\frac{y}{x^2}\right) = \cos x$
Step 4 — Integrate …
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KCET 2019Set A-11 markMCQ
$(2x+3y^2) dy = y dx (y > 0)$
›Reveal solutionSolution

The ODE is linear in $x$ as a function of $y$ , not the other way round — so write $\frac{dx}{dy} + P(y)x = Q(y)$ and compute $\text{I.F.} = e^{\int P(y)\,dy}$ .

Step 1 — Try the usual orientation and see it fail
$(2x + 3y^2)\,dy = y\,dx \;\Longrightarrow\; \frac{dy}{dx} = \frac{y}{2x + 3y^2}$
This is not of the form $\frac{dy}{dx} + P(x)y = Q(x)$ — the $y$ 's do not separate out linearly. So we flip the roles.

Step 2 — Treat $x$ as the dependent variable
$\frac{dx}{dy} = \frac{2x + 3y^2}{y} = \frac{2x}{y} + 3y$
Rearranging into standard linear-in- $x$ form:
$\frac{dx}{dy} - \frac{2}{y}\,x = 3y$
So
$P(y) = -\frac{2}{y}, \qquad Q(y) = 3y$
Step 3 — Compute the integrating factor
$\text{I.F.} = e^{\int P(y)\,dy} = e^{\int -\frac{2}{y}\,dy} = e^{-2\log y} = e^{\log y^{-2}} = \frac{1}{y^2}$
(The condition $y > 0$ given in the question is exactly what lets us write $\int \frac{dy}{y} = \log y$ without absolute values.)

Step 4 — Confirm it does its job

Multiplying through by $1/y^2$ : …
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KCET 2018Set A-11 markMCQ
$\dfrac{d^2 y}{dx^2} = \sqrt[3]{1 + \left(\dfrac{dy}{dx}\right)^2}$
›Reveal solutionSolution
Make the equation a polynomial in the derivatives (cube it), then read off: order 2, degree 3 — and answer in the order the question asks (degree first).

Step 1 — The definitions.

Order = the order of the highest derivative appearing in the equation.

Degree = the power of that highest-order derivative, after the equation has been made free of radicals and fractions in its derivatives (i.e. after it is a polynomial in the derivatives).

The degree is simply not defined until the radical is cleared — that is the whole point of this question.

Step 2 — Write the equation.
$\frac{d^{2}y}{dx^{2}} = \sqrt[3]{1+\left(\frac{dy}{dx}\right)^{2}} = \left[1+\left(\frac{dy}{dx}\right)^{2}\right]^{1/3}$
As printed, the right side carries a fractional power — so we cannot read the degree yet.

Step 3 — Clear the radical by cubing both sides.
$\left(\frac{d^{2}y}{dx^{2}}\right)^{3} = 1+\left(\frac{dy}{dx}\right)^{2}$
The equation is now a polynomial in $\dfrac{dy}{dx}$ and $\dfrac{d^{2}y}{dx^{2}}$ .

Step 4 — Read off order and degree. …
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KCET 2022Set C-41 markMCQ
$\dfrac{dy}{dx} + \dfrac{y}{x} = x^2$
›Reveal solutionSolution

Solve the linear ODE with an integrating factor, then evaluate the combination $2y(2) - y(1)$ — the arbitrary constant cancels, which is exactly why the question is answerable without an initial condition.

Step 1 — Recognise the standard linear form.
$\frac{dy}{dx} + P(x)\,y = Q(x), \qquad P(x) = \frac{1}{x}, \quad Q(x) = x^{2}.$
Step 2 — Integrating factor.
$\text{I.F.} = e^{\int P\,dx} = e^{\int \frac{1}{x}dx} = e^{\log x} = x.$
Step 3 — Multiply through and integrate.

Multiplying the ODE by $x$ makes the left side an exact derivative:
$x\frac{dy}{dx} + y = x^{3} \quad\Longrightarrow\quad \frac{d}{dx}\big(xy\big) = x^{3}.$
Integrating both sides:
$xy = \frac{x^{4}}{4} + C \quad\Longrightarrow\quad y = \frac{x^{3}}{4} + \frac{C}{x}.$
Step 4 — Evaluate at $x = 2$ and $x = 1$ .

$y(2) = \frac{8}{4} + \frac{C}{2} = 2 + \frac{C}{2}, \qquad y(1) = \frac{1}{4} + C.$ …
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Q.Form the differential equation of all circles which pass through origin and whose centres lie on $y$ -axis.

Concept understanding — Eliminating Arbitrary Constant

Eliminating Arbitrary Constants – The Core Idea

Why Differentiate?

The Precise Statement

A Second Example (Two Constants)

Why This Matters …

Set up the family

Differentiate

Eliminate $a$ …

Method: Forming the DE of a geometric family of circles

Steps

Common Mistakes

Step 1 — Put the equation in linear form

Step 2 — Integrating factor

Step 3 — Multiply through and recognise the product rule

Step 4 — Integrate …

Step 1 — Try the usual orientation and see it fail

Step 2 — Treat $x$ as the dependent variable

Step 3 — Compute the integrating factor

Step 4 — Confirm it does its job

Q.Form the differential equation of all circles which pass through origin and whose centres lie on yyy-axis.

Set up the family

Differentiate

Eliminate aaa …

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Q.Form the differential equation of all circles which pass through origin and whose centres lie on $y$ -axis.

Eliminate $a$ …

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