Find the equation of a curve passing through origin if the slope of the tangent to the curve at any point $(x,\,y)$ is equal to the square of the difference of the abscissa and ordinate of the point.

Find the equation of a curve passing through origin if the slope of the tangent to the curve at any point (x,y

NCERT Exemplar · Q31

(x,\,y)

Karnataka PUCLong· 5mImportance★★★★★

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Concept understanding — Differential Equation Modeling

Differential Equation Modeling: From Intuition to Precision

Imagine you're watching a cup of hot coffee cool down on your desk. It starts hot, then gradually loses heat to the room. But here's the key question: how fast does it cool at any given moment? The answer isn't a fixed number — it depends on how hot the coffee is right now. The hotter it is, the faster it cools. This is the core idea behind differential equation modeling: the rate of change of a quantity depends on the quantity itself.

The Intuition First

Suppose the room is at $20^\circ$ C and your coffee starts at $90^\circ$ C. Newton's Law of Cooling says the rate at which the coffee cools is proportional to the temperature difference between the coffee and the room. So when the coffee is $90^\circ$ C, the difference is $70^\circ$ C and it cools fast; when it's $40^\circ$ C, the difference is only $20^\circ$ C and it cools slowly. The rate changes as the temperature changes.

If we let $T(t)$ be the temperature at time $t$ , then "rate of change of temperature" is $\frac{dT}{dt}$ , and the statement above becomes:

\frac{dT}{dt} = -k(T - 20)

The minus sign is because the temperature is decreasing; the constant $k$ depends on the cup, the liquid, etc. This is a differential equation — an equation that involves a function and its derivative.

The Precise Statement

A differential equation is any equation that contains an unknown function and one or more of its derivatives. The goal is to find the function itself (here, $T(t)$ ) that satisfies the equation.

\frac{dy}{dx} = f(x, y)

Modeling means taking a real-world situation and translating it into a differential equation:

Identify the quantity you want to study (population, temperature, drug concentration, voltage).
State the rate law in words: "The population grows at a rate proportional to its size."
Translate into math: $\frac{dP}{dt} = kP$ .
Add initial conditions: $P(0) = P_0$ (the starting value).

Why This Matters

Without differential equations, you'd have to guess the whole future behavior of a system. With them, you get a precise rule that governs every instant of change. The solution to $\frac{dP}{dt} = kP$ is $P(t) = P_0 e^{kt}$ — exponential growth. The solution to the cooling equation is $T(t) = 20 + 70 e^{-kt}$ — exponential decay toward room temperature.

Note

Most real-world models are not this simple. But every model starts the same way: observe how something changes, express that change in terms of the thing itself, and write it as a differential equation.

A Simple Example to Try

A bacteria culture doubles every hour. If you start with 100 bacteria:

Step 1: Quantity = population $P(t)$ . …

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Modelling gives $\frac{dy}{dx}=(x-y)^2$ ; with $v=x-y$ it separates to $\frac{1+v}{1-v}=e^{2x}$ , so $y=x-\frac{e^{2x}-1}{e^{2x}+1}=x-\tanh x$ .

Translate the words

'Abscissa' is $x$ , 'ordinate' is $y$ , and the tangent slope is $\frac{dy}{dx}$ . 'Slope equals the square of their difference' means

\frac{dy}{dx}=(x-y)^2.

A substitution to separate it

The right side depends only on $x-y$ , so let $v=x-y$ . Then $\frac{dv}{dx}=1-\frac{dy}{dx}$ , and the equation becomes

1-\frac{dv}{dx}=v^2\quad\Rightarrow\quad\frac{dv}{dx}=1-v^2,

which is separable.

Integrate

\int\frac{dv}{1-v^2}=\int dx.

With $\dfrac{1}{1-v^2}=\dfrac12\Big(\dfrac{1}{1-v}+\dfrac{1}{1+v}\Big)$ ,

\frac12\log\left|\frac{1+v}{1-v}\right|=x+C.

Use the origin

At $(0,0)$ , $v=0$ , so $\frac12\log 1=0+C\Rightarrow C=0$ . Then …

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

KCET 2021Set A-11 markMCQ
$xdy - ydx = 0$
›Reveal solutionSolution

$x\,dy-y\,dx=0$ separates to $\dfrac{dy}{y}=\dfrac{dx}{x}$ , whose solution $y=cx$ is the family of lines through the origin.

Step 1 — Separate the variables.
$x\,dy-y\,dx=0 \;\Longrightarrow\; x\,dy=y\,dx \;\Longrightarrow\; \frac{dy}{y}=\frac{dx}{x}\qquad (x\neq 0,\ y\neq 0).$
The equation is variable-separable (and equivalently homogeneous of degree 0, since $\dfrac{dy}{dx}=\dfrac{y}{x}$ depends only on $y/x$ ).

Step 2 — Integrate both sides.
$\int\frac{dy}{y}=\int\frac{dx}{x}\;\Longrightarrow\; \log|y| = \log|x| + \log|c| .$
Step 3 — Exponentiate.
$|y| = |c|\,|x| \;\Longrightarrow\; \boxed{y = cx}$
Step 4 — Identify the curve. $y=cx$ is a one-parameter family of straight lines of slope $c$ , every one of which passes through $(0,0)$ . …
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COMEDK 2022Set 20221 markMCQ
${{{d^2}y} \over {d{x^2}}} = 0$
›Reveal solutionSolution

This two-parameter family y = ax + b is precisely the family of all straight lines in the plane (non-vertical ones; a circle/parabola/ellipse would need a non-zero second derivative somewhere).

Concept: Integrate twice.

d²y/dx² = 0 → dy/dx = a (constant) → y = ax + b, with a, b arbitrary constants. …
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KCET 2025Set A-11 markMCQ
$\left(\frac{d^2y}{dx^2}\right)^2 + \left(\frac{dy}{dx}\right)^3 + x^4 = 0,$
›Reveal solutionSolution
Order = highest derivative present ( $=2$ ); degree = power of that highest derivative once the equation is polynomial in derivatives ( $=2$ ); so $a-b=0$ .

Step 1 — Recall the two definitions (they are often confused).

Order = the order of the highest-order derivative occurring in the equation.

Degree = the exponent of that highest-order derivative, after the equation has been made free of radicals and fractional powers of derivatives — i.e. once it is a polynomial in the derivatives.

Step 2 — Inspect the equation.
$\left(\frac{d^2y}{dx^2}\right)^{2} + \left(\frac{dy}{dx}\right)^{3} + x^{4} = 0$
There are no radicals, no fractional powers and no derivatives inside transcendental functions — it is already a polynomial in $\dfrac{dy}{dx}$ and $\dfrac{d^2y}{dx^2}$ . So both definitions apply directly.

Step 3 — Order.

The derivatives present are $\dfrac{dy}{dx}$ (order 1) and $\dfrac{d^2y}{dx^2}$ (order 2). The highest is the second derivative, so

$a = 2 .$ …
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KCET 2022Set C-41 markMCQ
$\left(1 + y_1^2\right)^{2/3} = y_2$
›Reveal solutionSolution
The key idea is to rewrite the equation so the highest derivative appears without a fractional exponent, then read off the order and degree. The sum is 5.

The problem gives a differential equation in a form where the highest derivative $y_2$ (the second derivative) is already isolated on one side. But the left side has a fractional exponent, which makes the degree ambiguous. The degree of a differential equation is defined only when the equation is a polynomial in the derivatives — meaning every derivative must appear with a whole-number exponent. So our first job is to clear that fractional power.

Identify the order. The highest derivative present is $y_2 = \frac{d^2y}{dx^2}$ . That makes the order 2. This is immediate and doesn't change no matter how we rewrite the equation.

Remove the fractional exponent to find the degree. The equation is

$(1 + y_1^2)^{2/3} = y_2$
where $y_1 = \frac{dy}{dx}$ . To make the left side a polynomial in $y_1$ and $y_2$ , raise both sides to the power 3:
$\left[(1 + y_1^2)^{2/3}\right]^3 = y_2^3$
which simplifies to
$(1 + y_1^2)^2 = y_2^3.$
Now expand if you like — but you don't need to. The equation is already a polynomial in the derivatives: $y_2$ appears with exponent 3, and $y_1$ appears with exponent 2 (inside the square). The highest exponent on the highest derivative $y_2$ is 3. That is the degree. …
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Q.Find the equation of a curve passing through origin if the slope of the tangent to the curve at any point $(x,\,y)$ is equal to the square of the difference of the abscissa and ordinate of the point.

Concept understanding — Differential Equation Modeling

Differential Equation Modeling: From Intuition to Precision

The Intuition First

The Precise Statement

Why This Matters

A Simple Example to Try

Translate the words

A substitution to separate it

Integrate

Use the origin

Method: Modelling a slope condition, then substituting $v=x-y$

Steps

Common Mistakes

Q.Find the equation of a curve passing through origin if the slope of the tangent to the curve at any point (x, y)(x,\,y)(x,y) is equal to the square of the difference of the abscissa and ordinate of the point.

Translate the words

A substitution to separate it

Integrate

Use the origin

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Q.Find the equation of a curve passing through origin if the slope of the tangent to the curve at any point $(x,\,y)$ is equal to the square of the difference of the abscissa and ordinate of the point.

Unlock everything free for 14 days