(x) The solution of the differential equation cot ydx=xdy is ___.

NCERT Exemplar · Q43

\cot y\,dx=x\,dy

Uttarakhand UbseShort· 1mImportance★★★★★

Appeared in past exams:COMEDK 2022· Set 2022· 1mreworded

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Concept understanding — Separation Of Variables

Separation of Variables: From Intuition to Precision

Imagine you're baking a cake. The recipe says "mix the dry ingredients separately, then add the wet ones." You keep things that belong together together, and things that don't apart — until the right moment. Separation of Variables does exactly that for certain kinds of equations.

The Core Intuition

Some equations involve two different kinds of change happening at once. Think of a cup of hot coffee cooling down. The rate at which it cools depends on:

The temperature difference between the coffee and the room (a function of time)
The surface area of the cup (a function of shape, not time)

These two influences are tangled together in one equation. Separation of Variables is the mathematical trick that untangles them — it lets you handle the time part first, then the space part separately.

The Precise Statement

Separation of Variables applies to ordinary differential equations (ODEs) of the form:

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

where the right-hand side is a product of a function of $x$ alone and a function of $y$ alone. The method works in three clean steps:

\frac{dy}{dx} = f(x) \cdot g(y) \quad \Longrightarrow \quad \frac{1}{g(y)} \, dy = f(x) \, dx

Step 1: Separate. Multiply both sides by $dx$ and divide by $g(y)$ (assuming $g(y) \neq 0$ ). This moves all $y$ 's to one side and all $x$ 's to the other.

Step 2: Integrate. Put an integral sign on both sides:

\int \frac{1}{g(y)} \, dy = \int f(x) \, dx

Step 3: Solve. Evaluate both integrals and solve for $y$ explicitly if possible.

Watch out

You cannot separate if the equation is not in product form. For example, $\frac{dy}{dx} = x + y$ cannot be separated — the sum $x+y$ is not a product $f(x)g(y)$ .

Why This Works

The justification is the chain rule in reverse. From $\frac{dy}{dx} = f(x)g(y)$ , rewrite it as:

\frac{1}{g(y)} \frac{dy}{dx} = f(x)

Now integrate both sides with respect to $x$ :

\int \frac{1}{g(y)} \frac{dy}{dx} \, dx = \int f(x) \, dx

The left side is a substitution waiting to happen: $\frac{dy}{dx} dx = dy$ , so you get $\int \frac{1}{g(y)} dy$ . That's the entire trick — the chain rule dressed up.

A Concrete Example

Solve $\frac{dy}{dx} = 2xy$ , with $y(0) = 3$ .

Step 1: Separate. Divide both sides by $y$ (assuming $y \neq 0$ ):

\frac{1}{y} \, dy = 2x \, dx

Step 2: Integrate.

\int \frac{1}{y} \, dy = \int 2x \, dx \quad\Longrightarrow\quad \log|y| = x^2 + C

Step 3: Solve for $y$ .

|y| = e^{x^2 + C} = e^C \cdot e^{x^2}

Let $A = \pm e^C$ (absorbing the absolute value): $y = A e^{x^2}$ . Now use $y(0) = 3$ : $3 = A e^{0} = A$ , so $A = 3$ .

Final answer: $\boxed{y = 3e^{x^2}}$ …

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This is a first-order separable ODE disguised by the cotangent term. Rewriting it as $\frac{dx}{x} = \tan y\,dy$ and integrating both sides gives $\log|x| = -\log|\cos y| + C$ , which simplifies to $x \cos y = k$ as the general solution.

The key insight here is that the equation $\cot y\,dx = x\,dy$ looks like it might need an integrating factor, but it's actually separable — the variables $x$ and $y$ can be cleanly split. The cotangent function, $\cot y = \frac{\cos y}{\sin y}$ , is the only complication, but it becomes a simple trigonometric integral once we rearrange.

Let's work through it step by step.

Separate the variables. Start with $\cot y\,dx = x\,dy$ . Divide both sides by $x \cot y$ (assuming $x \neq 0$ and $\cot y \neq 0$ — we'll handle special cases later):

\frac{dx}{x} = \frac{dy}{\cot y}

Since $\frac{1}{\cot y} = \tan y$ , this becomes:

\frac{dx}{x} = \tan y\,dy

Now the variables are separated: everything with $x$ is on the left, everything with $y$ is on the right.

Integrate both sides.

\int \frac{dx}{x} = \int \tan y\,dy

The left integral is standard: $\int \frac{dx}{x} = \log|x| + C_1$ .

For the right side, recall that $\tan y = \frac{\sin y}{\cos y}$ . Let $u = \cos y$ , then $du = -\sin y\,dy$ , so $\sin y\,dy = -du$ . Thus:

\int \tan y\,dy = \int \frac{\sin y}{\cos y}\,dy = \int \frac{-du}{u} = -\log|u| + C_2 = -\log|\cos y| + C_2

So the integrated equation is:

\log|x| = -\log|\cos y| + C

where $C = C_2 - C_1$ is an arbitrary constant.

Tip

The integral $\int \tan y\,dy$ is worth memorizing: it equals $-\log|\cos y| + C$ (or equivalently $\log|\sec y| + C$ ). This saves time in exams.

Simplify the result. Combine the logarithms using properties: $-\log|\cos y| = \log\left|\frac{1}{\cos y}\right| = \log|\sec y|$ . So:

\log|x| = \log|\sec y| + C

Exponentiate both sides (let $e^C = k$ , a new constant, with $k > 0$ for now):

$|x| = k\,|\sec y|$ …

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Q.(x) The solution of the differential equation $\cot y\,dx=x\,dy$ is ______.

Concept understanding — Separation Of Variables

Separation of Variables: From Intuition to Precision

The Core Intuition

The Precise Statement

Why This Works

A Concrete Example

Method: Separation When the Derivative Is Written as $\frac{dx}{dy}$

Steps

Common Mistakes

Q.(x) The solution of the differential equation cot⁡y dx=x dy\cot y\,dx=x\,dycotydx=xdy is ______.

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Q.(x) The solution of the differential equation $\cot y\,dx=x\,dy$ is ______.

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