The solution of the differential equation $\cos x\sin y\,dx+\sin x\cos y\,dy=0$ is: (A) $\frac{\sin x}{\sin y}=c$ (B) $\sin x\sin y=c$ (C) $\sin x+\sin y=c$ (D) $\cos x\cos y=c$

The solution of the differential equation cos xsin ydx+sin xcos ydy=0 is: (A) (sin x)/(sin y)=c (B) sin xsin y

NCERT Exemplar · Q79

\cos x\sin y\,dx+\sin x\cos y\,dy=0

Uttarakhand UbseMCQ· 1mImportance★★★★★

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

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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 separable first-order differential equation. By separating the variables and integrating, we find that the general solution is $\sin x \sin y = c$ , which corresponds to option (B).

The key insight here is that the equation is written in a form where the coefficient of $dx$ is a product of a function of $x$ and a function of $y$ , and the same is true for the $dy$ term. That is the hallmark of a separable differential equation — we can rearrange it so that all $x$ 's are on one side with $dx$ and all $y$ 's on the other with $dy$ .

Let’s see why this works. The equation is:

\cos x \sin y \, dx + \sin x \cos y \, dy = 0

If we move the $dy$ term to the other side, we get:

\cos x \sin y \, dx = - \sin x \cos y \, dy

Now, divide both sides by $\sin y \sin x$ (assuming these are not zero — we’ll handle the constant solutions separately). This gives:

\frac{\cos x}{\sin x} \, dx = - \frac{\cos y}{\sin y} \, dy

Notice that each side is now a function of only one variable. That’s the separation.

Watch out

A common mistake is to forget the negative sign when moving terms. Always check the sign carefully — here the $dy$ term was originally positive on the left, so moving it to the right introduces a minus sign.

Now we integrate both sides:

\int \frac{\cos x}{\sin x} \, dx = - \int \frac{\cos y}{\sin y} \, dy

Each integral is of the form $\int \frac{f'(x)}{f(x)} dx$ , which gives $\log|f(x)| + C$ . So:

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

Combine the logarithms:

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

\log|\sin x \sin y| = C

Exponentiate both sides:

|\sin x \sin y| = e^C

Since $e^C$ is just a positive constant, we can rename it as $c$ (where $c > 0$ ). But the absolute value means $\sin x \sin y = \pm c$ , which is equivalent to $\sin x \sin y = k$ for some constant $k$ (which could be positive or negative). So the general solution is: …

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Q.The solution of the differential equation $\cos x\sin y\,dx+\sin x\cos y\,dy=0$ is:
(A) $\frac{\sin x}{\sin y}=c$
(B) $\sin x\sin y=c$
(C) $\sin x+\sin y=c$
(D) $\cos x\cos y=c$

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 of variables (differential form)

Steps

Common Mistakes

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