The general solution of excos ydx-exsin ydy=0 is: (A) excos y=k (B) exsin y=k (C) ex=kcos y (D) ex=ksin y

NCERT Exemplar · Q72

e^x\cos y\,dx-e^x\sin y\,dy=0

Assam AhsecMCQ· 1mImportance★★★★★

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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 an exact differential equation. By separating variables and integrating, we find that the general solution is $e^x \cos y = k$ , which corresponds to option (A).

We start with the equation:

e^x \cos y \, dx - e^x \sin y \, dy = 0

The key idea is Separation of Variables. This method works when we can rearrange a differential equation so that all terms involving $x$ (and $dx$ ) are on one side, and all terms involving $y$ (and $dy$ ) are on the other. Once separated, we integrate both sides independently.

Notice that $e^x$ is a common factor in both terms. We can factor it out:

e^x (\cos y \, dx - \sin y \, dy) = 0

Since $e^x$ is never zero for any real $x$ , we can safely divide both sides by $e^x$ without losing any solutions. This gives:

\cos y \, dx - \sin y \, dy = 0

Now, we rearrange to separate the variables. Move the $dy$ term to the other side:

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

To separate, divide both sides by $\cos y$ (provided $\cos y \neq 0$ — we'll check that case separately) and also by $dx$ (which is just algebraic manipulation):

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

The right-hand side simplifies: $\frac{\sin y}{\cos y} = \tan y$ . So we have:

dx = \tan y \, dy

Now the variables are separated: $x$ on the left, $y$ on the right. Integrate both sides:

\int dx = \int \tan y \, dy

The left integral is simply $x + C_1$ . For the right integral, recall that $\int \tan y \, dy = -\log|\cos y| + C_2$ . Combining constants, we get:

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

where $C$ is an arbitrary constant.

Tip

The integral $\int \tan y \, dy$ is a standard result. If you forget it, rewrite $\tan y = \frac{\sin y}{\cos y}$ and use the substitution $u = \cos y$ , $du = -\sin y \, dy$ , giving $\int \frac{-du}{u} = -\log|u| = -\log|\cos y|$ .

Now, solve for a nicer form. Bring the logarithmic term to the left:

x + \log|\cos y| = C

Exponentiate both sides (using $e$ as the base):

e^{x + \log|\cos y|} = e^C

Since $e^{a+b} = e^a e^b$ , this becomes:

e^x \cdot e^{\log|\cos y|} = e^C

But $e^{\log|\cos y|} = |\cos y|$ . So:

$e^x |\cos y| = e^C$ …

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

AHSEC Higher Secondary (HS) Final Examination 2023Set ANNUAL4 marks
$xy\dfrac{dy}{dx} = (x+2)(y+2)$
›Reveal solutionSolution

Separate variables, integrate both sides, then use the point $(1,-1)$ to find the constant.

Given $xy\dfrac{dy}{dx} = (x+2)(y+2)$ . Separate variables:

$\frac{y}{y+2}\,dy = \frac{x+2}{x}\,dx.$

Simplify each side: $\dfrac{y}{y+2} = 1 - \dfrac{2}{y+2}$ , and $\dfrac{x+2}{x} = 1 + \dfrac{2}{x}$ .

Integrate:

$\int\left(1-\frac{2}{y+2}\right)dy = \int\left(1+\frac2x\right)dx$

$y - 2\log|y+2| = x + 2\log|x| + C.$

Apply the condition: passes through $(1,-1)$ , i.e. $x=1,y=-1$ :

$-1 - 2\log|1| = 1 + 2\log|1| + C \implies -1 - 0 = 1 + 0 + C \implies C = -2.$

Therefore the required solution curve is

$y - 2\log|y+2| = x + 2\log|x| - 2.$

OR: Find a particular solution of $\dfrac{dy}{dx} + y\cot x = 4x\csc x$ , $y\left(\dfrac{\pi}{2}\right)=0$ .

This is linear in $y$ : $\dfrac{dy}{dx} + P(x)y = Q(x)$ with $P=\cot x$ , $Q = 4x\csc x$ .

…

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Q.The general solution of $e^x\cos y\,dx-e^x\sin y\,dy=0$ is:
(A) $e^x\cos y=k$
(B) $e^x\sin y=k$
(C) $e^x=k\cos y$
(D) $e^x=k\sin y$

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: Solving an Exact/Separable Equation by Spotting a Total Differential

Steps

Common Mistakes

Q.The general solution of excos⁡y dx−exsin⁡y dy=0e^x\cos y\,dx-e^x\sin y\,dy=0excosydx−exsinydy=0 is: (A) excos⁡y=ke^x\cos y=kexcosy=k (B) exsin⁡y=ke^x\sin y=kexsiny=k (C) ex=kcos⁡ye^x=k\cos yex=kcosy (D) ex=ksin⁡ye^x=k\sin yex=ksiny

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Q.The general solution of $e^x\cos y\,dx-e^x\sin y\,dy=0$ is:
(A) $e^x\cos y=k$
(B) $e^x\sin y=k$
(C) $e^x=k\cos y$
(D) $e^x=k\sin y$

Unlock everything free for 14 days