The solution of the differential equation $\frac{dy}{dx}=e^{x-y}+x^2 e^{-y}$ is: (A) $y=e^{x-y}-x^2 e^{-y}+c$ (B) $e^y-e^x=\frac{x^3}{3}+c$ (C) $e^x+e^y=\frac{x^3}{3}+c$ (D) $e^x-e^y=\frac{x^3}{3}+c$

The solution of the differential equation (dy)/(dx)=ex-y+x² e-y is: (A) y=ex-y-x² e-y+c (B) ey-ex=(x³)/(3)+c (

NCERT Exemplar · Q96

\frac{dy}{dx}=e^{x-y}+x^2 e^{-y}

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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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Factor out $e^{-y}$ , separate, and integrate: $e^{y}-e^{x}=\frac{x^3}{3}+c$ — option (B).

1. Spot the hidden product

The term $e^{x-y}$ is really $e^{x}e^{-y}$ , so

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

The right side is a function of $x$ times a function of $y$ — the signal for separation of variables.

2. Separate

Multiply both sides by $e^{y}\,dx$ :

e^{y}\,dy=(e^{x}+x^2)\,dx.

3. Integrate each side

$\int e^{y}\,dy=\int(e^{x}+x^2)\,dx\;\Rightarrow\;e^{y}=e^{x}+\frac{x^3}{3}+c.$ …

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

GUJCET 2024Set 131 markMCQ
$\frac{x\,dy - y\,dx}{y} = 0$
›Reveal solutionSolution

The equation reduces to $\frac{dy}{y}=\frac{dx}{x}$ , whose solution is $y = cx$ .

Steps. From $\dfrac{x\,dy - y\,dx}{y}=0$ we get $x\,dy - y\,dx = 0 \Rightarrow x\,dy = y\,dx$ , so

$\frac{dy}{y}=\frac{dx}{x}.$ …
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GUJCET 2026Set x1 markMCQ
$\frac{dy}{dx} = e^{x+y}$
›Reveal solutionSolution

Separate variables in $\frac{dy}{dx}=e^x e^y$ to get $e^x+e^{-y}=C$ .

Write $\frac{dy}{dx}=e^{x+y}=e^x e^y$ and separate:
$e^{-y}\,dy=e^{x}\,dx$
Integrating both sides:

$-e^{-y}=e^{x}+c_1$ …
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GUJCET 2025Set 031 markMCQ
$\frac{dy}{dx} = e^{x-y}$
›Reveal solutionSolution
[!TLDR]

B increases linearly inside the wire up to $r=a$ , then decreases as $1/r$ outside — graph (B).

Concept

Using Ampere's law for a long straight wire of radius $a$ carrying uniformly distributed current $I$ :

Inside ( $r<a$ ): $B = \dfrac{\mu_0 I r}{2\pi a^2}$ — proportional to $r$ .

Outside ( $r>a$ ): $B = \dfrac{\mu_0 I}{2\pi r}$ — proportional to $1/r$ .

Solution …
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GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
$\dfrac{dy}{dx} = e^{x-y}$
(a) $e^y - e^x = C$
(b) $e^x + e^{-y} = C$
(c) $e^y + e^x = C$
(d) $e^{-x} + e^y = C$
›Reveal solutionSolution

This equation separates directly once $e^{x-y}$ is split into $e^x\cdot e^{-y}$ .

$\dfrac{dy}{dx}=e^x\cdot e^{-y} \Rightarrow e^ydy=e^xdx$ .

…

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GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQ
$\frac{ydx - xdy}{y} = 0$
(a) $y = Cx^2$
(b) $y = Cx$
(c) $x = Cy^2$
(d) $xy = C$
›Reveal solutionSolution

Separate the variables directly.

$y\,dx-x\,dy=0 \Rightarrow \frac{dx}{x}=\frac{dy}{y}$ .

…

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GSEB Higher Secondary Certificate (HSC) Examination 2022Set ANNUAL1 markMCQ
$\frac{dy}{dx} = \frac{1 + y^2}{1 + x^2}$
(a) $\tan^{-1} y = \tan^{-1} x + C$
(b) $\sin^{-1} y = \sin^{-1} x + C$
(c) $\log|y^2 + 1| = \log|1 + x^2| + C$
(d) $\cos^{-1} y = \cos^{-1} x + C$
›Reveal solutionSolution

Separate variables: $\tfrac{dy}{1+y^2} = \tfrac{dx}{1+x^2}$ .

$\dfrac{dy}{dx} = \dfrac{1+y^2}{1+x^2} \Rightarrow \int\dfrac{dy}{1+y^2} = \int\dfrac{dx}{1+x^2}.$

…

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GUJCET 2021Set 151 markMCQ
$\frac{dy}{dx} = e^{x-y}$
›Reveal solutionSolution

The equation is variable-separable via e^(x-y) = e^x e^-y.

Concept. $\dfrac{dy}{dx}=e^{x-y}=e^x e^{-y}\Rightarrow e^y\,dy=e^x\,dx$ . …
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GSEB Higher Secondary Certificate (HSC) Examination 2019Set ANNUAL1 markMCQ
$2x\frac{dy}{dx} - y = 0$
(a) Parabola
(b) Straight line
(c) Circle
(d) Ellipse
›Reveal solutionSolution

Separate variables, integrate, and apply $y(1)=2$ to identify the curve.

$2x\frac{dy}{dx}=y \Rightarrow \frac{dy}{y}=\frac{dx}{2x}$ . Integrating: $\log y = \frac{1}{2}\log x + C \Rightarrow y = k\sqrt{x}$ .

…

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←Previous questionQ95 — The general solution of the differential equation (ex+1)ydy=(y+1)exdx …→Next questionQ97 — The solution of the differential equation (dy)/(dx)+(2xy)/(1+x²)=(1)/(…

Q.The solution of the differential equation $\frac{dy}{dx}=e^{x-y}+x^2 e^{-y}$ is:
(A) $y=e^{x-y}-x^2 e^{-y}+c$
(B) $e^y-e^x=\frac{x^3}{3}+c$
(C) $e^x+e^y=\frac{x^3}{3}+c$
(D) $e^x-e^y=\frac{x^3}{3}+c$

Concept understanding — Separation Of Variables

Separation of Variables: From Intuition to Precision

The Core Intuition

The Precise Statement

Why This Works

A Concrete Example

1. Spot the hidden product

2. Separate

3. Integrate each side

Method: Factor out $e^{-y}$ to reveal a separable equation

Steps

Common Mistakes

Concept

Solution …

1. Spot the hidden product

2. Separate

3. Integrate each side

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