Solve the following differential equation: $\left(1 + e^{y/x}\right) dy + e^{y/x}\left(1 - \dfrac{y}{x}\right) dx = 0 \quad (x \ne 0)$.

> [!TLDR] > The given differential equation is homogeneous, which we solve by substituting $y=vx$ to transform it into a separable equation. The final solution is $\boxed{x e^{y/x} + y = K}$. The given differential equation is $\left(1 + e^{y/x}\right) dy + e^{y/x}\left(1 - \dfrac{y}{x}\right) dx = 0$. Our first step in solving any differential equation is to identify its type. Let's rearrange it into the standard form $\frac{dy}{dx} = f(x,y)$: $\left(1 + e^{y/x}\right) dy = - e^{y/x}\left(1 - \dfrac{y}{x}\right) dx$ $\frac{dy}{dx} = - \frac{e^{y/x}\left(1 - \dfrac{y}{x}\right)}{1 + e^{y/x}}$ Notice that the right-hand side, $f(x,y)$, depends only on the ratio $y/x$. This is the defining characteristic of a homogeneous differential equation. > [!FORMULA] > A first-order differential equation $\frac{dy}{dx} = f(x,y)$ is homogeneous if $f(x,y)$ can be expressed as a function of $\frac{y}{x}$ alone, i.e., $f(x,y) = g\left(\frac{y}{x}\right)$. The intuition behind solving homogeneous equations is that if the function $f(x,y)$ only cares about the ratio $y/x$, we can simplify the problem by introducing a new variable that *is* this ratio. This substitution effectively "normalizes" the variables, allowing us to separate them. We use the substitution $y = vx$, where $v$ is a function of $x$. Differentiating $y = vx$ with respect to $x$ using the product rule gives: $\frac{dy}{dx} = v \cdot 1 + x \frac{dv}{dx}$ Now, we substitute $y=vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into our differential equation. 1. **Substitute $y=vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the equation.** The original equation is $\frac{dy}{dx} = - \frac{e^{y/x}\left(1 - \dfrac{y}{x}\right)}{1 + e^{y/x}}$. Substituting $y/x = v$: $v + x \frac{dv}{dx} = - \frac{e^{v}(1 - v)}{1 + e^{v}}$ 2. **Separate the variables $v$ and $x$.** First, isolate $x \frac{dv}{dx}$: $x \frac{dv}{dx} = - \frac{e^{v}(1 - v)}{1 + e^{v}} - v$ To combine the terms on the right,

Solve the following differential equation: (1 + ey/x) d — CBSE Class 12 Mathematics 2020 PYQ

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\left(1 + e^{y/x}\right) dy + e^{y/x}\left(1 - \dfrac{y}{x}\right) dx = 0 \quad (x \ne 0)

CBSECBSE Class XII Board 2020Subjective· 4mImportance★★★★★

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The given differential equation is homogeneous, which we solve by substituting $y=vx$ to transform it into a separable equation. The final solution is $\boxed{x e^{y/x} + y = K}$ .

The given differential equation is $\left(1 + e^{y/x}\right) dy + e^{y/x}\left(1 - \dfrac{y}{x}\right) dx = 0$ .

Our first step in solving any differential equation is to identify its type. Let's rearrange it into the standard form $\frac{dy}{dx} = f(x,y)$ :

$\left(1 + e^{y/x}\right) dy = - e^{y/x}\left(1 - \dfrac{y}{x}\right) dx$

$\frac{dy}{dx} = - \frac{e^{y/x}\left(1 - \dfrac{y}{x}\right)}{1 + e^{y/x}}$

Notice that the right-hand side, $f(x,y)$ , depends only on the ratio $y/x$ . This is the defining characteristic of a homogeneous differential equation.

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

The intuition behind solving homogeneous equations is that if the function $f(x,y)$ only cares about the ratio $y/x$ , we can simplify the problem by introducing a new variable that is this ratio. This substitution effectively "normalizes" the variables, allowing us to separate them.

We use the substitution $y = vx$ , where $v$ is a function of $x$ .

Differentiating $y = vx$ with respect to $x$ using the product rule gives:

$\frac{dy}{dx} = v \cdot 1 + x \frac{dv}{dx}$

Now, we substitute $y=vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into our differential equation.

Substitute $y=vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the equation.

The original equation is $\frac{dy}{dx} = - \frac{e^{y/x}\left(1 - \dfrac{y}{x}\right)}{1 + e^{y/x}}$ .

Substituting $y/x = v$ :

$v + x \frac{dv}{dx} = - \frac{e^{v}(1 - v)}{1 + e^{v}}$
Separate the variables $v$ and $x$ .

First, isolate $x \frac{dv}{dx}$ :

$x \frac{dv}{dx} = - \frac{e^{v}(1 - v)}{1 + e^{v}} - v$

To combine the terms on the right, find a common denominator:

$x \frac{dv}{dx} = \frac{-e^{v}(1 - v) - v(1 + e^{v})}{1 + e^{v}}$

$x \frac{dv}{dx} = \frac{-e^{v} + v e^{v} - v - v e^{v}}{1 + e^{v}}$

$x \frac{dv}{dx} = \frac{-e^{v} - v}{1 + e^{v}}$

$x \frac{dv}{dx} = - \frac{e^{v} + v}{1 + e^{v}}$

Now, we can separate the variables by moving all $v$ terms to one side with $dv$ and all $x$ terms to the other side with $dx$ :

$\frac{1 + e^{v}}{e^{v} + v} dv = - \frac{1}{x} dx$
Integrate both sides.

$\int \frac{1 + e^{v}}{e^{v} + v} dv = \int - \frac{1}{x} dx$

For the left-hand side integral, let $u = e^{v} + v$ . Then, differentiating $u$ with respect to $v$ gives $du = (e^{v} + 1) dv$ . This is exactly the numerator.

So, the left integral becomes:

$\int \frac{1}{u} du = \ln|u| + C_1 = \ln|e^{v} + v| + C_1$

For the right-hand side integral:

$\int - \frac{1}{x} dx = - \ln|x| + C_2$

Combining these results, we get:

$\ln|e^{v} + v| = - \ln|x| + C$ (where $C = C_2 - C_1$ is an arbitrary constant)
Simplify the expression and substitute back $v = y/x$ . …

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

CBSE 2026Set 65/3/11 markMCQ
$\frac{dy}{dx} = F(x, y)$
(i) $F(x, y) = 3x + 2y$
(ii) $F(x, y) = \sin \frac{y}{x} + \log y - \log x$
(iii) $F(x, y) = e^{y/x} + 1$
(iv) $F(x, y) = \sqrt{x^2 + y^2} - y$ (A)
(i) and
(ii) (B) (i),
(ii) and
(iii) (C) (ii),
(iii) and
(iv) (D)
(ii) and (iii)
›Reveal solutionSolution
A differential equation $\frac{dy}{dx} = F(x, y)$ is homogeneous exactly when $F$ is homogeneous of degree zero — meaning $F(tx, ty) = F(x, y)$ for all $t > 0$ , which is equivalent to $F$ being expressible purely as a function of $y/x$ . Only options (ii) and (iii) satisfy this, so the correct choice is (D).

The idea is simple: a homogeneous differential equation is one where the right-hand side $F(x, y)$ doesn't change if you scale both $x$ and $y$ by the same factor. Why does that matter? Because if $F$ has that property, you can substitute $y = vx$ and turn the equation into one in $v$ and $x$ alone — a separable equation you can actually solve. The test is clean: check whether $F(tx, ty) = F(x, y)$ for any $t > 0$ .

Let's go through each option.

Option (i): $F(x, y) = 3x + 2y$ Replace $x$ with $tx$ and $y$ with $ty$ :

$F(tx, ty) = 3(tx) + 2(ty) = t(3x + 2y) = t \cdot F(x, y)$
This is $t$ times the original, not equal to it — unless $t = 1$ . So $F$ is homogeneous of degree 1, not degree 0. It also cannot be written as a function of $y/x$ alone (try it: $3x + 2y = x(3 + 2(y/x))$ still has an $x$ factor outside). So this is not homogeneous for the purpose of $\frac{dy}{dx} = F(x, y)$ .

Option (ii): $F(x, y) = \sin\frac{y}{x} + \log y - \log x$ First simplify the log terms: $\log y - \log x = \log\frac{y}{x}$ . So

$F(x, y) = \sin\frac{y}{x} + \log\frac{y}{x}$
This is already written purely in terms of $y/x$ . That's a dead giveaway — it's homogeneous of degree 0. Check formally:
$F(tx, ty) = \sin\frac{ty}{tx} + \log\frac{ty}{tx} = \sin\frac{y}{x} + \log\frac{y}{x} = F(x, y)$
The $t$ cancels completely. So this is homogeneous.

Option (iii): $F(x, y) = e^{y/x} + 1$ Again, this is already a function of $y/x$ alone. …
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CBSE 2024Set 65/2/11 markMCQ
$\dfrac{dy}{dx} = F(x, y)$
›Reveal solutionSolution
A differential equation $\frac{dy}{dx} = F(x, y)$ is homogeneous if $F(x, y)$ is a homogeneous function of degree zero. This means $F(\lambda x, \lambda y) = F(x, y)$ for any non-zero $\lambda$ . Option (A) contains a term $\cos x$ , which prevents $F(x, y)$ from being homogeneous of degree zero, making it the correct answer.

To determine if a differential equation $\frac{dy}{dx} = F(x, y)$ is homogeneous, we need to understand what a homogeneous function is.

A function $F(x, y)$ is called a homogeneous function of degree $n$ if, for any non-zero constant $\lambda$ , the following condition holds:
$F(\lambda x, \lambda y) = \lambda^n F(x, y)$
For a differential equation $\frac{dy}{dx} = F(x, y)$ to be classified as a homogeneous differential equation, the function $F(x, y)$ must be a homogeneous function of degree zero. This means that when we replace $x$ with $\lambda x$ and $y$ with $\lambda y$ , the function $F(x, y)$ must remain unchanged:
$F(\lambda x, \lambda y) = \lambda^0 F(x, y) = F(x, y)$
This property is crucial because it allows us to transform the differential equation into a separable form by substituting $y = vx$ (or $x = vy$ ). If $F(x, y)$ is homogeneous of degree zero, it can always be expressed as a function of $\frac{y}{x}$ (or $\frac{x}{y}$ ). For example, if $F(\lambda x, \lambda y) = F(x, y)$ , we can choose $\lambda = \frac{1}{x}$ (assuming $x \neq 0$ ), then $F(x, y) = F\left(\frac{1}{x} \cdot x, \frac{1}{x} \cdot y\right) = F\left(1, \frac{y}{x}\right)$ , which is clearly a function of $\frac{y}{x}$ .

Let's examine each given option to see which $F(x, y)$ is not homogeneous of degree zero.

Option (A): $F(x, y) = \cos x - \sin\left(\dfrac{y}{x}\right)$ We test for homogeneity of degree zero by replacing $x$ with $\lambda x$ and $y$ with $\lambda y$ :

$F(\lambda x, \lambda y) = \cos(\lambda x) - \sin\left(\dfrac{\lambda y}{\lambda x}\right)$ $F(\lambda x, \lambda y) = \cos(\lambda x) - \sin\left(\dfrac{y}{x}\right)$
For this to be equal to $F(x, y)$, we would need $\cos(\lambda x) = \cos x$. This is generally not true for arbitrary $\lambda \neq 1$. For instance, if $\lambda = 2$, then $\cos(2x) \neq \cos x$. Therefore, $F(x, y) = \cos x - \sin\left(\dfrac{y}{x}\right)$ is **not** a homogeneous function of degree zero. This means the differential equation $\frac{dy}{dx} = \cos x - \sin\left(\dfrac{y}{x}\right)$ is not homogeneous.

2. Option (B): $F(x, y) = \dfrac{y}{x}$

Replace $x$ with $\lambda x$ and $y$ with $\lambda y$ :
$F(\lambda x, \lambda y) = \dfrac{\lambda y}{\lambda x} = \dfrac{y}{x}$
This is equal to $F(x, y)$. Thus, $F(x, y) = \dfrac{y}{x}$ is a homogeneous function of degree zero.

3. Option (C): $F(x, y) = \dfrac{x^2 + y^2}{xy}$

Replace $x$ with $\lambda x$ and $y$ with $\lambda y$ : …
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CBSE 2024Set 65/3/11 markMCQ
$x \log x\, \dfrac{dy}{dx} + y = 2 \log x$
›Reveal solutionSolution
The given equation can be rearranged into the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$ , making it a first order linear differential equation. The correct option is (C).

Let’s understand why this equation fits the first order linear category, and why it does not fit the others.

A differential equation is called first order linear if it can be written in the form:
$\frac{dy}{dx} + P(x) \, y = Q(x)$
where $P(x)$ and $Q(x)$ are functions of $x$ only. The key idea is that $y$ and $\frac{dy}{dx}$ appear only to the first power, and there is no product like $y \cdot \frac{dy}{dx}$ or $y^2$ .

Now, look at the given equation:
$x \log x \, \frac{dy}{dx} + y = 2 \log x$

Isolate $\frac{dy}{dx}$ Divide the entire equation by $x \log x$ (provided $x > 0$ and $x \neq 1$ , which is the natural domain for $\log x$ ):

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

Simplify the right-hand side Since $\frac{2 \log x}{x \log x} = \frac{2}{x}$ (cancelling $\log x$ ), we get:

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

Identify the form This is exactly $\frac{dy}{dx} + P(x) y = Q(x)$ with:

$P(x) = \frac{1}{x \log x}, \quad Q(x) = \frac{2}{x}$
Both are functions of $x$ only, and $y$ appears linearly. So it is a first order linear differential equation.

Now, why are the other options wrong?

Watch out

Common confusion

Variable separable: For separability, we need to write it as $f(y) \, dy = g(x) \, dx$ . Here, $y$ and $\frac{dy}{dx}$ are mixed — you cannot separate $y$ from $x$ completely because of the term $\frac{1}{x \log x} y$ . So it is not separable. …
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CBSE 2023Set ANNUAL1 markMCQ
$\dfrac{dy}{dx} = g\left(\dfrac{y}{x}\right)$
(a) $y = vx$
(b) $v = xy$
(c) $x = vy$
(d) $y = v$
›Reveal solutionSolution

A homogeneous equation $\dfrac{dy}{dx}=g\left(\dfrac{y}{x}\right)$ is solved by the standard substitution $y=vx$ .

For a homogeneous differential equation of the form

$\frac{dy}{dx} = g\left(\frac{y}{x}\right)$

the standard method is to substitute $y = vx$ , where $v$ is a function of $x$ . Then

$\frac{dy}{dx} = v + x\frac{dv}{dx}$ …

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CBSE 2019Set ANNUAL1 markMCQ
$Q. Which of the following is a homogeneous differential equation?$
(a) $x^2 y \, dx - (x^3 + y^3) dy = 0$
(b) $(xy) dx - (x^4 + y^4) dy = 0$
(c) $(2x + y - 3) dy - (x + 2y - 3) dx = 0$
(d) $(x - y) dy = (x^2 + y + 1) dx$
›Reveal solutionSolution

Option (a) is homogeneous (both coefficient functions are degree $3$ ).

…

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Q.Solve the following differential equation: $\left(1 + e^{y/x}\right) dy + e^{y/x}\left(1 - \dfrac{y}{x}\right) dx = 0 \quad (x \ne 0)$ .

Concept understanding — Homogeneous Differential Equation

Homogeneous Differential Equations

The idea: only the ratio $y/x$ matters

Why we care: it becomes separable

Q.Solve the following differential equation: (1+ey/x)dy+ey/x(1−yx)dx=0(x≠0)\left(1 + e^{y/x}\right) dy + e^{y/x}\left(1 - \dfrac{y}{x}\right) dx = 0 \quad (x \ne 0)(1+ey/x)dy+ey/x(1−xy​)dx=0(x=0).

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Q.Solve the following differential equation: $\left(1 + e^{y/x}\right) dy + e^{y/x}\left(1 - \dfrac{y}{x}\right) dx = 0 \quad (x \ne 0)$ .

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