Solve the differential equation $\frac{dy}{dx}+2xy=y$.

Solve the differential equation (dy)/(dx)+2xy=y.

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\frac{dy}{dx}+2xy=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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This is a first-order linear ODE solved by separation of variables. The general solution is $y = Ce^{x - x^2}$ , where $C$ is an arbitrary constant.

The equation $\frac{dy}{dx} + 2xy = y$ looks like it might need an integrating factor — but before jumping into that, notice something: the $y$ on the right-hand side can be brought over to the left. That gives us a chance to factor $y$ out entirely. When you can write the derivative in terms of $y$ times something, separation of variables is often the cleanest path.

Let’s rewrite it:

\frac{dy}{dx} + 2xy - y = 0 \quad \Rightarrow \quad \frac{dy}{dx} + y(2x - 1) = 0.

Now it’s clear: the derivative of $y$ plus $y$ times a function of $x$ equals zero. That’s a separable equation.

Separate the variables. Move the $y$ term to the other side:

\frac{dy}{dx} = -y(2x - 1).

Divide both sides by $y$ (assuming $y \neq 0$ for now; we’ll check the zero case later) and multiply by $dx$ :

\frac{1}{y}\, dy = -(2x - 1)\, dx.

Integrate both sides. The left integrates to $\log|y|$ , the right is a simple polynomial:

\int \frac{1}{y}\, dy = \int -(2x - 1)\, dx.

\log|y| = -x^2 + x + C_1,

where $C_1$ is the constant of integration.

Solve for $y$ . Exponentiate both sides to remove the logarithm:

|y| = e^{-x^2 + x + C_1} = e^{C_1} e^{x - x^2}.

Let $C = \pm e^{C_1}$ (or simply an arbitrary constant, since $e^{C_1} > 0$ and the $\pm$ absorbs the absolute value). Then:

y = C e^{x - x^2}.

Check the special case $y = 0$ . …

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

Showing the 12 most recent of 44 on this concept.

CBSE 2026Set 65/2/11 markMCQ
$\frac{dy}{dx} = \frac{\sqrt{y}}{\sqrt{x}}$
›Reveal solutionSolution
This is a separable first-order ODE. Separate variables, integrate, and simplify to get $\sqrt{y} - \sqrt{x} = C$ , which is option (C).

The key idea: whenever you see $\frac{dy}{dx}$ expressed as a ratio of functions of $y$ and $x$ alone, you can "separate" them — move all $y$ terms to one side and all $x$ terms to the other — then integrate each side independently. That's exactly what we have here: $\frac{dy}{dx} = \frac{\sqrt{y}}{\sqrt{x}}$ .

Notice that $\sqrt{y}$ depends only on $y$ , and $\sqrt{x}$ only on $x$ . So the equation is already in separable form — we just need to rearrange it properly.

Separate the variables. Multiply both sides by $dx$ and divide by $\sqrt{y}$ :

$\frac{dy}{\sqrt{y}} = \frac{dx}{\sqrt{x}}$
This is valid as long as $x > 0$ and $y > 0$ (so the square roots are defined and non-zero).

Integrate both sides. Each side is a standard power integral:

$\int y^{-1/2}\, dy = \int x^{-1/2}\, dx$ $\frac{y^{1/2}}{1/2} = \frac{x^{1/2}}{1/2} + C_1$
which simplifies to:
$2\sqrt{y} = 2\sqrt{x} + C_1$

Simplify the constant. Divide through by 2:

$\sqrt{y} = \sqrt{x} + \frac{C_1}{2}$
Let $C = \frac{C_1}{2}$ (just renaming the arbitrary constant). Then:
$\sqrt{y} - \sqrt{x} = C$
Watch out

A common mistake is to forget the constant of integration or to combine the two integration constants incorrectly. When you integrate both sides, you get a constant on each side — but they can be merged into a single constant. Also, don't lose the factor of 2 from the power rule: $\int u^{-1/2}\, du = 2u^{1/2}$ , not $u^{1/2}$ . …
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CBSE 2026Set 65/3/11 markMCQ
$\frac{dy}{dx} = e^{x+y}$
›Reveal solutionSolution
Separate variables in $\frac{dy}{dx} = e^{x+y}$ to find the general solution $e^x + e^{-y} = C$ ; the particular solution $e^x + e^{-y} = -2$ is impossible because the left side is always positive while the right is negative.

The differential equation $\frac{dy}{dx} = e^{x+y}$ is separable. The key insight is to rewrite the exponential sum in the exponent as a product: $e^{x+y} = e^x \cdot e^y$ . This lets us collect all $x$ -terms with $dx$ and all $y$ -terms with $dy$ .

Solving the differential equation

Separate the variables.

$\frac{dy}{dx} = e^x \cdot e^y$
Rearranging:
$\frac{dy}{e^y} = e^x \, dx$
or equivalently,
$e^{-y} \, dy = e^x \, dx$

Integrate both sides.

$\int e^{-y} \, dy = \int e^x \, dx$
The left side gives $-e^{-y}$ and the right gives $e^x$ :
$-e^{-y} = e^x + C_1$
where $C_1$ is an arbitrary constant.

Rearrange to standard form. Multiply through by $-1$ :

$e^{-y} = -e^x - C_1$
or equivalently,
$e^x + e^{-y} = -C_1$
Renaming $-C_1$ as $C$ (still an arbitrary constant):
$e^x + e^{-y} = C$
This is the general solution, so Reason (R) is true.

Checking the particular solution

Now examine the proposed particular solution $e^x + e^{-y} = -2$ .

For this to be valid, we need $C = -2$ in the general solution. But notice:

$e^x > 0$ for all real $x$ …
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CBSE 2026Set CX1 markMCQ
$\dfrac{dy}{dx}=e^{x+y}$
(a) $e^{-y}=e^{x}+c$
(b) $e^{x}+e^{-y}=c$
(c) $e^{-x}-e^{-y}=c$
(d) $e^{-x}+e^{-y}=c$
›Reveal solutionSolution

The equation is variable-separable; separating and integrating gives $e^{x}+e^{-y}=c$ — option (b).

Why separate? Since $e^{x+y}=e^{x}\,e^{y}$ , the right side factors into an $x$ -part and a $y$ -part, so the variables separate cleanly.

$\dfrac{dy}{dx}=e^{x}\,e^{y}\;\Rightarrow\;e^{-y}\,dy=e^{x}\,dx$

…

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CBSE 2026Set A1 markMCQ
$\frac{dy}{dx} = e^{x+y}$
(a) $e^x + e^{-y} = k$
(b) $e^x + e^y = k$
(c) $e^{-x} + e^y = k$
(d) $e^{-x} + e^{-y} = k$
›Reveal solutionSolution

Separate variables in $\tfrac{dy}{dx}=e^{x+y}$ : $\int e^{-y}dy=\int e^x dx\Rightarrow e^x+e^{-y}=k$ .

Write $\dfrac{dy}{dx}=e^{x+y}=e^x\cdot e^y$ and separate:

$e^{-y}\,dy=e^x\,dx$ .

…

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CBSE 2026Set A1 markMCQ
$x\frac{dy}{dx} = \cot y$
(a) $x\cos y = k$
(b) $x\tan y = k$
(c) $x\sec y = k$
(d) $x\sin y = k$
›Reveal solutionSolution

Separate variables: $\int\tan y\,dy=\int\tfrac{dx}{x}\Rightarrow \log|\sec y|=\log|x|+c\Rightarrow x\cos y=k$ .

From $x\dfrac{dy}{dx}=\cot y$ , separate:

$\dfrac{dy}{\cot y}=\dfrac{dx}{x}\Rightarrow \tan y\,dy=\dfrac{dx}{x}$ .

…

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CBSE 2026Set ANNUAL1 mark
$\frac{dy}{dx} = \frac{1+y^2}{1+x^2}$
›Reveal solutionSolution

Separate variables and integrate both sides using the standard integral of $1/(1+t^2)$ .

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

…

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CBSE 2026Set ANNUAL1 markMCQ
$\dfrac{dy}{dx} = e^{x+y}$
(a) $e^x + e^{-y} = c$
(b) $e^x + e^y = c$
(c) $e^{-x} + e^y = c$
(d) $e^{-x} + e^{-y} = c$
›Reveal solutionSolution

Separate variables using $e^{x+y}=e^x\cdot e^y$ and integrate both sides.

$\dfrac{dy}{dx}=e^{x+y}=e^x\cdot e^y$

Separating variables: $e^{-y}\,dy=e^x\,dx$

…

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CBSE 2026Set ANNUAL1 mark
$Q.Find the general solution of the differential equation \frac{dy}{dx} = (1 + x^2)(1 + y^2).$
›Reveal solutionSolution
$\tan^{-1}y=x+\dfrac{x^3}{3}+c$ .

Concept. A separable differential equation $\dfrac{dy}{dx}=g(x)h(y)$ is solved by collecting all $y$ -terms on one side and all $x$ -terms on the other, then integrating both sides.

Steps.

$\dfrac{dy}{dx}=(1+x^2)(1+y^2)$ .

Separate: $\dfrac{dy}{1+y^2}=(1+x^2)\,dx$ . …
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CBSE 2026Set ANNUAL1 markMCQ
$(x^{2}+1)\dfrac{dy}{dx}=1,\ y(1)=\dfrac{\pi}{2}$
(a) $y=\tan^{-1}x+\dfrac{\pi}{3}$
(b) $y=\tan^{-1}x$
(c) $y=\tan^{-1}x+\dfrac{\pi}{6}$
(d) $y=\tan^{-1}x+\dfrac{\pi}{4}$
›Reveal solutionSolution

Integrating gives $y=\tan^{-1}x+C$ ; the condition fixes $C=\tfrac{\pi}{4}$ .

Step 1: $(x^{2}+1)\dfrac{dy}{dx}=1\Rightarrow dy=\dfrac{dx}{x^{2}+1}$ .

Step 2: Integrating, $y=\tan^{-1}x+C$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$\dfrac{y\,dx - x\,dy}{y} = 0$
(a) $y = cx$
(b) $x = cy^2$
(c) $xy = c$
(d) $y = cx^2$
›Reveal solutionSolution

Separate variables and integrate.

$\dfrac{y\,dx-x\,dy}{y}=0 \implies y\,dx = x\,dy \implies \dfrac{dx}{x}=\dfrac{dy}{y}$

…

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CBSE 2025Set ANNUAL1 markMCQ
$\dfrac{dy}{dx} = e^{x+y}$
(a) $e^x + e^y = C$
(b) $e^{-x} + e^y = C$
(c) $e^x + e^{-y} = C$
(d) $e^{-x} + e^{-y} = C$
›Reveal solutionSolution

Separate variables ( $e^{-y}dy = e^x dx$ ) and integrate both sides.

$\dfrac{dy}{dx} = e^{x+y} = e^x\cdot e^y$

Separating variables:

$e^{-y}\,dy = e^x\,dx$

Integrating both sides: …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The general solution of the differential equation dy/dx = e^(x+y) is$
(a) eˣ+eʸ=c
(b) eˣ+e⁻ʸ=c
(c) e⁻ˣ+eʸ=c
(d) e⁻ˣ+e⁻ʸ=c
›Reveal solutionSolution

Separate variables using $e^{x+y}=e^x\cdot e^y$ , then integrate both sides.

Given $\dfrac{dy}{dx} = e^{x+y} = e^x\cdot e^y$ . Separate variables:

$e^{-y}\,dy = e^x\,dx$

Integrate both sides:

$\int e^{-y}\,dy = \int e^x\,dx \;\Rightarrow\; -e^{-y} = e^x + C_1$

…

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←Previous questionQ4 — Solve the differential equation (x²-1)(dy)/(dx)+2xy=(1)/(x²-1).→Next questionQ6 — Find the general solution of (dy)/(dx)+ay=emx.

Q.Solve the differential equation $\frac{dy}{dx}+2xy=y$ .

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: Separating a first-order equation by factoring the derivative

Steps

Common Mistakes

Solving the differential equation

Checking the particular solution

Q.Solve the differential equation dydx+2xy=y\frac{dy}{dx}+2xy=ydxdy​+2xy=y.

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Q.Solve the differential equation $\frac{dy}{dx}+2xy=y$ .

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