Solve the differential equation $(1+y^2)\tan^{-1}x\,dx+2y(1+x^2)\,dy=0$.

Solve the differential equation (1+y²)tan⁻¹xdx+2y(1+x²)dy=0.

NCERT Exemplar · Q23

(1+y^2)\tan^{-1}x\,dx+2y(1+x^2)\,dy=0

Telangana TsbieShort· 3mImportance★★★★★

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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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Separating variables gives $(\tan^{-1}x)^2+2\log(1+y^2)=C$ .

Why separable

Divide the equation by $(1+x^2)(1+y^2)$ : the $dx$ -term then carries only $x$ and the $dy$ -term only $y$ , so each variable can sit on its own side.

Separate

(1+y^2)\tan^{-1}x\,dx=-2y(1+x^2)\,dy

\frac{\tan^{-1}x}{1+x^2}\,dx=-\frac{2y}{1+y^2}\,dy.

Integrate each side

Left: with $u=\tan^{-1}x$ , $du=\dfrac{dx}{1+x^2}$ , so $\displaystyle\int u\,du=\frac{(\tan^{-1}x)^2}{2}$ . …

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

TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
$\dfrac{dy}{dx} = \dfrac{2y^2 + 1}{2y^3 - 4xy + y}$
›Reveal solutionSolution

Treat $x$ as a function of $y$ ; the equation becomes linear in $x$ , giving $4xy^2 + 2x = y^4 + y^2 + c$ — option (A).

Step-by-step solution

Given $\dfrac{dy}{dx} = \dfrac{2y^2 + 1}{2y^3 - 4xy + y}$ , invert to make $x$ the dependent variable:
$\frac{dx}{dy} = \frac{2y^3 - 4xy + y}{2y^2 + 1}.$
Separate the $x$ -term:
$\frac{dx}{dy} + \frac{4y}{2y^2+1}\,x = \frac{2y^3 + y}{2y^2+1} = \frac{y(2y^2+1)}{2y^2+1} = y.$
This is linear in $x$ . Integrating factor:
$\mu = e^{\int \frac{4y}{2y^2+1}\,dy} = e^{\log(2y^2+1)} = 2y^2 + 1.$
Then …
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TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
$\left( x \sin \frac{y}{x} \right) dy = \left( y \sin \frac{y}{x} - x \right) dx$
›Reveal solutionSolution

Homogeneous DE; put $y=vx$ , separate variables, and integrate to get $\cos\!\big(\tfrac{y}{x}\big)=\log|x|+c$ .

Write in standard form.
$\left(x\sin\frac{y}{x}\right)dy=\left(y\sin\frac{y}{x}-x\right)dx \;\Rightarrow\; \frac{dy}{dx}=\frac{y\sin\frac{y}{x}-x}{x\sin\frac{y}{x}}.$
This is homogeneous (degree 0 in $x,y$ ).

Substitute $y=vx$ , so $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$ and $\dfrac{y}{x}=v$ :
$v+x\frac{dv}{dx}=\frac{vx\sin v-x}{x\sin v}=v-\frac{1}{\sin v}.$
Separate variables. …
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TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
$x^2 dy - (xy - y^2) dx = 0$
›Reveal solutionSolution

The equation is homogeneous; the substitution $y = vx$ integrates to $y\log x = x + cy$ — option (C).

Rewrite the equation $x^2\,dy - (xy - y^2)\,dx = 0$ as
$\frac{dy}{dx} = \frac{xy - y^2}{x^2} = \frac{y}{x} - \left(\frac{y}{x}\right)^2.$
This is homogeneous. Put $y = vx$ , so $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$ :
$v + x\frac{dv}{dx} = v - v^2 \quad\Rightarrow\quad x\frac{dv}{dx} = -v^2.$
Separate variables:
$\frac{dv}{v^2} = -\frac{dx}{x} \quad\Rightarrow\quad -\frac{1}{v} = -\log x + c_1.$
Since $\dfrac{1}{v} = \dfrac{x}{y}$ : …
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TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
$\dfrac{dy}{dx} = \dfrac{xy + x - 2y - 2}{xy - 2x + y - 2}$
›Reveal solutionSolution

The equation separates into (y−2)/(y+1)dy = (x−2)/(x+1)dx, integrating to x − y + 3log|(y+1)/(x+1)| = c.

Numerator xy+x−2y−2 = (x−2)(y+1); denominator xy−2x+y−2 = (x+1)(y−2). Separating variables: ∫(y−2)/(y+1)dy = ∫(x−2)/(x+1)dx. Each integrand is 1 − 3/(·+1), giving y − 3ln|y+1| = x − 3ln|x+1| + c. Rearrangin …
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Q.Solve the differential equation $(1+y^2)\tan^{-1}x\,dx+2y(1+x^2)\,dy=0$ .

Concept understanding — Separation Of Variables

Separation of Variables: From Intuition to Precision

The Core Intuition

The Precise Statement

Why This Works

A Concrete Example

Why separable

Separate

Integrate each side

Method: Separable DE with recognisable $du$ patterns

Steps

Common Mistakes

Step-by-step solution

Q.Solve the differential equation (1+y2)tan⁡−1x dx+2y(1+x2) dy=0(1+y^2)\tan^{-1}x\,dx+2y(1+x^2)\,dy=0(1+y2)tan−1xdx+2y(1+x2)dy=0.

Why separable

Separate

Integrate each side

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Q.Solve the differential equation $(1+y^2)\tan^{-1}x\,dx+2y(1+x^2)\,dy=0$ .

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