(vii) The solution of $(1+x^2)\frac{dy}{dx}+2xy-4x^2=0$ is ______.

> [!TLDR] > This is a first-order linear differential equation. We rewrite it in standard form, find the integrating factor $1+x^2$, and integrate to get the general solution: $y(1+x^2) = \frac{4x^3}{3} + C$. The key to solving any first-order linear differential equation is recognising its standard form and using an integrating factor. Let's see why that works here. The given equation is $(1+x^2)\frac{dy}{dx} + 2xy - 4x^2 = 0$. This is linear in $y$ because $y$ and $\frac{dy}{dx}$ appear only to the first power, and there's no product like $y\frac{dy}{dx}$. The standard form for such an equation is $\frac{dy}{dx} + P(x)y = Q(x)$, where the coefficient of $\frac{dy}{dx}$ is 1. Once we have that, the integrating factor $\mu(x) = e^{\int P\,dx}$ makes the left side a perfect derivative, so we can integrate both sides directly. Let's work through it. 1. **Rewrite in standard form.** Divide every term by $1+x^2$ (which is never zero for real $x$, so no worries about losing solutions): $$ \frac{dy}{dx} + \frac{2x}{1+x^2}\,y = \frac{4x^2}{1+x^2}. $$ Here $P(x) = \frac{2x}{1+x^2}$ and $Q(x) = \frac{4x^2}{1+x^2}$. 2. **Find the integrating factor.** $$ \mu(x) = e^{\int P\,dx} = e^{\int \frac{2x}{1+x^2}\,dx}. $$ The integral $\int \frac{2x}{1+x^2}\,dx$ is a standard logarithmic form: let $u = 1+x^2$, then $du = 2x\,dx$, so the integral becomes $\int \frac{du}{u} = \log|1+x^2|$. Since $1+x^2 > 0$, we drop the absolute value: $$ \int \frac{2x}{1+x^2}\,dx = \log(1+x^2). $$ Hence, $$ \mu(x) = e^{\log(1+x^2)} = 1+x^2. $$ > [!TIP] > Notice that the integrating factor turned out to be exactly the original coefficient of $\frac{dy}{dx}$. This often happens when $P(x)$ is of the form $\frac{f'(x)}{f(x)}$ — a neat shortcut to spot. 3. **Multiply the standard-form equation by $\mu(x)$.** $$ (1+x^2)\frac{dy}{dx} + 2x\,y = 4x^2. $$ The left side is now the derivative of $\mu(x) \cdot y$: $$ \frac{d}{dx}\

(vii) The solution of (1+x²)(dy)/(dx)+2xy-4x²=0 is ___.

NCERT Exemplar · Q40

(1+x^2)\frac{dy}{dx}+2xy-4x^2=0

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Concept understanding — Integrating Factor Method

Integrating Factor Method

Some first-order differential equations refuse to separate — you cannot get all the $y$ 's on one side and all the $x$ 's on the other. The integrating factor method is the standard trick for a special (and very common) family of these: the linear first-order equation. The idea is beautifully simple: multiply the whole equation by one cleverly chosen function, and the messy left-hand side collapses into a single derivative that we can integrate directly.

The Standard Form

An equation is linear of first order if it can be written as

\frac{dy}{dx} + P\,y = Q

where $P$ and $Q$ are functions of $x$ alone (or constants). Notice $y$ and $\frac{dy}{dx}$ appear only to the first power, and never multiplied together — that is what "linear" means here.

Tip

Always rearrange into this exact shape first. The coefficient of $\frac{dy}{dx}$ must be $1$ before you read off $P$ and $Q$ .

The Integrating Factor

The magic multiplier is

\text{I.F.} = e^{\int P\,dx}.

Why this one? Multiply the equation by $e^{\int P\,dx}$ :

e^{\int P\,dx}\frac{dy}{dx} + P\,e^{\int P\,dx}\,y = Q\,e^{\int P\,dx}.

By the product rule, the left-hand side is exactly $\dfrac{d}{dx}\!\left(y\cdot e^{\int P\,dx}\right)$ , because the derivative of $e^{\int P\,dx}$ is $P\,e^{\int P\,dx}$ . So the equation becomes

\frac{d}{dx}\big(y\cdot \text{I.F.}\big) = Q\cdot \text{I.F.}

The left side is now a single derivative — that is the whole point of choosing this factor.

The Solution

Integrate both sides with respect to $x$ :

y\cdot \text{I.F.} = \int \big(Q\cdot \text{I.F.}\big)\,dx + C.

This is the general solution. In words: (solution) × (integrating factor) = integral of (Q × integrating factor), plus a constant.

A Quick Illustration

For $\dfrac{dy}{dx} + \dfrac{1}{x}\,y = x$ , we read $P = \tfrac1x$ , $Q = x$ . Then $\int P\,dx = \log x$ , so $\text{I.F.} = e^{\log x} = x$ . The solution is

y\cdot x = \int x\cdot x\,dx + C = \frac{x^3}{3} + C.

--- …

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This is a first-order linear differential equation. We rewrite it in standard form, find the integrating factor $1+x^2$ , and integrate to get the general solution: $y(1+x^2) = \frac{4x^3}{3} + C$ .

The key to solving any first-order linear differential equation is recognising its standard form and using an integrating factor. Let's see why that works here.

The given equation is $(1+x^2)\frac{dy}{dx} + 2xy - 4x^2 = 0$ . This is linear in $y$ because $y$ and $\frac{dy}{dx}$ appear only to the first power, and there's no product like $y\frac{dy}{dx}$ . The standard form for such an equation is $\frac{dy}{dx} + P(x)y = Q(x)$ , where the coefficient of $\frac{dy}{dx}$ is 1. Once we have that, the integrating factor $\mu(x) = e^{\int P\,dx}$ makes the left side a perfect derivative, so we can integrate both sides directly.

Let's work through it.

Rewrite in standard form. Divide every term by $1+x^2$ (which is never zero for real $x$ , so no worries about losing solutions):

\frac{dy}{dx} + \frac{2x}{1+x^2}\,y = \frac{4x^2}{1+x^2}.

Here $P(x) = \frac{2x}{1+x^2}$ and $Q(x) = \frac{4x^2}{1+x^2}$ .

Find the integrating factor.

\mu(x) = e^{\int P\,dx} = e^{\int \frac{2x}{1+x^2}\,dx}.

The integral $\int \frac{2x}{1+x^2}\,dx$ is a standard logarithmic form: let $u = 1+x^2$ , then $du = 2x\,dx$ , so the integral becomes $\int \frac{du}{u} = \log|1+x^2|$ . Since $1+x^2 > 0$ , we drop the absolute value:

\int \frac{2x}{1+x^2}\,dx = \log(1+x^2).

Hence,

\mu(x) = e^{\log(1+x^2)} = 1+x^2.

Tip

Notice that the integrating factor turned out to be exactly the original coefficient of $\frac{dy}{dx}$ . This often happens when $P(x)$ is of the form $\frac{f'(x)}{f(x)}$ — a neat shortcut to spot.

Multiply the standard-form equation by $\mu(x)$ .

(1+x^2)\frac{dy}{dx} + 2x\,y = 4x^2.

The left side is now the derivative of $\mu(x) \cdot y$ :

$\frac{d}{dx}\big[(1+x^2)y\big] = 4x^2.$ …

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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 15 on this concept.

KEAM 2026Set eng-2026-04174 marksMCQ
$x^3 \dfrac{dy}{dx} + 3x^2 y = \cos x$
›Reveal solutionSolution

Recognize the exact derivative $\dfrac{d}{dx}(x^3y)$ .

Since $\dfrac{d}{dx}(x^3y)=x^3\dfrac{dy}{dx}+3x^2y$ , the equation $x^3\dfrac{dy}{dx}+3x^2y=\cos x$ becomes

$\frac{d}{dx}(x^3y)=\cos x.$ …
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KEAM 2026Set eng-2026-04174 marksMCQ
$2\, dy = (y + \cos x)\, dx$
›Reveal solutionSolution

Put in standard linear form and read off the integrating factor.

From $2\,dy=(y+\cos x)\,dx$ :

$2\frac{dy}{dx}=y+\cos x\ \Rightarrow\ \frac{dy}{dx}-\frac12 y=\frac12\cos x.$ …
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KEAM 2026Set eng-2026-04184 marksMCQ
$\dfrac{dy}{dx} + \dfrac{2y}{20 - x} = 10$
›Reveal solutionSolution

The integrating factor is $e^{\int P\,dx}$ with $P=\dfrac{2}{20-x}$ , which evaluates to $\dfrac{1}{(20-x)^2}$ .

For $\dfrac{dy}{dx}+P\,y=Q$ with $P=\dfrac{2}{20-x}$ ,

$\int P\,dx=\int\dfrac{2}{20-x}\,dx=-2\log|20-x|.$ …
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KEAM 2026Set eng-2026-04224 marksMCQ
$(y + x^2)\, dx = x\, dy, x > 0$
›Reveal solutionSolution

Rewrite as a first-order linear ODE $y'-y/x=x$ , use integrating factor $1/x$ , then apply the point $(1,0)$ .

From $(y+x^2)dx=x\,dy$ : $x\frac{dy}{dx}=y+x^2\Rightarrow \frac{dy}{dx}-\frac{y}{x}=x$ .

Integrating factor $=e^{-\int dx/x}=\frac1x$ , so $\frac{d}{dx}\Big(\frac{y}{x}\Big)=1$ . …
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KEAM 2025Set eng-2025-04234 marksMCQ
$\sin x\, dy = \dfrac{1}{2}(\sin 2x + 2y\cos x)\, dx$
›Reveal solutionSolution
The equation reduces to y' - (cot x) y = cos x, giving integrating factor e^{-integral cot x dx} = csc x.

Concept and Intuition

Rewrite the equation in standard linear form dy/dx + P(x)y = Q(x); the integrating factor is e^{integral P dx}.

Step-by-Step Solution

sin x dy = (1/2)(sin2x + 2y cos x) dx; using sin2x = 2 sin x cos x, the RHS = (sin x cos x + y cos x) dx.

So sin x (dy/dx) = sin x cos x + y cos x, i.e. dy/dx = cos x + y cot x. …
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KEAM 2025Set eng-2025-04254 marksMCQ
$(1+x^2)dy=(1-2xy)dx$
›Reveal solutionSolution

Put in standard linear form $\dfrac{dy}{dx}+Py=Q$ and compute $e^{\int P\,dx}$ .

Rewrite $(1+x^{2})dy=(1-2xy)dx$ as
$\dfrac{dy}{dx}+\dfrac{2x}{1+x^{2}}\,y=\dfrac{1}{1+x^{2}}.$
Here $P=\dfrac{2x}{1+x^{2}}$ , so …
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KEAM 2025Set eng-2025-04274 marksMCQ
$\dfrac{dy}{dx} - 2y = 2x - 3$
›Reveal solutionSolution
[!TLDR]

With $P(x) = -2$ , the integrating factor $e^{\int P\,dx} = e^{-2x}$ .

Concept

A linear first-order differential equation in the form $\frac{dy}{dx} + P(x)y = Q(x)$ is solved by multiplying through by the integrating factor $\mu(x) = e^{\int P(x)\,dx}$ — the standard method in the NCERT/CBSE differential-equations chapter.

Solution

The equation is

$\frac{dy}{dx} - 2y = 2x - 3.$ …
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KEAM 2024Set eng-2024-06054 marksMCQ
$\left(1 + 2e^{-x}\right)\frac{dy}{dx} - 2e^{-x}y = 1 + e^{-x}$
›Reveal solutionSolution

$\int P\,dx=\log(1+2e^{-x})$ , so the integrating factor is $1+2e^{-x}$ .

Divide by $(1+2e^{-x})$ :
$\frac{dy}{dx}-\frac{2e^{-x}}{1+2e^{-x}}\,y=\frac{1+e^{-x}}{1+2e^{-x}},$
so $P(x)=-\dfrac{2e^{-x}}{1+2e^{-x}}$ . With $w=1+2e^{-x}$ , $dw=-2e^{-x}dx$ : …
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KEAM 2024Set eng-2024-06064 marksMCQ
$x\frac{dy}{dx}+2y=xe^{x}$
›Reveal solutionSolution

Dividing by $x$ gives $P(x)=\frac{2}{x}$ , so the integrating factor is $x^2$ .

Divide $x\frac{dy}{dx}+2y=xe^x$ by $x$ : $\frac{dy}{dx}+\frac{2}{x}y=e^x$ . …
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KEAM 2024Set eng-2024-06074 marksMCQ
$2y\tan x+\dfrac{dy}{dx}=5\sin x$
›Reveal solutionSolution

Solve the linear ODE with integrating factor $\sec^2 x$ .

Rewrite as $\frac{dy}{dx}+2\tan x\,y=5\sin x$ . Integrating factor:
$\mu=e^{\int 2\tan x\,dx}=e^{2\log\sec x}=\sec^2 x.$
Then

$\frac{d}{dx}\left(y\sec^2 x\right)=5\sin x\sec^2 x=5\frac{\sin x}{\cos^2 x}.$ …
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KEAM 2024Set eng-2024-06074 marksMCQ
$(3\sin x\cos x)\,dy=(1+3y\sin^2 x)\,dx$
›Reveal solutionSolution

Put in linear form; the coefficient of $y$ gives IF $=\cos x$ .

From $(3\sin x\cos x)dy=(1+3y\sin^2 x)dx$ :
$\frac{dy}{dx}=\frac{1+3y\sin^2 x}{3\sin x\cos x}=\frac{1}{3\sin x\cos x}+\frac{3\sin^2 x}{3\sin x\cos x}y.$
The $y$ -coefficient is $\frac{\sin x}{\cos x}=\tan x$ , so …
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KEAM 2024Set eng-2024-06094 marksMCQ
$\dfrac{dy}{dx} + y = 5$
›Reveal solutionSolution

The equation is separable: $\frac{dy}{5-y}=dx$ , which integrates to $-\log|5-y|=x+C$ .

Write $\frac{dy}{dx}=5-y$ , so $\frac{dy}{5-y}=dx$ . …
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←Previous questionQ39 — (vi) The solution of the differential equation x(dy)/(dx)+2y=x² is ___…→Next questionQ41 — (viii) The solution of the differential equation ydx+(x+xy)dy=0 is ___…

Q.(vii) The solution of $(1+x^2)\frac{dy}{dx}+2xy-4x^2=0$ is ______.

Concept understanding — Integrating Factor Method

Integrating Factor Method

The Standard Form

The Integrating Factor

The Solution

A Quick Illustration

Method: Recognising the Coefficient Itself as the Integrating Factor

Steps

Common Mistakes

Concept and Intuition

Step-by-Step Solution

Concept

Solution

Q.(vii) The solution of (1+x2)dydx+2xy−4x2=0(1+x^2)\frac{dy}{dx}+2xy-4x^2=0(1+x2)dxdy​+2xy−4x2=0 is ______.

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Q.(vii) The solution of $(1+x^2)\frac{dy}{dx}+2xy-4x^2=0$ is ______.

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