Solve: $y+\frac{d}{dx}(xy)=x(\sin x+\log x)$.

Solve: y+(d)/(dx)(xy)=x(sin x+log x).

NCERT Exemplar · Q25

y+\frac{d}{dx}(xy)=x(\sin x+\log x)

Gujarat GsebLong· 5mImportance★★★★★

Appeared in past exams:MHT-CET 2025· Set pcm-2025-04-22-M· 2mexact

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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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Expanding $\frac{d}{dx}(xy)$ turns the equation into the linear ODE $\frac{dy}{dx}+\frac{2}{x}y=\sin x+\log x$ (I.F. $x^2$ ), giving $y=-\cos x+\frac{2\sin x}{x}+\frac{2\cos x}{x^2}+\frac{x}{3}\log x-\frac{x}{9}+\frac{C}{x^2}$ .

Unpack the left side

The term $\frac{d}{dx}(xy)$ is a product-rule derivative: $\frac{d}{dx}(xy)=x\frac{dy}{dx}+y$ . Substituting,

y+x\frac{dy}{dx}+y=x(\sin x+\log x)\quad\Rightarrow\quad x\frac{dy}{dx}+2y=x(\sin x+\log x).

Put it in linear form

Divide by $x$ :

\frac{dy}{dx}+\frac{2}{x}y=\sin x+\log x.

Here $P=\frac{2}{x}$ , so the integrating factor is

\text{I.F.}=e^{\int \frac{2}{x}\,dx}=e^{2\log x}=x^2.

Collapse to a single derivative

Multiplying through by $x^2$ ,

\frac{d}{dx}(x^2y)=x^2\sin x+x^2\log x.

Integrate each piece

$\int x^2\sin x\,dx$ (integration by parts twice): $=-x^2\cos x+2x\sin x+2\cos x$ . …

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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 14 on this concept.

GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQ
$e^x\,dy + (ye^x+2x)\,dx = 0$
(a) $y\cdot e^x + x^2 = C$
(b) $y\cdot e^x - x^2 = C$
(c) $x\cdot e^x + y^2 = C$
(d) $x\cdot e^x - y^2 = C$
›Reveal solutionSolution

Divide through by $e^x$ to get a linear ODE in $y$ , then solve with an integrating factor of $1$ (equation is already exact after simplification).

$e^x dy+(ye^x+2x)dx=0\Rightarrow \dfrac{dy}{dx}+y=-2xe^{-x}$ . Integrating factor $=e^{\int1\,dx}=e^x$ .

…

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GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQ
$(\tan^{-1}y - x)\,dy = (1+y^2)\,dx$
(a) $e^{\tan^{-1}x}$
(b) $e^{\tan^{-1}y}$
(c) $e^{-\tan^{-1}x}$
(d) $e^{-\tan^{-1}y}$
›Reveal solutionSolution

Rewrite as a linear equation in $x$ (treating $y$ as the independent variable) and find its integrating factor.

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

…

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GUJCET 2025Set 031 markMCQ
$x \cdot \frac{dy}{dx} + 2y = x^2,\ (x \neq 0)$
›Reveal solutionSolution

Put in the form $\tfrac{dy}{dx}+P(x)y=Q(x)$ and use $\text{IF}=e^{\int P\,dx}$ .

Dividing by $x$ : $\tfrac{dy}{dx}+\tfrac{2}{x}y=x$ , so $P=\tfrac{2}{x}$ and …
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GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
$(1 - y^2)\dfrac{dx}{dy} + yx = ay$
(a) $\dfrac{1}{y^2 - 1}$
(b) $\dfrac{1}{1 - y^2}$
(c) $\dfrac{1}{\sqrt{y^2 - 1}}$
(d) $\dfrac{1}{\sqrt{1 - y^2}}$
›Reveal solutionSolution

Write the equation in standard linear form $\dfrac{dx}{dy}+Px=Q$ and compute $\text{I.F.}=e^{\int P\,dy}$ .

Divide by $(1-y^2)$ : $\dfrac{dx}{dy}+\dfrac{y}{1-y^2}x=\dfrac{ay}{1-y^2}$ , so $P=\dfrac{y}{1-y^2}$ .

Let $u=1-y^2$ , $du=-2y\,dy$ : $\displaystyle\int\dfrac{y}{1-y^2}dy=-\dfrac12\ln|1-y^2|$ .

…

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GUJCET 2024Set 131 markMCQ
$(\tan^{-1} y - x)\,dy = (1 + y^2)\,dx$
›Reveal solutionSolution

Treated as linear in $x$ , the integrating factor is $e^{\int \frac{dy}{1+y^2}} = e^{\tan^{-1}y}$ .

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

$\frac{dx}{dy} = \frac{\tan^{-1}y - x}{1+y^2} \ \Rightarrow\ \frac{dx}{dy} + \frac{x}{1+y^2} = \frac{\tan^{-1}y}{1+y^2}.$ …
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GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQ
$x\frac{dy}{dx} + 2y = x^2$
(a) $2\log x$
(b) $\log x$
(c) $\frac{2}{x}$
(d) $x^2$
›Reveal solutionSolution

Write the equation in standard linear form $\frac{dy}{dx}+Py=Q$ and use $IF=e^{\int P\,dx}$ .

Divide by $x$ : $\frac{dy}{dx}+\frac{2}{x}y=x$ , so $P=\frac{2}{x}$ .

…

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GUJCET 2023Set 091 markMCQ
$\frac{dy}{dx} + y\tan x = \sec x$
›Reveal solutionSolution
[!TLDR] Max $R$ = longest length ÷ smallest area, i.e. current along the 10 cm length entering the $1\times\tfrac12$ cm faces.

Concept

Resistance of a uniform conductor is $R = \dfrac{\rho L}{A}$ , where $L$ is the length along the current direction and $A$ is the perpendicular cross-sectional area. To maximise $R$ , choose the connection giving the greatest $L$ and the smallest $A$ .

Solution

The rod is $10\,\text{cm}\times1\,\text{cm}\times\tfrac12\,\text{cm}$ . The three ways to connect across opposite faces:

Across $10\times1$ cm faces: $L = \tfrac12$ cm, $A = 10$ cm $^2$ → $R \propto 0.5/10 = 0.05$ . …
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GUJCET 2022Set 081 markMCQ
$x\frac{dy}{dx}-y=x^2$
›Reveal solutionSolution

Divide by $x$ to get $\dfrac{dy}{dx}-\dfrac{y}{x}=x$ ; then $\text{IF}=e^{\int(-1/x)dx}=\tfrac1x$ .

Concept. $x\dfrac{dy}{dx}-y=x^2\Rightarrow\dfrac{dy}{dx}-\dfrac{1}{x}y=x$ . Here $P(x)=-\tfrac1x$ , so …
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GSEB Higher Secondary Certificate (HSC) Examination 2022Set ANNUAL1 markMCQ
$x\frac{dy}{dx} + 2y = x^2 \log x$
(a) $e^{2x}$
(b) $x^2$
(c) $e^x$
(d) $x$
›Reveal solutionSolution

Put in standard linear form and compute IF $= e^{\int P\,dx}$ .

$x\dfrac{dy}{dx}+2y = x^2\log x \Rightarrow \dfrac{dy}{dx} + \dfrac2x\,y = x\log x.$

…

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GUJCET 2020Set 071 markMCQ
$(\tan^{-1} y - x)\,dy = (1 + y^2)\,dx$
›Reveal solutionSolution

Treat $x$ as the dependent variable; the integrating factor is $e^{\int\frac{dy}{1+y^2}}=e^{\tan^{-1}y}$ .

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

$\frac{dx}{dy}=\frac{\tan^{-1}y-x}{1+y^2}\ \Rightarrow\ \frac{dx}{dy}+\frac{1}{1+y^2}\,x=\frac{\tan^{-1}y}{1+y^2}.$ …
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GSEB Higher Secondary Certificate (HSC) Examination 2020Set ANNUAL1 markMCQ
$y\,dx-(x+2y^2)\,dy=0$
(a) $-1/y$
(b) $-y$
(c) $y$
(d) $1/y$
›Reveal solutionSolution

Rearrange into the linear form $\dfrac{dx}{dy}+P(y)x=Q(y)$ (treating $x$ as the dependent variable), then compute $I.F.=e^{\int P\,dy}$ .

$y\,dx-(x+2y^2)dy=0 \Rightarrow y\,dx=(x+2y^2)dy \Rightarrow \dfrac{dx}{dy}=\dfrac{x}{y}+2y \Rightarrow \dfrac{dx}{dy}-\dfrac1y\,x=2y$ . …

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GUJCET 2019Set 171 markMCQ
$\dfrac{dy}{dx}(1+x) - xy = 1-x$
›Reveal solutionSolution
Reduce to $\dfrac{dy}{dx}-\dfrac{x}{1+x}y=\dfrac{1-x}{1+x}$ ; I.F. $=e^{\int P\,dx}$ .

Concept. Divide by $(1+x)$ : $P(x) = -\dfrac{x}{1+x}$ .

Steps.

$\displaystyle\int \frac{x}{1+x}dx = \int\left(1-\frac{1}{1+x}\right)dx = x-\log(1+x).$ …
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Q.Solve: $y+\frac{d}{dx}(xy)=x(\sin x+\log x)$ .

Concept understanding — Integrating Factor Method

Integrating Factor Method

The Standard Form

The Integrating Factor

The Solution

A Quick Illustration

Unpack the left side

Put it in linear form

Collapse to a single derivative

Integrate each piece

Method: Expanding a product-derivative, then applying the integrating factor

Steps

Common Mistakes

Concept

Solution

Q.Solve: y+ddx(xy)=x(sin⁡x+log⁡x)y+\frac{d}{dx}(xy)=x(\sin x+\log x)y+dxd​(xy)=x(sinx+logx).

Unpack the left side

Put it in linear form

Collapse to a single derivative

Integrate each piece

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Q.Solve: $y+\frac{d}{dx}(xy)=x(\sin x+\log x)$ .

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