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)

Odisha ChseLong· 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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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

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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