(iv) $\frac{dy}{dx}+\frac{y}{x\log x}=\frac{1}{x}$ is an equation of the type ______.

(iv) (dy)/(dx)+(y)/(xlog x)=(1)/(x) is an equation of the type ___.

NCERT Exemplar · Q37

\frac{dy}{dx}+\frac{y}{x\log x}=\frac{1}{x}

Manipur CohsemShort· 1mImportance★★★★★

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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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The equation fits $\frac{dy}{dx}+P(x)\,y=Q(x)$ , so it is a first-order linear differential equation.

We are asked to classify

\frac{dy}{dx}+\frac{y}{x\log x}=\frac{1}{x}.

What makes an equation "linear"

A first-order equation is linear when it can be written as

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

where $P$ and $Q$ depend on $x$ only, and $y$ together with $\frac{dy}{dx}$ appear to the first power and are never multiplied by each other.

Match the pattern

Read off the coefficients directly:

P(x)=\frac{1}{x\log x},\qquad Q(x)=\frac{1}{x}.

Both are functions of $x$ alone, and $y$ occurs only linearly. So the equation is exactly of the linear type.

How such an equation is solved

The integrating factor is …

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Q.(iv) $\frac{dy}{dx}+\frac{y}{x\log x}=\frac{1}{x}$ is an equation of the type ______.

Concept understanding — Integrating Factor Method

Integrating Factor Method

The Standard Form

The Integrating Factor

The Solution

A Quick Illustration

What makes an equation "linear"

Match the pattern

How such an equation is solved

Method: Identifying the Type of a First-Order Differential Equation

Steps

Common Mistakes

Q.(iv) dydx+yxlog⁡x=1x\frac{dy}{dx}+\frac{y}{x\log x}=\frac{1}{x}dxdy​+xlogxy​=x1​ is an equation of the type ______.

What makes an equation "linear"

Match the pattern

How such an equation is solved

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Q.(iv) $\frac{dy}{dx}+\frac{y}{x\log x}=\frac{1}{x}$ is an equation of the type ______.

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