(i) Integrating factor of the differential equation of the form $\frac{dx}{dy}+p_1 x=Q_1$ is given by $e^{\int p_1\,dy}$. (State True or False.)

(i) Integrating factor of the differential equation of the form (dx)/(dy)+p₁ x=Q₁ is given by e∫ p₁dy. (State

NCERT Exemplar · Q45

\frac{dx}{dy}+p_1 x=Q_1

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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 statement is True — the integrating factor for a first-order linear ODE in $x$ as a function of $y$ is indeed $e^{\int p_1\,dy}$ , exactly analogous to the $y$ -as-function-of- $x$ case.

The question tests whether you recognise the standard form of a first-order linear differential equation and its integrating factor — but with the roles of $x$ and $y$ swapped.

1. Recall the standard form

When we write a first-order linear ODE with $y$ as a function of $x$ , the form is:

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

and the integrating factor (I.F.) is $e^{\int P(x)\,dx}$ .

The logic: multiplying both sides by this factor turns the left-hand side into the derivative of $(y \cdot \text{I.F.})$ , making the equation directly integrable.

2. What happens when $x$ is the dependent variable?

The given form is:

\frac{dx}{dy} + p_1\,x = Q_1

Here $x$ is a function of $y$ , and $p_1$ and $Q_1$ are functions of $y$ (or constants). This is exactly the same structure — just with $x$ and $y$ swapped.

So the integrating factor becomes:

\text{I.F.} = e^{\int p_1\,dy}

Tip

Don’t memorise two separate rules. The integrating factor is always $e^{\int (\text{coefficient of dependent variable})\,d(\text{independent variable})}$ . Here the independent variable is $y$ , so we integrate with respect to $y$ .

3. Why does this work?

Multiply the equation by $e^{\int p_1\,dy}$ :

e^{\int p_1\,dy}\,\frac{dx}{dy} + p_1\,e^{\int p_1\,dy}\,x = Q_1\,e^{\int p_1\,dy}

Notice that the left-hand side is exactly: …

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Q.(i) Integrating factor of the differential equation of the form $\frac{dx}{dy}+p_1 x=Q_1$ is given by $e^{\int p_1\,dy}$ . (State True or False.)

Concept understanding — Integrating Factor Method

Integrating Factor Method

The Standard Form

The Integrating Factor

The Solution

A Quick Illustration

1. Recall the standard form

2. What happens when $x$ is the dependent variable?

3. Why does this work?

Method: Integrating Factor for the $\frac{dx}{dy}$ -Linear Form

Steps

Q.(i) Integrating factor of the differential equation of the form dxdy+p1x=Q1\frac{dx}{dy}+p_1 x=Q_1dydx​+p1​x=Q1​ is given by e∫p1 dye^{\int p_1\,dy}e∫p1​dy. (State True or False.)

1. Recall the standard form

2. What happens when xxx is the dependent variable?

3. Why does this work?

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Q.(i) Integrating factor of the differential equation of the form $\frac{dx}{dy}+p_1 x=Q_1$ is given by $e^{\int p_1\,dy}$ . (State True or False.)

2. What happens when $x$ is the dependent variable?

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