The integrating factor of the differential equation $\frac{dy}{dx}+y=\frac{1+y}{x}$ is: (A) $\frac{x}{e^x}$ (B) $\frac{e^x}{x}$ (C) $xe^x$ (D) $e^x$

> [!TLDR] > The differential equation is first rewritten in standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$. After simplification, $P(x) = 1 - \frac{1}{x}$, so the integrating factor is $\mu = e^{\int (1 - 1/x)\,dx} = \frac{e^x}{x}$. The correct option is **(B)**. The Integrating Factor (IF) method is the go-to tool for solving first-order linear differential equations of the form $\frac{dy}{dx} + P(x)y = Q(x)$. The idea is simple: multiply the entire equation by a specially chosen function $\mu(x)$ so that the left-hand side becomes the exact derivative of $\mu(x) y$. That function is $\mu = e^{\int P\,dx}$. But here, the given equation isn't in that clean form yet. It has a $y$ term on the right-hand side mixed with a constant. So the first job is to rearrange it into the standard linear shape. 1. **Rewrite the equation** Start with: $$\frac{dy}{dx} + y = \frac{1+y}{x}$$ The right-hand side is $\frac{1}{x} + \frac{y}{x}$. So: $$\frac{dy}{dx} + y = \frac{1}{x} + \frac{y}{x}$$ 2. **Bring all $y$ terms to the left** Subtract $\frac{y}{x}$ from both sides: $$\frac{dy}{dx} + y - \frac{y}{x} = \frac{1}{x}$$ Factor $y$: $$\frac{dy}{dx} + y\left(1 - \frac{1}{x}\right) = \frac{1}{x}$$ Now it's in the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$, with: $$P(x) = 1 - \frac{1}{x}, \quad Q(x) = \frac{1}{x}$$ 3. **Find the integrating factor** The formula is $\mu(x) = e^{\int P(x)\,dx}$. Compute: $$\int P(x)\,dx = \int \left(1 - \frac{1}{x}\right)dx = x - \log|x| + C$$ We only need one integrating factor, so take $C = 0$: $$\mu = e^{x - \log x} = e^x \cdot e^{-\log x} = e^x \cdot \frac{1}{x} = \frac{e^x}{x}$$ > [!TIP] > Remember: $e^{-\log x} = \frac{1}{x}$ because $e^{\log(1/x)} = 1/x$. This is a common simplification that saves time. 4. **Verify (optional but good practice)** Multiply the standard-form equation by $\mu$: …

The integrating factor of the differential equation (dy)/(dx)+y=(1+y)/(x) is: (A) (x)/(ex) (B) (ex)/(x) (C) xe

NCERT Exemplar · Q77

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

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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 differential equation is first rewritten in standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$ . After simplification, $P(x) = 1 - \frac{1}{x}$ , so the integrating factor is $\mu = e^{\int (1 - 1/x)\,dx} = \frac{e^x}{x}$ . The correct option is (B).

The Integrating Factor (IF) method is the go-to tool for solving first-order linear differential equations of the form $\frac{dy}{dx} + P(x)y = Q(x)$ . The idea is simple: multiply the entire equation by a specially chosen function $\mu(x)$ so that the left-hand side becomes the exact derivative of $\mu(x) y$ . That function is $\mu = e^{\int P\,dx}$ .

But here, the given equation isn't in that clean form yet. It has a $y$ term on the right-hand side mixed with a constant. So the first job is to rearrange it into the standard linear shape.

Rewrite the equation Start with:

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

The right-hand side is $\frac{1}{x} + \frac{y}{x}$ . So:

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

Bring all $y$ terms to the left Subtract $\frac{y}{x}$ from both sides:

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

Factor $y$ :

\frac{dy}{dx} + y\left(1 - \frac{1}{x}\right) = \frac{1}{x}

Now it's in the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$ , with:

P(x) = 1 - \frac{1}{x}, \quad Q(x) = \frac{1}{x}

Find the integrating factor The formula is $\mu(x) = e^{\int P(x)\,dx}$ . Compute:

\int P(x)\,dx = \int \left(1 - \frac{1}{x}\right)dx = x - \log|x| + C

We only need one integrating factor, so take $C = 0$ :

\mu = e^{x - \log x} = e^x \cdot e^{-\log x} = e^x \cdot \frac{1}{x} = \frac{e^x}{x}

Tip

Remember: $e^{-\log x} = \frac{1}{x}$ because $e^{\log(1/x)} = 1/x$ . This is a common simplification that saves time.

Verify (optional but good practice) Multiply the standard-form equation by $\mu$ : …

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Q.The integrating factor of the differential equation $\frac{dy}{dx}+y=\frac{1+y}{x}$ is:
(A) $\frac{x}{e^x}$
(B) $\frac{e^x}{x}$
(C) $xe^x$
(D) $e^x$

Concept understanding — Integrating Factor Method

Integrating Factor Method

The Standard Form

The Integrating Factor

The Solution

A Quick Illustration

Method: Integrating factor for a linear first-order equation (after rearranging)

Steps

Common Mistakes

Q.The integrating factor of the differential equation dydx+y=1+yx\frac{dy}{dx}+y=\frac{1+y}{x}dxdy​+y=x1+y​ is: (A) xex\frac{x}{e^x}exx​ (B) exx\frac{e^x}{x}xex​ (C) xexxe^xxex (D) exe^xex

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Q.The integrating factor of the differential equation $\frac{dy}{dx}+y=\frac{1+y}{x}$ is:
(A) $\frac{x}{e^x}$
(B) $\frac{e^x}{x}$
(C) $xe^x$
(D) $e^x$

Unlock everything free for 14 days