(xi) The integrating factor of $\frac{dy}{dx}+y=\frac{1+y}{x}$ is ______.

> [!TLDR] > The equation is not in standard linear form — rewriting it as $\frac{dy}{dx} + \left(1 - \frac{1}{x}\right)y = \frac{1}{x}$ reveals the integrating factor $e^{\int (1 - 1/x)\,dx} = \frac{e^x}{x}$. The Integrating Factor (IF) method is designed for first-order linear differential equations of the form $$\frac{dy}{dx} + P(x)\,y = Q(x).$$ The idea is to multiply through by a function $\mu(x)$ that turns the left-hand side into the derivative of $\mu(x) y$. That function is $\mu(x) = e^{\int P(x)\,dx}$. Here, the given equation is $$\frac{dy}{dx} + y = \frac{1+y}{x}.$$ It looks almost linear, but the right-hand side mixes $y$ with $x$. We must first rearrange it into the standard form. 1. **Rewrite the equation.** Expand the right-hand side: $$\frac{dy}{dx} + y = \frac{1}{x} + \frac{y}{x}.$$ 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 $y$ from the two middle terms: $$\frac{dy}{dx} + \left(1 - \frac{1}{x}\right)y = \frac{1}{x}.$$ Now it is in the standard linear form with $$P(x) = 1 - \frac{1}{x}, \quad Q(x) = \frac{1}{x}.$$ 3. **Find the integrating factor.** Compute $\int P(x)\,dx$: $$\int \left(1 - \frac{1}{x}\right)dx = x - \log|x| + C.$$ We only need one antiderivative (the constant is absorbed later), so take $$\int P(x)\,dx = x - \log|x|.$$ Then the integrating factor is $$\mu(x) = e^{\,x - \log|x|} = e^x \cdot e^{-\log|x|} = e^x \cdot \frac{1}{|x|}.$$ Since we usually work with positive $x$ in such problems (or take $x>0$ for simplicity), we drop the absolute value: $$\mu(x) = \frac{e^x}{x}.$$ > [!TIP] > A common shortcut: $e^{\,x - \log x} = \frac{e^x}{x}$ directly, because $e^{-\log x} = 1/x$ for $x>0$. …

(xi) The integrating factor of (dy)/(dx)+y=(1+y)/(x) is ___.

NCERT Exemplar · Q44

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

Rajasthan RbseShort· 1mImportance★★★★★

Appeared in past exams:MHT-CET 2024· Set pcm-2024-05-09-E· 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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The equation is not in standard linear form — rewriting it as $\frac{dy}{dx} + \left(1 - \frac{1}{x}\right)y = \frac{1}{x}$ reveals the integrating factor $e^{\int (1 - 1/x)\,dx} = \frac{e^x}{x}$ .

The Integrating Factor (IF) method is designed for first-order linear differential equations of the form

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

The idea is to multiply through by a function $\mu(x)$ that turns the left-hand side into the derivative of $\mu(x) y$ . That function is $\mu(x) = e^{\int P(x)\,dx}$ .

Here, the given equation is

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

It looks almost linear, but the right-hand side mixes $y$ with $x$ . We must first rearrange it into the standard form.

Rewrite the equation. Expand the right-hand side:

\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 $y$ from the two middle terms:

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

Now it is in the standard linear form with

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

Find the integrating factor. Compute $\int P(x)\,dx$ :

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

We only need one antiderivative (the constant is absorbed later), so take

\int P(x)\,dx = x - \log|x|.

Then the integrating factor is

\mu(x) = e^{\,x - \log|x|} = e^x \cdot e^{-\log|x|} = e^x \cdot \frac{1}{|x|}.

Since we usually work with positive $x$ in such problems (or take $x>0$ for simplicity), we drop the absolute value:

\mu(x) = \frac{e^x}{x}.

Tip

A common shortcut: $e^{\,x - \log x} = \frac{e^x}{x}$ directly, because $e^{-\log x} = 1/x$ for $x>0$ . …

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Q.(xi) The integrating factor of $\frac{dy}{dx}+y=\frac{1+y}{x}$ is ______.

Concept understanding — Integrating Factor Method

Integrating Factor Method

The Standard Form

The Integrating Factor

The Solution

A Quick Illustration

Method: Finding the Integrating Factor After Collecting the $y$ -Terms

Steps

Common Mistakes

Q.(xi) The integrating factor of dydx+y=1+yx\frac{dy}{dx}+y=\frac{1+y}{x}dxdy​+y=x1+y​ is ______.

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Q.(xi) The integrating factor of $\frac{dy}{dx}+y=\frac{1+y}{x}$ is ______.

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