Integrating factor of $x\frac{dy}{dx}-y=x^4-3x$ is: (A) $x$ (B) $\log x$ (C) $\frac{1}{x}$ (D) $-x$

Integrating factor of x(dy)/(dx)-y=x⁴-3x is: (A) x (B) log x (C) (1)/(x) (D) -x

NCERT Exemplar · Q65

x\frac{dy}{dx}-y=x^4-3x

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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 given differential equation is linear in $y$ but not in standard form. Dividing by $x$ gives $\frac{dy}{dx} - \frac{1}{x}y = x^3 - 3$ , so the integrating factor is $e^{\int -\frac{1}{x}dx} = \frac{1}{x}$ . The correct option is (C).

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

Here, the equation is $x\frac{dy}{dx} - y = x^4 - 3x$ . Notice the coefficient of $\frac{dy}{dx}$ is $x$ , not $1$ . So before we can identify $P(x)$ , we must first rewrite the equation in the standard form.

Rewrite in standard form Divide every term by $x$ (assuming $x \neq 0$ , which is fine for the integrating factor itself):

\frac{dy}{dx} - \frac{1}{x}y = x^3 - 3.

Now it matches $\frac{dy}{dx} + P(x)y = Q(x)$ with $P(x) = -\frac{1}{x}$ and $Q(x) = x^3 - 3$ .

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

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

We only need one antiderivative (the constant is irrelevant), so take:

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

For the purpose of solving, we usually drop the absolute value and take $\mu(x) = \frac{1}{x}$ (the sign is absorbed later if needed). So the integrating factor is $\frac{1}{x}$ . …

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Q.Integrating factor of $x\frac{dy}{dx}-y=x^4-3x$ is:
(A) $x$
(B) $\log x$
(C) $\frac{1}{x}$
(D) $-x$

Concept understanding — Integrating Factor Method

Integrating Factor Method

The Standard Form

The Integrating Factor

The Solution

A Quick Illustration

Method: Integrating Factor with a Negative $P(x)$

Steps

Common Mistakes

Q.Integrating factor of xdydx−y=x4−3xx\frac{dy}{dx}-y=x^4-3xxdxdy​−y=x4−3x is: (A) xxx (B) log⁡x\log xlogx (C) 1x\frac{1}{x}x1​ (D) −x-x−x

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Q.Integrating factor of $x\frac{dy}{dx}-y=x^4-3x$ is:
(A) $x$
(B) $\log x$
(C) $\frac{1}{x}$
(D) $-x$

Unlock everything free for 14 days