The solution of $x\frac{dy}{dx}+y=e^x$ is: (A) $y=\frac{e^x}{x}+\frac{k}{x}$ (B) $y=xe^x+cx$ (C) $y=xe^x+k$ (D) $x=\frac{e^y}{y}+\frac{k}{y}$

The solution of x(dy)/(dx)+y=ex is: (A) y=(ex)/(x)+(k)/(x) (B) y=xex+cx (C) y=xex+k (D) x=(ey)/(y)+(k)/(y)

NCERT Exemplar · Q80

x\frac{dy}{dx}+y=e^x

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Appeared in past exams:AP EAPCET 2025· Set eng-2025-05-21-FN· 1mrewordedKEAM 2025· Set eng-2025-0429· 4mexact

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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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This is a first-order linear ODE solved by rewriting it as $\frac{d}{dx}(xy) = e^x$ , then integrating both sides. The general solution is $y = \frac{e^x}{x} + \frac{k}{x}$ , which matches option (A).

The key insight here is that the left-hand side of the equation, $x\frac{dy}{dx} + y$ , looks like the derivative of a product. When you see a combination of a function and its derivative multiplied by the independent variable, your first instinct should be to check if it’s the result of the product rule in reverse.

Specifically, recall that $\frac{d}{dx}(x \cdot y) = x\frac{dy}{dx} + y \cdot 1$ . That’s exactly the left side of our equation. So the ODE is already in a “perfect differential” form — no need for an integrating factor.

Let’s work through it step by step.

Recognize the derivative form The given equation is:

x\frac{dy}{dx} + y = e^x

Notice that the left side is precisely $\frac{d}{dx}(xy)$ . So we can rewrite the entire equation as:

\frac{d}{dx}(xy) = e^x

Integrate both sides Integrating with respect to $x$ gives:

xy = \int e^x \, dx = e^x + k

where $k$ is the constant of integration.

Solve for $y$ Divide through by $x$ (assuming $x \neq 0$ ):

y = \frac{e^x}{x} + \frac{k}{x}

Match with the options …

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Q.The solution of $x\frac{dy}{dx}+y=e^x$ is:
(A) $y=\frac{e^x}{x}+\frac{k}{x}$
(B) $y=xe^x+cx$
(C) $y=xe^x+k$
(D) $x=\frac{e^y}{y}+\frac{k}{y}$

Concept understanding — Integrating Factor Method

Integrating Factor Method

The Standard Form

The Integrating Factor

The Solution

A Quick Illustration

Method: Spot a product-rule derivative before reaching for the I.F.

Steps

Common Mistakes

Q.The solution of xdydx+y=exx\frac{dy}{dx}+y=e^xxdxdy​+y=ex is: (A) y=exx+kxy=\frac{e^x}{x}+\frac{k}{x}y=xex​+xk​ (B) y=xex+cxy=xe^x+cxy=xex+cx (C) y=xex+ky=xe^x+ky=xex+k (D) x=eyy+kyx=\frac{e^y}{y}+\frac{k}{y}x=yey​+yk​

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Q.The solution of $x\frac{dy}{dx}+y=e^x$ is:
(A) $y=\frac{e^x}{x}+\frac{k}{x}$
(B) $y=xe^x+cx$
(C) $y=xe^x+k$
(D) $x=\frac{e^y}{y}+\frac{k}{y}$

Unlock everything free for 14 days