Find the general solution of $\frac{dy}{dx}+ay=e^{mx}$.

Find the general solution of (dy)/(dx)+ay=emx.

NCERT Exemplar · Q6

\frac{dy}{dx}+ay=e^{mx}

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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 using the integrating factor method. The general solution is $y = \frac{e^{mx}}{a+m} + Ce^{-ax}$ when $a \neq -m$ , and $y = x e^{-ax} + Ce^{-ax}$ when $a = -m$ .

The equation $\frac{dy}{dx} + a y = e^{mx}$ is a classic first-order linear ordinary differential equation. The structure is: derivative of $y$ plus a constant times $y$ equals an exponential. The key idea is to multiply both sides by an integrating factor — a function that turns the left-hand side into the derivative of a product. This works because the left side already looks like the result of a product rule if we choose the right factor.

Why does this work? For any linear ODE of the form $\frac{dy}{dx} + P(x) y = Q(x)$ , the integrating factor is $\mu(x) = e^{\int P(x) dx}$ . Here $P(x) = a$ (constant), so $\mu(x) = e^{\int a \, dx} = e^{ax}$ . Multiplying through gives $\frac{d}{dx}(y e^{ax}) = e^{(a+m)x}$ , which is straightforward to integrate.

Let’s go step by step.

Identify the standard form and find the integrating factor.

The equation is already in the form $\frac{dy}{dx} + a y = e^{mx}$ , with $P(x) = a$ and $Q(x) = e^{mx}$ .

The integrating factor is $\mu(x) = e^{\int a \, dx} = e^{ax}$ .
Multiply the entire equation by $\mu(x)$ .

e^{ax} \frac{dy}{dx} + a e^{ax} y = e^{ax} \cdot e^{mx} = e^{(a+m)x}.

Recognize the left-hand side as a derivative. Notice that $\frac{d}{dx} \left( y e^{ax} \right) = e^{ax} \frac{dy}{dx} + a e^{ax} y$ . So the equation becomes:

\frac{d}{dx} \left( y e^{ax} \right) = e^{(a+m)x}.

Integrate both sides with respect to $x$ .

y e^{ax} = \int e^{(a+m)x} \, dx + C.

The integral depends on whether $a+m = 0$ or not — this is the critical branching point.

Case 1: $a + m \neq 0$

\int e^{(a+m)x} \, dx = \frac{e^{(a+m)x}}{a+m} + C.

 So $y e^{ax} = \frac{e^{(a+m)x}}{a+m} + C$.

Case 2: $a + m = 0$ (i.e., $m = -a$ ) Then $e^{(a+m)x} = e^{0} = 1$ , so

\int 1 \, dx = x + C.

 So $y e^{ax} = x + C$.

5. Solve for $y$ in each case.

For $a+m \neq 0$ : …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 15 on this concept.

KEAM 2026Set eng-2026-04174 marksMCQ
$x^3 \dfrac{dy}{dx} + 3x^2 y = \cos x$
›Reveal solutionSolution

Recognize the exact derivative $\dfrac{d}{dx}(x^3y)$ .

Since $\dfrac{d}{dx}(x^3y)=x^3\dfrac{dy}{dx}+3x^2y$ , the equation $x^3\dfrac{dy}{dx}+3x^2y=\cos x$ becomes

$\frac{d}{dx}(x^3y)=\cos x.$ …
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KEAM 2026Set eng-2026-04174 marksMCQ
$2\, dy = (y + \cos x)\, dx$
›Reveal solutionSolution

Put in standard linear form and read off the integrating factor.

From $2\,dy=(y+\cos x)\,dx$ :

$2\frac{dy}{dx}=y+\cos x\ \Rightarrow\ \frac{dy}{dx}-\frac12 y=\frac12\cos x.$ …
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KEAM 2026Set eng-2026-04184 marksMCQ
$\dfrac{dy}{dx} + \dfrac{2y}{20 - x} = 10$
›Reveal solutionSolution

The integrating factor is $e^{\int P\,dx}$ with $P=\dfrac{2}{20-x}$ , which evaluates to $\dfrac{1}{(20-x)^2}$ .

For $\dfrac{dy}{dx}+P\,y=Q$ with $P=\dfrac{2}{20-x}$ ,

$\int P\,dx=\int\dfrac{2}{20-x}\,dx=-2\log|20-x|.$ …
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KEAM 2026Set eng-2026-04224 marksMCQ
$(y + x^2)\, dx = x\, dy, x > 0$
›Reveal solutionSolution

Rewrite as a first-order linear ODE $y'-y/x=x$ , use integrating factor $1/x$ , then apply the point $(1,0)$ .

From $(y+x^2)dx=x\,dy$ : $x\frac{dy}{dx}=y+x^2\Rightarrow \frac{dy}{dx}-\frac{y}{x}=x$ .

Integrating factor $=e^{-\int dx/x}=\frac1x$ , so $\frac{d}{dx}\Big(\frac{y}{x}\Big)=1$ . …
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KEAM 2025Set eng-2025-04234 marksMCQ
$\sin x\, dy = \dfrac{1}{2}(\sin 2x + 2y\cos x)\, dx$
›Reveal solutionSolution
The equation reduces to y' - (cot x) y = cos x, giving integrating factor e^{-integral cot x dx} = csc x.

Concept and Intuition

Rewrite the equation in standard linear form dy/dx + P(x)y = Q(x); the integrating factor is e^{integral P dx}.

Step-by-Step Solution

sin x dy = (1/2)(sin2x + 2y cos x) dx; using sin2x = 2 sin x cos x, the RHS = (sin x cos x + y cos x) dx.

So sin x (dy/dx) = sin x cos x + y cos x, i.e. dy/dx = cos x + y cot x. …
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KEAM 2025Set eng-2025-04254 marksMCQ
$(1+x^2)dy=(1-2xy)dx$
›Reveal solutionSolution

Put in standard linear form $\dfrac{dy}{dx}+Py=Q$ and compute $e^{\int P\,dx}$ .

Rewrite $(1+x^{2})dy=(1-2xy)dx$ as
$\dfrac{dy}{dx}+\dfrac{2x}{1+x^{2}}\,y=\dfrac{1}{1+x^{2}}.$
Here $P=\dfrac{2x}{1+x^{2}}$ , so …
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KEAM 2025Set eng-2025-04274 marksMCQ
$\dfrac{dy}{dx} - 2y = 2x - 3$
›Reveal solutionSolution
[!TLDR]

With $P(x) = -2$ , the integrating factor $e^{\int P\,dx} = e^{-2x}$ .

Concept

A linear first-order differential equation in the form $\frac{dy}{dx} + P(x)y = Q(x)$ is solved by multiplying through by the integrating factor $\mu(x) = e^{\int P(x)\,dx}$ — the standard method in the NCERT/CBSE differential-equations chapter.

Solution

The equation is

$\frac{dy}{dx} - 2y = 2x - 3.$ …
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KEAM 2024Set eng-2024-06054 marksMCQ
$\left(1 + 2e^{-x}\right)\frac{dy}{dx} - 2e^{-x}y = 1 + e^{-x}$
›Reveal solutionSolution

$\int P\,dx=\log(1+2e^{-x})$ , so the integrating factor is $1+2e^{-x}$ .

Divide by $(1+2e^{-x})$ :
$\frac{dy}{dx}-\frac{2e^{-x}}{1+2e^{-x}}\,y=\frac{1+e^{-x}}{1+2e^{-x}},$
so $P(x)=-\dfrac{2e^{-x}}{1+2e^{-x}}$ . With $w=1+2e^{-x}$ , $dw=-2e^{-x}dx$ : …
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KEAM 2024Set eng-2024-06064 marksMCQ
$x\frac{dy}{dx}+2y=xe^{x}$
›Reveal solutionSolution

Dividing by $x$ gives $P(x)=\frac{2}{x}$ , so the integrating factor is $x^2$ .

Divide $x\frac{dy}{dx}+2y=xe^x$ by $x$ : $\frac{dy}{dx}+\frac{2}{x}y=e^x$ . …
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KEAM 2024Set eng-2024-06074 marksMCQ
$2y\tan x+\dfrac{dy}{dx}=5\sin x$
›Reveal solutionSolution

Solve the linear ODE with integrating factor $\sec^2 x$ .

Rewrite as $\frac{dy}{dx}+2\tan x\,y=5\sin x$ . Integrating factor:
$\mu=e^{\int 2\tan x\,dx}=e^{2\log\sec x}=\sec^2 x.$
Then

$\frac{d}{dx}\left(y\sec^2 x\right)=5\sin x\sec^2 x=5\frac{\sin x}{\cos^2 x}.$ …
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KEAM 2024Set eng-2024-06074 marksMCQ
$(3\sin x\cos x)\,dy=(1+3y\sin^2 x)\,dx$
›Reveal solutionSolution

Put in linear form; the coefficient of $y$ gives IF $=\cos x$ .

From $(3\sin x\cos x)dy=(1+3y\sin^2 x)dx$ :
$\frac{dy}{dx}=\frac{1+3y\sin^2 x}{3\sin x\cos x}=\frac{1}{3\sin x\cos x}+\frac{3\sin^2 x}{3\sin x\cos x}y.$
The $y$ -coefficient is $\frac{\sin x}{\cos x}=\tan x$ , so …
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KEAM 2024Set eng-2024-06094 marksMCQ
$\dfrac{dy}{dx} + y = 5$
›Reveal solutionSolution

The equation is separable: $\frac{dy}{5-y}=dx$ , which integrates to $-\log|5-y|=x+C$ .

Write $\frac{dy}{dx}=5-y$ , so $\frac{dy}{5-y}=dx$ . …
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←Previous questionQ5 — Solve the differential equation (dy)/(dx)+2xy=y.→Next questionQ7 — Solve the differential equation (dy)/(dx)+1=ex+y.

Q.Find the general solution of $\frac{dy}{dx}+ay=e^{mx}$ .

Concept understanding — Integrating Factor Method

Integrating Factor Method

The Standard Form

The Integrating Factor

The Solution

A Quick Illustration

Method: Linear DE with constant coefficient and exponential forcing

Steps

Common Mistakes

Concept and Intuition

Step-by-Step Solution

Concept

Solution

Q.Find the general solution of dydx+ay=emx\frac{dy}{dx}+ay=e^{mx}dxdy​+ay=emx.

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Q.Find the general solution of $\frac{dy}{dx}+ay=e^{mx}$ .

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