Solve using integrating factor method. Find IF = eˣ.

Solve dy/dx + y = eˣ — Linear First Order ODE

Question

Solve the differential equation:

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

Find the general solution.

Solution — Step by Step

This is a linear first-order ODE of the form $\frac{dy}{dx} + P(x)\cdot y = Q(x)$ .

Here, $P(x) = 1$ and $Q(x) = e^x$ . The moment we see this form, the integrating factor method is our go-to weapon.

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

\mu = e^{\int 1\,dx} = e^x

We multiply both sides of the ODE by this factor. Why? Because it turns the left side into an exact derivative — specifically, $\frac{d}{dx}(\mu \cdot y)$ .

Multiplying both sides by $e^x$ :

e^x\frac{dy}{dx} + e^x y = e^x \cdot e^x = e^{2x}

The left side is exactly $\frac{d}{dx}(e^x \cdot y)$ . This is the whole point of the integrating factor — it makes the left side a perfect derivative.

\frac{d}{dx}(e^x y) = e^{2x}

Integrating both sides with respect to $x$ :

e^x y = \int e^{2x}\,dx = \frac{e^{2x}}{2} + C

Don’t forget the constant of integration — in board exams and JEE both, missing $C$ costs you marks.

Divide throughout by $e^x$ :

y = \frac{e^{2x}}{2e^x} + Ce^{-x} = \frac{e^x}{2} + Ce^{-x}

General solution: $y = \dfrac{e^x}{2} + Ce^{-x}$

Why This Works

The integrating factor trick is built on one key observation: if you have $\frac{dy}{dx} + P\cdot y$ , this is almost the result of the product rule applied to $e^{\int P\,dx} \cdot y$ . Multiplying by $e^{\int P\,dx}$ completes that pattern and collapses the messy left side into a single derivative.

After that, everything reduces to plain integration. The equation goes from “ugly ODE” to “just integrate this function” — that’s why the method works every single time for linear first-order equations.

For this specific question, the right side $e^{2x}$ integrates cleanly, giving a tidy answer. In JEE problems, they usually engineer this neatness — if your integral on the right looks horrible, double-check your IF.

Alternative Method (Variation of Parameters)

We can also get the particular integral by assuming $y_p = ke^x$ for some constant $k$ .

Substituting into the ODE: $ke^x + ke^x = e^x \Rightarrow 2ke^x = e^x \Rightarrow k = \frac{1}{2}$ .

The complementary solution is found from $\frac{dy}{dx} + y = 0$ , giving $y_c = Ce^{-x}$ .

So the general solution is $y = y_c + y_p = Ce^{-x} + \frac{e^x}{2}$ — same answer, different route.

For JEE Main, the integrating factor method is faster and more reliable. Use variation of parameters only when the right side is a polynomial or trig function — for exponentials, IF gets you there in fewer steps.

Common Mistake

The classic error is computing $\int e^{2x}\,dx = e^{2x}$ (forgetting to divide by 2). After multiplying both sides by the IF $e^x$ , the right side becomes $e^{2x}$ , not $e^x$ . Students who rush through write $\int e^{2x}\,dx = e^{2x} + C$ and end up with $y = \frac{e^x}{2} + Ce^{-x}$ becoming $y = e^x + Ce^{-x}$ — off by a factor of 2, wrong answer.

Always slow down on the integration step: $\int e^{ax}\,dx = \frac{e^{ax}}{a} + C$ .

Question

Solution — Step by Step

Why This Works

Alternative Method (Variation of Parameters)

Common Mistake

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