Solve the differential equation dy/dx + y = eˣ

Question

Solve the differential equation:

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

(NCERT Class 12, Exercise 9.6)

Solution — Step by Step

This is of the form $\frac{dy}{dx} + P(x) \cdot y = Q(x)$ , where $P(x) = 1$ and $Q(x) = e^x$ .

This is a linear first-order differential equation, so we use the integrating factor method.

\text{IF} = e^{\int P(x)\,dx} = e^{\int 1\,dx} = e^x

The integrating factor is $e^x$ .

Multiply through by $e^x$ :

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

The left side is the derivative of $y \cdot e^x$ (that’s exactly why the IF works):

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

Integrate both sides:

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

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

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

where $C$ is an arbitrary constant.

Why This Works

The integrating factor method converts a non-exact equation into an exact one. When we multiply by $e^{\int P\,dx}$ , the left side magically becomes the derivative of a product — $\frac{d}{dx}(y \cdot \text{IF})$ . This lets us integrate both sides directly.

Why does $e^{\int P\,dx}$ do the trick? Because $\frac{d}{dx}(y \cdot e^{\int P\,dx}) = e^{\int P\,dx}\left(\frac{dy}{dx} + Py\right)$ , which is exactly the left side of our equation multiplied by the IF. The product rule gives us this for free.

Alternative Method — Complementary Function + Particular Integral

For those familiar with the CF + PI approach (common in JEE):

CF: Solve $\frac{dy}{dx} + y = 0$ , which gives $y = Ce^{-x}$ .

PI: For $Q(x) = e^x$ , try $y_p = Ae^x$ . Substituting: $Ae^x + Ae^x = e^x$ , so $2A = 1$ , giving $A = 1/2$ .

General solution: $y = Ce^{-x} + \frac{e^x}{2}$ . Same answer.

The CF + PI method is faster when $Q(x)$ is a standard form ( $e^{ax}$ , $\sin bx$ , polynomial). For CBSE boards, stick with the IF method — it’s what the marking scheme expects. For JEE, knowing both approaches saves time on different question types.

Common Mistake

Students often forget to divide by the IF at the end. After integrating, you have $y \cdot e^x = \frac{e^{2x}}{2} + C$ . The final answer is $y = \frac{e^x}{2} + Ce^{-x}$ , not $y = \frac{e^{2x}}{2} + C$ . Always isolate $y$ — leaving the IF attached to $y$ in your final answer will cost you marks.

Question

Solution — Step by Step

Why This Works

Alternative Method — Complementary Function + Particular Integral

Common Mistake

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