Q.Solve the following differential equation: $\frac{dy}{dx} + y = 1 (y \neq 1)$

Concept: Initial Value Problem — we solve a first-order linear ODE using an integrating factor.

Step 1: Identify the integrating factor.

The standard form is $\frac{dy}{dx} + P(x)y = Q(x)$. Here $P(x) = 1$, so the integrating factor is

$\mu(x) = e^{\int 1 \, dx} = e^x$.

Step 2: Multiply through and integrate.

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

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

Integrate: $e^x y = \int e^x \, dx = e^x + C$.

Step 3: Solve for $y$.

$y = 1 + Ce^{-x}$.

The equation $\frac{dy}{dx} + y = 1$ is a classic first-order linear differential equation. The key idea is that we can multiply both sides by a cleverly chosen function — the integrating factor — which turns the left-hand side into the derivative of a product. This lets us integrate directly.

Why does this work? Because the left side already looks like the derivative of $y$ plus something times $y$. If we can make that "something" match the derivative of the multiplying factor, we get a perfect derivative. Here, the coefficient of $y$ is $1$, so the integrating factor is $e^{\int 1\,dx} = e^x$.

Let’s walk through it step by step.

1. Identify the standard form.

   The equation is already in the form $\frac{dy}{dx} + P(x)y = Q(x)$, with $P(x) = 1$ and $Q(x) = 1$.

2. Compute the integrating factor.

   The integrating factor $\mu(x)$ is given by $\mu(x) = e^{\int P(x)\,dx} = e^{\int 1\,dx} = e^x$.

3. Multiply the entire equation by $\mu(x)$.

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

5. Integrate both sides with respect to $x$.

This gives:

where $C$ is the constant of integration.

6. Solve for $y$. Divide both sides by $e^x$ (which is never zero):

Method: Solving a linear first-order equation with an integrating factor

Use this for any equation of the form $\dfrac{dy}{dx}+P(x)\,y = Q(x)$. (Simple constant-coefficient cases like $y'+y=1$ can also be separated, but the integrating factor is the general tool.)

Steps

Step 1: Put the equation in standard linear form

Identify $P(x)$ and $Q(x)$ from $\dfrac{dy}{dx}+P(x)y=Q(x)$. For $y'+y=1$, $P=1$ and $Q=1$.

Step 2: Build the integrating factor

Multiplying through by $\mu$ makes the left side an exact derivative $\dfrac{d}{dx}\big[\mu\,y\big]$.

Step 3: Integrate and solve for $y$

Common Mistakes

Mistake 1: Misidentifying $P(x)$ and $Q(x)$

Why it's wrong: the integrating factor is $e^{\int P\,dx}$, so a wrong $P$ gives a wrong factor and a wrong answer. Correct approach: write the equation strictly as $y'+P(x)y=Q(x)$ before reading off $P=1$, $Q=1$.

Mistake 2: Forgetting that the solution splits into a particular part plus a decaying term

Why it's wrong: the full solution is $y=1+Ce^{-x}$; dropping the $Ce^{-x}$ leaves only the constant equilibrium and loses generality. Correct approach: keep the homogeneous term with its arbitrary constant.

Mistake 3: Overlooking the $y\neq1$ condition

Why it's wrong: $y=1$ corresponds to $C=0$; the restriction rules that value out. Correct approach: note $C\neq0$ so the solution stays away from $y=1$.