Q.Solve the following differential equation: $\frac{dy}{dx} + 3y = e^{-2x}$

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

Step 1: Identify the integrating factor.

The standard form is $\frac{dy}{dx} + P(x)y = Q(x)$. Here $P(x) = 3$, so the integrating factor is

$\mu(x) = e^{\int 3\,dx} = e^{3x}$.

Step 2: Multiply through and integrate.

Multiplying the equation by $e^{3x}$:

$e^{3x}\frac{dy}{dx} + 3e^{3x}y = e^{x}$.

The left side is $\frac{d}{dx}(e^{3x}y)$. Hence

$\frac{d}{dx}(e^{3x}y) = e^{x}$.

Step 3: Integrate both sides.

$e^{3x}y = \int e^{x}\,dx = e^{x} + C$.

Thus $y = e^{-2x} + Ce^{-3x}$.

The equation $\frac{dy}{dx} + 3y = e^{-2x}$ is a classic first-order linear ordinary differential equation. It fits the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = 3$ and $Q(x) = e^{-2x}$.

The key insight: we cannot directly integrate because $y$ and its derivative are mixed. But we can multiply both sides by a cleverly chosen function — the integrating factor — that turns the left-hand side into the derivative of a product. This makes the equation integrable in one step.

Let’s work through it.

1. Find the integrating factor. For $\frac{dy}{dx} + P(x)y = Q(x)$, the integrating factor is $\mu(x) = e^{\int P(x)\,dx}$. Here $P(x) = 3$, so $\int 3\,dx = 3x$. Thus

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

Notice the left side is exactly $\frac{d}{dx}\left( e^{3x} y \right)$ because the derivative of $e^{3x}y$ is $e^{3x}\frac{dy}{dx} + 3e^{3x}y$ (product rule). So we have:

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

where $C$ is the constant of integration.

4. Solve for $y$. Divide through by $e^{3x}$:

The solution is a sum of two parts: the particular solution $e^{-2x}$ (which matches the forcing term’s form) and the complementary solution $Ce^{-3x}$ (which solves the homogeneous equation $\frac{dy}{dx} + 3y = 0$). The constant $C$ will be fixed if an initial condition is given.

Method: Integrating factor with a constant coefficient $P$

Whenever $\dfrac{dy}{dx}+P\,y=Q$ has $P$ a constant, the integrating factor is a plain exponential — the cleanest case of the method.

Steps

Step 1: Check the standard form.

Confirm the coefficient of $\dfrac{dy}{dx}$ is $1$ and identify the constant $P$ and the function $Q$.

Step 2: Write the integrating factor.

With $P$ constant, $\int P\,dx=Px$, so

Step 3: Collapse the left side and integrate.

Multiplying gives $\dfrac{d}{dx}\big(y\,e^{Px}\big)=Q\,e^{Px}$, so

Step 4: Divide back by the I.F.

Solve for $y$. When $Q$ is itself an exponential $e^{kx}$, the product $Q\,e^{Px}$ is one exponential and integrates in a single step.

Common Mistakes

Mistake 1: Mishandling $e^{3x}\cdot e^{-2x}$.

Why it's wrong: the right side becomes $e^{3x}e^{-2x}=e^{x}$, not $e^{-6x^2}$ or $e^{x^2}$; adding exponents is required. Correct approach: $e^{3x}\cdot e^{-2x}=e^{(3-2)x}=e^{x}$, then integrate to $e^x$.

Mistake 2: Dividing only some terms by the I.F.

Why it's wrong: from $e^{3x}y=e^{x}+C$, dividing by $e^{3x}$ gives $y=e^{-2x}+Ce^{-3x}$; forgetting to divide the constant leaves a wrong $y$. Correct approach: divide every term by $e^{3x}$.

Mistake 3: Treating a constant $P=3$ as if it needed a special integral.

Why it's wrong: $\int 3\,dx=3x$ exactly, so the I.F. is simply $e^{3x}$. Correct approach: for constant $P$, the exponent is $Px$.