Solve the first-order linear differential equation dy/dx + 2y = e^(-x).

Step-by-step solution

1. 1

   Confirm the equation is already in standard linear form. A first-order linear equation looks like $\dfrac{dy}{dx} + P(x)y = Q(x)$. Here $P(x)=2$ and $Q(x)=e^{-x}$, with the derivative coefficient equal to $1$ — no dividing through is needed before starting.

2. 2

   Build the integrating factor. The whole method exists to force the left-hand side into a single derivative. The factor that does it is

   $\mu(x) = e^{\int P(x)\,dx} = e^{\int 2\,dx} = e^{2x}$

   (the constant of integration is dropped because any nonzero multiple of $\mu$ works equally well).

3. 3

   Multiply through and collapse the left side. Multiplying every term by $e^{2x}$:

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

   The left side is exactly the product rule run backwards:

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

4. 4

   Integrate both sides once. Integrating the derivative on the left simply removes it, and the right side is elementary:

   $e^{2x}y = e^{x} + C$

   The arbitrary constant $C$ enters here, and this is the only place it can enter — a first-order equation carries exactly one.

5. 5

   Solve for $y$. Divide by $e^{2x}$, or equivalently multiply by $e^{-2x}$:

   $y = e^{-2x}\left(e^{x} + C\right) = e^{-x} + Ce^{-2x}$

6. 6

   Verify by substitution. With $y = e^{-x} + Ce^{-2x}$, the derivative is $y' = -e^{-x} - 2Ce^{-2x}$, so

   $y' + 2y = \left(-e^{-x} - 2Ce^{-2x}\right) + \left(2e^{-x} + 2Ce^{-2x}\right) = e^{-x}$

   The $C$ terms cancel for every $C$, confirming the general solution. The $e^{-x}$ piece is the particular solution and $Ce^{-2x}$ is the decaying homogeneous part.