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

Step-by-step solution

1. 1

   Identify the equation as first-order linear and read off $P$ and $Q$. It is already in the standard form $y'+P(x)y=Q(x)$ with

   $P(x)=2,\qquad Q(x)=e^{-x}$

   That form is what makes the integrating-factor method available; nothing has to be rearranged.

2. 2

   Build the integrating factor. The factor $\mu(x)=e^{\int P\,dx}$ is chosen precisely so that multiplying by it turns the left side into a single derivative:

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

   (The constant of integration is omitted here because any nonzero multiple of $\mu$ works equally well.)

3. 3

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

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

   By the product rule the left side is exactly $\dfrac{d}{dx}\left(e^{2x}y\right)$, so

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

4. 4

   Integrate both sides once. Antidifferentiating an exact derivative is immediate, and this is where the single arbitrary constant enters:

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

5. 5

   Divide by the integrating factor to isolate $y$. Since $e^{2x}$ is never zero, division is legal everywhere:

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

6. 6

   Substitute the answer back into the original equation. With $y=e^{-x}+Ce^{-2x}$ we get $y'=-e^{-x}-2Ce^{-2x}$, so

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

   The $C$ terms cancel for every $C$, confirming the whole one-parameter family solves the equation.