Solve the differential equation dy/dx = 2y - x.

Step-by-step solution

1. 1

   Put the equation in standard linear form. Move the $y$ term to the left so it reads $y'+P(x)y=Q(x)$:

   $\frac{dy}{dx}-2y=-x,\qquad P(x)=-2,\quad Q(x)=-x$

2. 2

   Build the integrating factor.

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

   Its purpose is to make the left-hand side a single derivative, which is what turns the ODE into a plain integration.

3. 3

   Multiply through and recognise the product derivative.

   $e^{-2x}\frac{dy}{dx}-2e^{-2x}y=-xe^{-2x}\quad\Longrightarrow\quad\frac{d}{dx}\left(ye^{-2x}\right)=-xe^{-2x}$

4. 4

   Integrate the right side by parts. Take $u=x$, $dv=-e^{-2x}dx$, so $du=dx$ and $v=\tfrac12 e^{-2x}$:

   $\int -xe^{-2x}\,dx=\frac{x}{2}e^{-2x}-\int\frac{1}{2}e^{-2x}\,dx=\frac{x}{2}e^{-2x}+\frac{1}{4}e^{-2x}+C$

5. 5

   Solve for $y$ and verify. Multiply through by $e^{2x}$:

   $y=Ce^{2x}+\frac{x}{2}+\frac{1}{4}$

   Check: $y'=2Ce^{2x}+\tfrac12$, while $2y-x=2Ce^{2x}+x+\tfrac12-x=2Ce^{2x}+\tfrac12$ $\checkmark$. The two agree for every $C$, and $\tfrac{x}{2}+\tfrac14$ is the particular solution while $Ce^{2x}$ is the homogeneous part.