Solve the differential equation dy/dx = 6 - y/2.

Step-by-step solution

1. 1

   Put it in standard linear form. Move the $y$ term left:

   $\frac{dy}{dx}+\frac12 y=6$

   This matches $y'+p(x)y=q(x)$ with the constants $p=\frac12$ and $q=6$, so the integrating-factor method applies directly.

2. 2

   Find the equilibrium first — it predicts the answer. A constant solution needs $\frac{dy}{dx}=0$, i.e. $6-\frac{y}{2}=0$, so $y=12$. That constant function really is a solution, and since the coefficient of $y$ is negative in the original form, solutions should decay toward $12$. Expect the answer to look like $12$ plus a decaying term.

3. 3

   Build the integrating factor.

   $\mu(x)=e^{\int \frac12\,dx}=e^{x/2}$

4. 4

   Multiply through and recognise a product derivative.

   $e^{x/2}\frac{dy}{dx}+\frac12 e^{x/2}y=6e^{x/2}\;\Longrightarrow\;\frac{d}{dx}\left(ye^{x/2}\right)=6e^{x/2}$

   Collapsing the left side into one derivative is the whole point of the factor $e^{x/2}$.

5. 5

   Integrate and solve for $y$. Since $\int 6e^{x/2}\,dx=12e^{x/2}+C$,

   $ye^{x/2}=12e^{x/2}+C\;\Longrightarrow\;y=12+Ce^{-x/2}$

   Separation of variables gives the same family, with $C$ absorbing the constant of integration.

6. 6

   Verify and read off the behaviour. Differentiating, $y'=-\frac{C}{2}e^{-x/2}$, while the right-hand side is

   $6-\frac{y}{2}=6-6-\frac{C}{2}e^{-x/2}=-\frac{C}{2}e^{-x/2}$

   The two match for every $C$. As $x\to\infty$ the exponential dies and $y\to 12$ regardless of the initial value, so $y=12$ (the case $C=0$) is a stable equilibrium.