The given differential equation is:

$xy'' + 2y' = 0$

Where:

• $y''$ represents the second derivative of $y$ with respect to $x$,
• $y'$ represents the first derivative of $y$ with respect to $x$.

Step-by-Step Solution:

1. Rearrange the equation:

   Divide the entire equation by $x$ (assuming $x \neq 0$) to simplify:

   $y'' + \frac{2}{x} y' = 0$

2. Solve the equation:

   This is a linear second-order homogeneous differential equation with variable coefficients. We can solve this using a method known as separation of variables.

   First, treat the equation as a first-order equation by defining $v = y'$, so that $y'' = v'$. The equation becomes:

   $v' + \frac{2}{x} v = 0$

   This is now a first-order linear equation in $v$. We can solve it by using an integrating factor.

3. Solve for $v$:

   The integrating factor for the equation $v' + \frac{2}{x} v = 0$ is $\mu(x) = e^{\int \frac{2}{x} \, dx} = x^2$.

   Multiply the entire equation by $x^2$:

   $x^2 v' + 2x v = 0$ Which simplifies to:

   $\frac{d}{dx}(x^2 v) = 0$

   Integrating both sides with respect to $x$, we get:

   $x^2 v = C$

   where $C$ is a constant of integration. Now, substitute $v = y'$:

   $x^2 y' = C$

   Solving for $y'$:

   $y' = \frac{C}{x^2}$

4. Integrate to find $y(x)$:

   Now, integrate $y'$ to find $y(x)$:

   $y(x) = \int \frac{C}{x^2} \, dx = -\frac{C}{x} + D$

   where $D$ is another constant of integration.

General Solution:

Thus, the general solution to the differential equation is:

$y(x) = -\frac{C}{x} + D$

where $C$ and $D$ are constants of integration.

---

Would you like a more detailed explanation of any part of this process?

Here are 5 related questions:

1. How do we handle equations with variable coefficients?
2. What is the method of integrating factors, and when is it used?
3. How does the solution change if we have different coefficients in the equation?
4. What happens when the equation involves non-homogeneous terms?
5. Can you solve similar equations involving higher-order derivatives?

Tip: Always check whether the equation is homogeneous or non-homogeneous to choose the appropriate solution technique.