** ( x + 3 ) y” + x y’ + ( ln|x| ) y = 0, y(1) = 0, y'(1) = 1 **

The aim of this question is to** qualitatively** find the **possible interval** of the differential **equation’s solution**.

For this we need to **convert any given differential equation** to the following **standard form**:

\[ y^{”} \ + \ p(x) y’ \ + \ q(x) y \ = \ g(x) \]

Then we have to **find the domain of the functions** $ p(x), \ q(x), \ and \ g(x) $. The **intersection of the domains** of these functions represents the **longest interval** of all the possible solutions to the differential equation.

## Expert Answer

**Given the differential equation:**

\[ ( x + 3 ) y^{”} + x y’ + ( ln|x| ) y = 0 \]

**Rearranging:**

\[ y^{”} + \dfrac{ x }{ x + 3 } y’ + \dfrac{ ln| x | }{ x + 3 } y = 0 \]

**Let:**

\[ p(x) = \dfrac{ x }{ x + 3 } \]

\[ q(x) = \dfrac{ ln|x| }{ x + 3 } \]

\[ g(x) = 0 \]

Then, the above equation takes the **form of the standard equation**:

\[ y^{”} + p(x) y’ + q(x) y = g(x) \]

**Incorporating** $ y(1) = 0 $ and $ y'(1) = 1$, **It can be noticed that:**

\[ p(x) = \dfrac{ x }{ x + 3 } \text{ is defined on the intervals } (-\infty , \ -3) \text{ and } (-3, \ \infty) \]

\[ q(x) = \dfrac{ ln|x| }{ x + 3 } \text{ is defined on the intervals } (-\infty , \ -3), \ (-3, \ 0) \text{ and } (0, \ \infty) \]

\[ g(x) = 0 \text{ is defined on the intervals } (-\infty, \ \infty) \]

If we check the intersection of all the above intervals, it can be concluded that the **longest interval of the solution is $ (0, \ \infty) $.**

## Numerical Result

$ (0, \ \infty) $ is the **longest interval** in which the given initial value problem is certain to have a unique twice differentiable solution.

## Example

Determine the **longest interval** in which the given **initial value problem** is certain to have a **unique twice differentiable** solution.

\[ \boldsymbol{ y^{”} \ + \ x y’ \ + \ ( ln|x| ) y \ = \ 0, \ y(1) \ = \ 0, \ y'(1) \ = \ 1 } \]

**Comparing with the standard equation:**

\[ y^{”} + p(x) y’ + q(x) y = g(x) \]

We have:

\[ p(x) = x \Rightarrow \text{ is defined on the interval } (0, \ \infty) \]

\[ q(x) = ln|x| \Rightarrow \text{ is defined on the interval } (-\infty, \ \infty) \]

\[ g(x) = 0 \]

**If we check the intersection of all the above intervals, it can be concluded that the longest interval of the solution is $ (0, \ \infty) $.**