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The given image shows an initial value problem (IVP) with a first-order differential equation:

\[
x \ln(x) \frac{dy}{dx} + y = x e^x, \quad y(2) = -4
\]

It also provides a proposed solution for \( y(x) \):

\[
y(x) = \frac{e^x - 4 \ln(2) - e^2}{\ln(x)}
\]

The question asks for the interval of existence of the solution, which depends on the nature of the solution and the domain where the differential equation and its solution are valid.

### Interval of existence:

The key constraint in the given solution \( y(x) = \frac{e^x - 4 \ln(2) - e^2}{\ln(x)} \) comes from the logarithmic function, \( \ln(x) \), in the denominator. The natural logarithm \( \ln(x) \) is defined for \( x > 0 \), and \( \ln(x) = 0 \) at \( x = 1 \), which causes a division by zero.

So, the interval of existence will exclude \( x = 1 \) and be constrained to positive \( x \)-values. Since the initial condition is given at \( x = 2 \), we conclude that the interval of existence for the solution is \( (0, 1) \cup (1, \infty) \).

#### Final Answer:
The interval of existence is:

\[
(0, 1) \cup (1, \infty)
\]

Would you like more details or clarification? Here are some related questions:

1. How do we solve a first-order linear differential equation like this one?
2. What is the method for finding the general solution of a differential equation with an initial condition?
3. Why does the logarithmic function create restrictions on the interval of existence?
4. How is the solution \( y(x) \) derived step by step for this problem?
5. What happens if we try to substitute \( x = 1 \) into the solution?

**Tip**: Always check for points where the denominator of a function could cause undefined behavior to determine the interval of existence.

Solve the initial value problem: x ln(x) (dy/dx) + y = x e^x, with y(2) = -4.

This page covers the step-by-step solution of the initial value problem x ln(x) (dy/dx) + y = x e^x, with y(2) = -4. It discusses the use of logarithmic and exponential functions in solving first-order linear differential equations and determines the interval of existence.

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