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Let's solve the given differential equation step by step.

### Given Differential Equation:
\[
\frac{dy}{dx} = \frac{5xy}{(\ln y)^2}
\]
We are also given the initial condition that the solution passes through the point \((0, e)\).

### Step 1: Separate the Variables
To solve this differential equation, it's beneficial to separate the variables \(y\) and \(x\). Let's rewrite the equation in a form that allows us to integrate:

\[
\frac{1}{y} (\ln y)^2 \, dy = 5x \, dx
\]

### Step 2: Integrate Both Sides
Now, integrate both sides with respect to their respective variables.

1. **Left-hand side**:

\[
\int \frac{(\ln y)^2}{y} \, dy
\]

Let's perform a substitution to make this integration easier. Set:

\[
u = \ln y \quad \Rightarrow \quad \frac{du}{dy} = \frac{1}{y} \quad \Rightarrow \quad dy = y \, du
\]

Substituting this in, we get:

\[
\int \frac{(\ln y)^2}{y} \, dy = \int u^2 \, du = \frac{u^3}{3} + C = \frac{(\ln y)^3}{3} + C
\]

2. **Right-hand side**:

\[
\int 5x \, dx = \frac{5x^2}{2} + C_1
\]

### Step 3: Combine the Integrals
Combining the results from both sides, we have:

\[
\frac{(\ln y)^3}{3} = \frac{5x^2}{2} + C
\]

### Step 4: Solve for the Constant \(C\)
We are given the point \((0, e)\), which we will use to determine the constant \(C\).

Substituting \(x = 0\) and \(y = e\):

\[
\frac{(\ln e)^3}{3} = \frac{5(0)^2}{2} + C
\]

Since \(\ln e = 1\):

\[
\frac{1^3}{3} = C \quad \Rightarrow \quad C = \frac{1}{3}
\]

### Step 5: Write the General Solution
Now substitute back the value of \(C\):

\[
\frac{(\ln y)^3}{3} = \frac{5x^2}{2} + \frac{1}{3}
\]

To solve for \(\ln y\):

\[
(\ln y)^3 = 3 \left( \frac{5x^2}{2} + \frac{1}{3} \right)
\]

\[
(\ln y)^3 = \frac{15x^2}{2} + 1
\]

Taking the cube root on both sides:

\[
\ln y = \left( \frac{15x^2}{2} + 1 \right)^{1/3}
\]

### Final Answer
\[
\ln y = \left( \frac{15x^2}{2} + 1 \right)^{1/3}
\]

Would you like more details on any part of this solution? 

Here are some follow-up questions you might be interested in:
1. How can we verify if a solution satisfies a differential equation?
2. What are some other methods to solve differential equations besides separation of variables?
3. How does the initial condition affect the uniqueness of a solution in differential equations?
4. What are the possible applications of such differential equations in physics or engineering?
5. How would the solution change if we had a different initial point?

**Tip:** When dealing with logarithmic differential equations, substitution methods (like \(u = \ln y\)) can greatly simplify the integration process.

Find the solution to the differential equation dy/dx = (5xy) / (ln(y))^2 which passes through the point (0, e). Express your answer as ln y = ?

Learn how to solve the differential equation dy/dx = (5xy) / (ln y)^2 using separation of variables and substitution techniques. This solution includes integrating both sides and applying the initial condition (0, e) to find the constant. Ideal for students studying calculus and differential equations at the undergraduate level.

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