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.