The problem asks to apply Euler's theorem and find the solution for a given equation. The equation provided is:

$x^2 \frac{\partial u}{\partial x} + 2xy \frac{\partial u}{\partial y} + y^2 \frac{\partial u}{\partial y}$

And the function to be used is:

$u = 5x^2 y \cos^{-1} \left( \frac{y}{x} \right)$

Steps to solve:

1. Euler's Theorem in this context relates to the property of homogeneous functions. For a function $u(x, y)$, Euler's Theorem states: $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = n u(x, y)$ where $u(x, y)$ is a homogeneous function of degree $n$.

2. We need to compute the partial derivatives $\frac{\partial u}{\partial x}$ and $\frac{\partial u}{\partial y}$, and then apply these to the given differential equation.

Let me go ahead and solve this problem step by step.The partial derivatives and the final expression for the given problem are:

1. Partial derivative of $u$ with respect to $x$: $\frac{\partial u}{\partial x} = 10xy \cdot \cos^{-1} \left( \frac{y}{x} \right) + \frac{5y^2}{\sqrt{1 - \frac{y^2}{x^2}}}$

2. Partial derivative of $u$ with respect to $y$: $\frac{\partial u}{\partial y} = 5x^2 \cdot \cos^{-1} \left( \frac{y}{x} \right) - \frac{5xy}{\sqrt{1 - \frac{y^2}{x^2}}}$

3. Final expression when substituted into the given equation: $20x^3 y \cdot \cos^{-1} \left( \frac{y}{x} \right) + 5x^2 y^2 \cdot \cos^{-1} \left( \frac{y}{x} \right) - \frac{5x^2 y^2}{\sqrt{1 - \frac{y^2}{x^2}}} - \frac{5xy^3}{\sqrt{1 - \frac{y^2}{x^2}}}$

This is the simplified result from applying Euler's theorem.

Would you like more details or further explanation?

Here are 5 related questions to explore further:

1. What is Euler's theorem and how is it used in homogeneous functions?
2. How do you calculate partial derivatives of implicit functions?
3. What is the significance of the inverse cosine function $\cos^{-1}(x)$ in physics and geometry?
4. How do you identify if a function is homogeneous, and what is its degree?
5. How do you simplify expressions involving trigonometric functions and square roots?

Tip: When dealing with inverse trigonometric functions, remember that the domain restrictions are crucial for ensuring valid solutions.