Find and use the initial value Green's function to solve $$ x^{2} y^{\prime \prime}+3 x y^{\prime}-15 y=x^{4} e^{x}, y(1)=1, y^{\prime}(1)=0 $$

Ordinary Differential Equations (ODEs) are equations that involve some function and its derivatives. They are a fundamental concept in mathematics and are used to model a wide array of phenomena in physics, engineering, biology, economics, and many other fields. An ODE is called 'ordinary' to distinguish it from 'partial' differential equations which involve partial derivatives of a function of multiple variables.

An example of an ODE is the equation used in our exercise: \[x^{2} y^{\prime \prime}+3 x y^{\prime}-15 y=x^{4} e^{x},\] where \(y\) is a function of a single variable \(x\), and \(y^{\prime}\) and \(y^{\prime \prime}\) are the first and second derivatives of \(y\) with respect to \(x\). The ODE contains terms with both the function \(y\) and its derivatives, and in this case, is set equal to another function \(x^{4} e^{x}\) that acts as a source term or driving function.

A Homogeneous ODE is a special type of differential equation that is set to zero on one side. In our exercise, the corresponding homogeneous ODE to the given problem is written as \[x^{2} y^{\prime \prime}+3 x y^{\prime}-15 y=0.\] The solution to this equation, denoted as \(y_h(x)\), does not include the non-zero right-hand side function \(x^{4} e^{x}\) present in the original problem.

Homogeneous ODEs are simpler to solve than their inhomogeneous counterparts because they do not have the additional complexity of the driving function. The solutions to homogeneous ODEs are important because they form the complementary or homogeneous part of the general solution to the inhomogeneous ODE.

When dealing with non-homogeneous ODEs, which are equations that equal a non-zero function, such as the original problem, we search for a Particular Solution, denoted as \(y_p(x)\). This solution directly addresses the non-zero part of the ODE and 'completes' the general solution when added to the homogeneous solution \(y_h(x)\).

The Particular Solution is often obtained by methods like undetermined coefficients or variation of parameters. However, in the context of this problem, we use a Green's function approach which involves integrating the product of the Green's function and the non-zero right-hand side over a certain interval. This approach can elegantly handle a wide range of boundary conditions and source terms.

Boundary conditions are additional pieces of information that specify values or behaviors of the solution to an ODE at specific points, which are necessary to uniquely determine the solution. For second-order ODEs, like the one in our exercise, you typically need two boundary conditions.

In this context, boundary conditions help us find the constants in the homogeneous solution and define the behavior of Green's function. The problem provides initial conditions (a type of boundary condition) at \(x=1\): \(y(1) = 1\) and \(y'(1) = 0\). These are used after finding the homogeneous and particular solutions to determine the unknown constants in the general solution, resulting in a solution that satisfies both the differential equation and the initial conditions given.