91Ó°ÊÓ

A fundamental set of solutions is a crucial concept in understanding differential equations. It provides a basis for constructing the general solution to the homogeneous differential equation. When dealing with a linear differential equation, one often starts by finding its homogeneous counterpart:

• In this exercise, the homogeneous equation is given by: \[ t y^{(4)} + 2 y^{\text{'''}} = 0 \].
• This equation simplifies the process by allowing us to determine the general solution for the homogeneous part.

The next step is to identify the fundamental set of solutions, which are functions that satisfy the homogeneous equation and are linearly independent. For our equation, the provided set of solutions is \( S = \{1, t, t \, \ln t, t^2\} \). This fundamental set leads to the general solution for the homogeneous equation:\[ y_h = c_1 + c_2 t + c_3 t \, \ln t + c_4 t^2 \].By combining these basic solutions with appropriate coefficients, we can construct any solution for the given homogeneous differential equation.

When facing a non-homogeneous differential equation, such as \( t y^{(4)} + 2 y^{\text{'''}} = \frac{45}{8} t^{-7/2} \), the method of variation of parameters is a powerful and general technique to find a particular solution.This method involves modifying the coefficients in the general solution of the homogeneous equation to make them functions of the independent variable (in this case, \( t \)). The particular solution is sought by assuming a form like:\[ y_p = u_1(t) + u_2(t)t + u_3(t)t \, \ln t + u_4(t)t^2. \]However, in this exercise, the particular solution can be more directly computed through integration. We set up the integrals aimed to solve:

• Compute \( \int \left( \frac{45}{8} t^{-3.5} \right) dt \).
• Next, compute \( \int t^{-3} \left( -\frac{15}{4} t^{-1.5} \right) dt \).

Finally, we determine:\[ y_p = t^2 \left( \frac{15}{14} t^{-3.5} \right) = \frac{15}{14} t^{-1.5}. \]This yields a specific solution which, when added to the homogeneous solution, accounts for the non-homogeneous part of the differential equation.

Initial conditions are essential in determining the specific solution to a differential equation that uniquely fits a given problem. They ensure that the solution is not just any arbitrary function but matches the provided constraints at a particular point:

• In our problem, initial conditions are given at \( t = 1 \): \( y(1) = 0 \), \( y'(1) = 0 \), \( y''(1) = 1 \), and \( y'''(1) = 0 \).
• To find these constants, we substitute the initial conditions into our combined general solution \( y = y_h + y_p \).

This yields a system of equations:

• For \( y(1) = 0 \): \[ c_1 + c_2 + c_4 + \frac{15}{14} = 0 \]
• For \( y'(1) = 0 \): \[ c_2 + c_3 + 2c_4 - \frac{15}{28} = 0 \]
• For \( y''(1) = 1 \): \[ 2c_4 - \frac{15}{8} = 1 \]
• For \( y'''(1) = 0 \): Consistent with previous conditions, thus not providing a new constraints.

By solving these equations sequentially, we determine the values for \( c_1, c_2, c_3, \) and \( c_4 \):

• Firstly, solve for \( c_4 \): \[ c_4= \frac{23}{16} \]
• Next for \( c_3 \): \[ c_3= -\frac{253}{56} \]
• Then for \( c_2 \): \[ c_2= -\frac{457}{28} \]
• Lastly for \( c_1 \): \[ c_1= -\frac{1613}{28} \]

Inserting these constants back, we obtain the specific solution to the original differential equation that satisfies the initial conditions.