91Ó°ÊÓ

When solving a non-homogeneous differential equation, it is crucial to first understand the associated homogeneous equation. The fundamental set of solutions is a group of linearly independent solutions to the homogeneous differential equation. These solutions can be combined to form the general solution of the homogeneous part.

In our case, the homogeneous differential equation is \(2t^3 y''' + t^2 y'' + t y' - y = 0\). The given fundamental set of solutions for this equation is \(S=\{t, t \ln t, \sqrt{t}\}\). This means that all independent solutions can be written as a combination of these elements.

Thus, the general solution to the homogeneous equation is:
\(y_h(t) = C_1 t + C_2 t \ln t + C_3 \sqrt{t}\)

Here, \(C_1, C_2,\) and \(C_3\) are arbitrary constants that will later be determined using initial conditions.

Initial conditions are values given for the function and its derivatives at a particular point, typically used to find the specific solution to a differential equation. They help determine the values of the constants in the general solution of a differential equation.

In our problem, the initial conditions provided are: \(y(1) = 0, y'(1) = 1, y''(1) = 0\). These conditions will allow us to find \(C_1, C_2, \text{and} C_3\).

We use the general solution:
\(y(t) = C_1 t + C_2 t \ln t + C_3 \sqrt{t} - t^2\)
and compute its first and second derivatives:
\(y'(t) = C_1 + C_2 (1 + \ln t) + C_3 \frac{1}{2} t^{-1/2} - 2t\)
\(y''(t) = C_2 t^{-1} - \frac{C_3}{4} t^{-3/2} - 2\)

Substituting our initial conditions \(t = 1\) into these expressions:
\(y(1) = C_1 + C_3 - 1 = 0\)
\(y'(1) = C_1 + \frac{C_3}{2} - 2 = 1\)
\(y''(1) = C_2 - \frac{C_3} {4} = 2\)

We solve the system of equations obtained to determine the constants.

To solve a non-homogeneous differential equation, we need to find a particular solution that satisfies the complete equation, including the non-homogeneous part. The given non-homogeneous equation is: \(2 t^3 y''' + t^2 y'' + t y' - y = -3 t^2\).

We often guess the form of the particular solution based on the form of the non-homogeneous term. Here, we assume: \(y_p(t) = At^2\).

Next, we calculate the derivatives of \(y_p(t) = At^2\):

1. \(y_p' = 2At\)
2. \(y_p'' = 2A\)
3. \(y_p''' = 0\)

Substituting these into the original equation:
\(2t^3(0) + t^2(2A) + t(2At) - At^2 = -3t^2\)
Simplifying, we get:
\(2A t^2 + 2A t^2 - A t^2 = -3 t^2\)

This further simplifies to:
\(3 A t^2 = -3 t^2\).

Solving for \(A\), we find: \(A = -1\). Therefore, the particular solution is:
\(y_p(t) = -t^2\).

Finally, we combine the homogeneous and particular solutions to form the general solution to the non-homogeneous equation:
\(y(t) = y_h(t) + y_p(t)\), which results in:
\(y(t) = C_1 t + C_2 t \ln t + C_3 \sqrt{t} - t^2\).