91Ó°ÊÓ

To start solving a second-order linear non-homogeneous differential equation, we first tackle its homogeneous counterpart. This involves removing the non-homogeneous part. For our problem, we change the given equation \[ y'' + 2t y' - 4y = 2 \] into: \[ y'' + 2t y' - 4y = 0 \].
This new equation is called the homogeneous equation. It is easier to handle as it does not include the '2' on the right-hand side.
This simplification lets us focus solely on the behavior determined by the terms involving \( y, y' \), and \( y'' \).
By comprehending the homogeneous equation, we lay the groundwork for solving the more complex non-homogeneous equation later. By solving the homogeneous first, we later add a particular solution to handle the non-homogeneous part.

Next, we handle the homogeneous equation by forming its characteristic equation.
For a differential equation of the form \[ y'' + 2ty' - 4y = 0 \], we assume a solution format like \( y = e^{rt} \).
We insert our assumption and its derivatives back into the homogeneous equation to identify the characteristic equation. Accordingly, we substitute:

• \( y = e^{rt} \)
• \( y' = re^{rt} \)
• \( y'' = r^2 e^{rt} \)

This substitution transforms \[ y'' + 2ty' - 4y = 0 \] into \[ r^2 e^{rt} + 2t r e^{rt} - 4 e^{rt} = 0 \].
We factor out the common term \( e^{rt} \) to get: \[ e^{rt} (r^2 + 2tr - 4) = 0 \].
Since \( e^{rt} eq 0 \), we focus on \[ r^2 + 2tr - 4 = 0 \].
This is the characteristic equation.
Solving the characteristic equation helps us find the solution that the homogeneous part of the differential equation follows.

Initial conditions are values given to help determine specific solutions to differential equations.
They tell us the state of the system at the start.
In our problem, the initial conditions are: \[ y(0) = 0 \] and \[ y'(0) = 0 \].
These conditions mean that at time t = 0, both the function y(t) and its first derivative y'(t) equal 0.
We use these initial conditions to solve for the arbitrary constants in our general solution.
They help tailor our solution to fit the specific problem we are dealing with.

A particular solution is what we add to the homogeneous solution to solve a non-homogeneous differential equation completely.
This specific solution fits the non-homogeneous part of the equation.
For our problem, we assume a particular solution in the form that matches the non-homogeneous term.
Here, we try a constant solution, so let's call it \( y_p = A \).
We plug this back into the non-homogeneous equation: \[ (y_p)'' + 2t(y_p)' - 4 y_p = 2 \].
Given \( y_p = A \), we see that \( (y_p)'' = 0 \) and \( (y_p)' = 0 \).
The equation becomes: \[ 0 + 0 - 4A = 2 \], which simplifies to \[ -4A = 2 \].
Solving for A, we get: \[ A = -\frac{1}{2} \].
This particular solution \( y_p = -\frac{1}{2} \) will be combined with the homogeneous solution to form the general solution.