91Ó°ÊÓ

A homogeneous differential equation is a specific type of differential equation where all terms are dependent on the unknown function and its derivatives. These equations have the form:

• \(a_n \frac{d^n y}{dt^n} + a_{n-1} \frac{d^{n-1} y}{dt^{n-1}} + ... + a_1 \frac{dy}{dt} + a_0 y = 0\)

In our exercise, the equation given is the homogeneous linear differential equation: \(y'' - y' - 2y = 0\). Every term involves the function \(y\) or its derivatives, with no independent function of \(t\) included.
What makes homogeneous equations special is their solution structure; they allow for a "superposition" of solutions. If you identify two solutions that fulfill the equation, any linear combination of these solutions will also be a solution. This property is useful in forming the general solution, which in our case is \(y(t) = C_1 e^{-t} + C_2 e^{2t}\), where \(C_1\) and \(C_2\) are constants.

To ensure the general solution of a differential equation, the solutions need to be linearly independent. Linear independence means that no solution can be expressed as a combination of others. For instance,

• The set \(\{y_1(t), y_2(t)\} = \{e^{-t}, e^{2t}\}\) is linearly independent if there are no constants \(a\) and \(b\), other than both being zero, such that \(ay_1(t) + by_2(t) = 0\) for all \(t\).

For our functions \(y_1(t) = e^{-t}\) and \(y_2(t) = e^{2t}\), they are considered linearly independent solutions. This can be proven if the Wronskian, which is a determinant involving the solutions and their derivatives, is non-zero:\[ W(t) = \begin{vmatrix} e^{-t} & e^{2t} \ -e^{-t} & 2e^{2t} \end{vmatrix} = e^{-t}(2e^{2t}) - (-e^{-t})(e^{2t}) = 3e^{t} eq 0\] Thus, the solutions are linearly independent, allowing the construction of a valid general solution.

Solution verification involves substituting potential solutions back into the differential equation to check their validity.
Here's how it works for our given functions:

• For \(y_1(t) = e^{-t}\), the first and second derivatives are \(y_1'(t) = -e^{-t}\) and \(y_1''(t) = e^{-t}\). When substituting these into the equation \(y'' - y' - 2y = 0\), you get:\[e^{-t} - (-e^{-t}) - 2e^{-t} = 0\]

The equation holds true, confirming \(y_1(t)\) as a solution.

• Similarly, for \(y_2(t) = e^{2t}\), its derivatives are \(y_2'(t) = 2e^{2t}\) and \(y_2''(t) = 4e^{2t}\). Substituting into the equation gives:\[4e^{2t} - 2e^{2t} - 2e^{2t} = 0\]

This substitution verifies \(y_2(t)\) is also a solution. Verifying solutions this way ensures that we don't mistakenly use incorrect functions in forming the general solution of the differential equation.