91Ó°ÊÓ

The characteristic equation is a critical part of solving linear differential equations, particularly those with constant coefficients. In our initial-value problem, the differential equation is given as:
\(y^{\backprime \backprime} - y = 0\).
To find the characteristic equation, assume a solution of the form:
\(y = e^{rt}\).
Substituting this assumed solution into the differential equation yields:
\(r^2e^{rt} - e^{rt} = 0\).
This equation can be factored as:
\(e^{rt}(r^2 - 1) = 0\).
Since \(e^{rt}\) is never zero, we can set
\(r^2 - 1 = 0\).
Solving this, we find the characteristic roots:
\(r = 1\) and \(r = -1\). These roots are crucial for forming the general solution.

With the characteristic roots \(r = 1\) and \(r = -1\), we can now form the general solution of the differential equation. The general solution is a linear combination of solutions that correspond to the characteristic roots. Here, it is written as:
\(y(t) = C_1e^{t} + C_2e^{-t}\).
In this equation, \(C_1\) and \(C_2\) are arbitrary constants that we need to determine by using the initial conditions provided.

The initial conditions are specific values given for the solution and its derivative at a particular point, usually at \(t=0\). These conditions help us find the specific values of the constants \(C_1\) and \(C_2\) in our general solution.
Given \(y(0) = 1\), we substitute \(t = 0\) into the general solution:
\(y(0) = C_1e^{0} + C_2e^{0} = C_1 + C_2 = 1\).
This gives us the first equation:
\(C_1 + C_2 = 1\).
Next, given \(y^{\backprime}(0) = 3\), we first find the derivative of our general solution:
\(y^{\backprime}(t) = C_1e^{t} - C_2e^{-t}\).
Substituting \(t = 0\), we get:
\(y^{\backprime}(0) = C_1e^{0} - C_2e^{0} = C_1 - C_2 = 3\).
This gives us the second equation:
\(C_1 - C_2 = 3\).

With the two equations derived from the initial conditions, we now have a system of linear equations:
\[ \begin{cases} C_1 + C_2 = 1 \ (1) \ C_1 - C_2 = 3 \ (2) \ \end{cases} \].
To solve this system, add equations (1) and (2) to eliminate \(C_2\):
\(C_1 + C_2 + C_1 - C_2 = 1 + 3\), which simplifies to:
\(2C_1 = 4\).
Solving for \(C_1\), we get:
\(C_1 = 2\).
Substitute \(C_1 = 2\) back into equation (1):
\(2 + C_2 = 1\), which simplifies to:
\(C_2 = -1\).
Thus, we have found \(C_1 = 2\) and \(C_2 = -1\), solving the system of equations.

With the values of \(C_1\) and \(C_2\) found, we can now write the particular solution to the differential equation. Substitute \(C_1 = 2\) and \(C_2 = -1\) into our general solution:
\(y(t) = 2e^{t} - e^{-t}\).
This is the particular solution that satisfies both the differential equation and the initial conditions provided in the problem.