91Ó°ÊÓ

In differential equations, initial conditions are values specified for the solution at a particular point. For example, in this exercise, the system of differential equations is given with initial conditions:

• \( x(0) = 0 \)
• \( y(0) = 1 \)

These initial conditions are crucial because they allow us to find the exact values of the constants in the general solution of the differential equation, leading to a particular solution. Without these conditions, we would only have a general form of the solution with unknown constants, which couldn't describe a specific situation completely.

A second-order differential equation involves the second derivative of the unknown function. In this exercise, we start with a system of first-order differential equations, but by differentiating and substituting, we reduce it to a single second-order equation:

• \( x' = y \)
• \( y' = 2x - y \)

Differentiating \( x' = y \) and substituting into \( y' = 2x - y \), we get:

\( x'' = y' = 2x - y \)
Using \( y = x' \), we substitute and rearrange to form:

\( x'' + x' - 2x = 0 \)
This second-order linear homogeneous differential equation is easier to solve using characteristic equations.

To solve a second-order linear homogeneous differential equation, we use the characteristic equation. The characteristic equation is derived from assuming a solution of the form \( x = e^{rt} \). For the equation \( x'' + x' - 2x = 0 \), substituting \( x = e^{rt} \) gives:

\( r^2 e^{rt} + r e^{rt} - 2 e^{rt} = 0 \)
\( e^{rt}(r^2 + r - 2) = 0 \)
From this, the characteristic equation is:
\( r^2 + r - 2 = 0 \)
Solving this quadratic equation using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = 1 \), and \( c = -2 \), we find:

\( r = \frac{-1 \pm 3}{2} \)
Thus, the roots are \( r_1 = 1 \) and \( r_2 = -2 \).

The general solution to the second-order differential equation uses the roots of the characteristic equation. For roots \( r_1 = 1 \) and \( r_2 = -2 \), the general solution is:

\( x(t) = A e^{t} + B e^{-2t} \)

To find the particular solution, we apply the initial conditions to determine the constants \( A \) and \( B \). Using the initial conditions \( x(0) = 0 \) and \( y(0) = 1 \) leads to:

For \( x(0) = 0 \):

\( A + B = 0 \)
So, \( A = -B \).

For \( y(0) = 1 \), using \( y = x' \):

\( y(0) = A - 2B = 1 \)

Since \( A = -B \):

\( -B - 2B = 1 \)
\( -3B = 1 \)
\( B = -\frac{1}{3} \)
\( A = \frac{1}{3} \)

Thus, the particular solution is:

\( x(t) = \frac{1}{3} e^{t} - \frac{1}{3} e^{-2t} \)
\( y(t) = x' = \frac{1}{3} e^{t} + \frac{2}{3} e^{-2t} \)