91Ó°ÊÓ

When dealing with second-order differential equations, like our example \(y'' - y' - 2y = 0\), finding the characteristic equation is a key step. This involves assuming a solution of the form \(y = e^{mx}\).

By differentiating this assumed solution, we get its first and second derivatives:

• First derivative: \(\frac{dy}{dx} = me^{mx}\)
• Second derivative: \(\frac{d^2y}{dx^2} = m^2e^{mx}\)

Substituting these derivatives back into the original differential equation results in an equation that includes the terms \(m^2e^{mx}\), \(me^{mx}\), and \(2e^{mx}\).

Since every term contains the factor \(e^{mx}\), which cannot be zero, factor it out to derive the characteristic equation: \(m^2 - m - 2 = 0\). This quadratic expression must be solved to find the values of \(m\), which characterize the general solutions of the differential equation.

To solve the characteristic equation \(m^2 - m - 2 = 0\), we use the quadratic formula, a tool to find the roots of any quadratic equation \(ax^2 + bx + c = 0\). Here, \(a = 1\), \(b = -1\), and \(c = -2\).

The quadratic formula is given by:
\[m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
Substituting the coefficients from our equation:

• \(b^2 = (-1)^2 = 1\)
• \(-4ac = -4(1)(-2) = 8\)

Applying these values, the equation becomes:
\[m = \frac{1 \pm \sqrt{1 + 8}}{2}\]Simplifying further gives us the solutions \(m = -1\) and \(m = 2\). These solutions form the basis for finding the general solution of the original differential equation.

With values of \(m\) from the quadratic formula, the general solution is formed. Each solution corresponds to each root of the characteristic equation. For our equation \(y'' - y' - 2y = 0\), the roots were \(m = -1\) and \(m = 2\).

Thus, the exponential solutions to the differential equation are:

• \(y_1(x) = e^{-x}\)
• \(y_2(x) = e^{2x}\)

These functions provide two linearly independent solutions to the differential equation. By combining them, you can create the general solution, often expressed as:

\[y(x) = C_1e^{-x} + C_2e^{2x}\]Where \(C_1\) and \(C_2\) are constants determined by initial or boundary conditions that would be given in a specific problem context.

Verification means confirming that your found solutions actually satisfy the original differential equation. Take each solution, such as \(y_1(x) = e^{-x}\), calculate its derivatives, and substitute them back into \(y'' - y' - 2y = 0\).

For verification of \(y_1(x) = e^{-x}\):

• Second derivative: \(\frac{d^2}{dx^2}(e^{-x}) = -e^{-x}\)
• First derivative: \(\frac{d}{dx}(e^{-x}) = -e^{-x}\)

Substitute into the differential equation:
\(-e^{-x} - (-e^{-x}) - 2(e^{-x}) = 0\)

The terms cancel out, confirming the solution is valid. Similarly, check \(y_2(x) = e^{2x}\):

• Second derivative: \(\frac{d^2}{dx^2}(e^{2x}) = 4e^{2x}\)
• First derivative: \(\frac{d}{dx}(e^{2x}) = 2e^{2x}\)

Checking, \(4e^{2x} - 2e^{2x} - 2e^{2x} = 0\), validates the solution. Both methods confirm these functions are correct solutions of the differential equation.