91Ó°ÊÓ

When dealing with linear homogeneous differential equations like the given one, a key step is forming the characteristic equation. This is a crucial concept as it helps determine the nature of the solutions.

To derive the characteristic equation, assume a solution of the form \( f(x) = e^{rx} \). When substituted into the differential equation, the exponential function naturally cancels out, leading to a polynomial equation. In this exercise, the characteristic equation derived is:

• \( r^2 - 2r + 5 = 0 \)

Solving this polynomial will indicate the types of roots we have and guide us to the general solution. This characteristic equation essentially converts the differential equation problem into an algebraic one.

Complex roots occur in the quadratic equation when the discriminant \( b^2 - 4ac \) is negative. In our example, the characteristic equation \( r^2 - 2r + 5 = 0 \) has a discriminant of \(-16\). This negative result indicates complex roots.

Calculating using the quadratic formula, the roots are:

• \( r = 1 + 2i \)
• \( r = 1 - 2i \)

Complex roots appear in conjugate pairs due to their structure, \( r = \, \alpha \pm \beta i \). Here, \( \alpha = 1 \) and \( \beta = 2 \). These roots lead to a general solution that involves both sine and cosine components, reflecting the oscillatory nature of the complex roots.

Boundary conditions are additional constraints given to find a unique solution to a differential equation. They specify the values of the solution, or its derivatives, at certain points. For the equation at hand, the boundary conditions provided are:

• \( f(0) = 1 \)
• \( f(\pi/4) = 0 \)

Applying these conditions to the general solution allows us to solve for the constants \( C_1 \) and \( C_2 \). Such constraints ensure that the found solution fits the specific problem scenario, distinguishing it from the infinite solutions of the homogeneous equation without conditions.

The general solution of a differential equation constitutes a fundamental part of solving such mathematical problems. For second-order linear differential equations with complex roots like this one, the general solution is expressed as:

\[ f(t) = e^{\alpha t} (C_1 \cos(\beta t) + C_2 \sin(\beta t)) \] In our example, substituting \( \alpha = 1 \) and \( \beta = 2 \) results in:

• \( f(t) = e^{t} (C_1 \cos(2t) + C_2 \sin(2t)) \)

This form allows flexibility through the constants \( C_1 \) and \( C_2 \), which are determined by applying boundary conditions. These constants adjust to tailor the general formula to the specific scenario being assessed. By imposing the given boundary conditions, the particular solution \( f(t) = e^t \cos(2t) \) is ultimately found, providing a specific answer to the differential equation question.