91Ó°ÊÓ

When dealing with linear differential equations, the characteristic equation plays a crucial role. It helps us understand the behavior of the solution without directly solving the differential equation. For a homogeneous equation like \( y'' + 4y = 0 \), we replace the derivatives with powers of \( r \), following the characteristic polynomial:

• For \( y'' \), use \( r^2 \)
• For \( y \), use a constant, here it is 4

Thus, the characteristic equation turns into: \( r^2 + 4 = 0 \). Solving this equation gives the roots, which are key to constructing the homogeneous solution. In this example, the roots are complex: \( r = \pm 2i \). This indicates oscillatory behavior, typical of trigonometric functions, which reflects in the solution format with sines and cosines.

An initial value problem means you have extra information that allows you to find specific constants in your general solution. Here, the additional conditions are given: \( y(0) = 1 \) and \( y'(0) = 2 \). These conditions specify the value of the solution and its derivative at a particular point, helping to find unique values for any constants left from solving the characteristic equation. Applying these initial conditions transforms a broad set of possible solutions into one specific solution tailored to fit these constraints, making the problem more concrete.

For differential equations such as \( y'' + 4y = 2x \), where you have a non-zero term on the right side, a particular solution addresses this non-homogeneous component. To find it, we propose a guess in the form \( y_p(x) = ax + b \), suited for the linear part of \( 2x \).Substituting \( y_p \) into the equation, we adjust \( a \) and \( b \) to satisfy the equation. Here, it leads to \( a = \frac{1}{2} \) and \( b = 0 \), giving \( y_p(x) = \frac{x}{2} \). The particular solution deals only with the non-homogeneous part, complementing the homogeneous solution.

The general solution combines both the homogeneous and particular solutions, offering a complete picture of the original differential equation's behavior. For this exercise, it looks like:\[ y(x) = C_1 \cos(2x) + C_2 \sin(2x) + \frac{x}{2} \].This formula includes:

• The trigonometric terms \( C_1 \cos(2x) + C_2 \sin(2x) \) representing the response to the homogeneous part \( y'' + 4y = 0 \)
• The linear term \( \frac{x}{2} \) specifically addressing the inhomogeneous \( 2x \) term.

By applying the initial conditions, we solve for the constants \( C_1 = 1 \) and \( C_2 = \frac{3}{4} \). The general solution, once these values are incorporated, provides a unique solution for the initial value problem: \( y(x) = \cos(2x) + \frac{3}{4} \sin(2x) + \frac{x}{2} \).