91Ó°ÊÓ

To solve a second-order linear homogeneous differential equation, we start by finding the characteristic equation. This equation helps us determine the behavior of the solutions. For our example, y'' + 2y' + 5y = 0,we convert this into the characteristic equation using the general form ar^2 + br + c = 0.Here, a = 1, b = 2, and c = 5. By substituting these values, we get the characteristic equation:r^2 + 2r + 5 = 0.This quadratic equation will help us find the roots which are essential for constructing the general solution.

When dealing with second-order differential equations, the quadratic formula is a powerful tool to solve the characteristic equation. The quadratic formula is given by\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].In our problem, we have a = 1, b = 2, and c = 5. Plugging in these values, we get:\[ r = \frac{-2 \pm \sqrt{4 - 20}}{2} = \frac{-2 \pm \sqrt{-16}}{2} \].Inside the square root, \( \sqrt{-16} \) becomes \( 4i \), leading to:\[ r = \frac{-2 \pm 4i}{2} = -1 \pm 2i \].These roots, \( r = -1 + 2i \) and \( r = -1 - 2i \) , will guide us in forming the general solution.

The roots of the characteristic equation can sometimes be complex, as we have seen with r = -1 \pm 2i.Complex roots indicate that the solutions will involve exponential functions multiplied by trigonometric functions. For any complex root of the form \( \alpha \pm \beta i \),the general solution to the differential equation can be written as:\[ y(x) = e^{\alpha x}(C_1\cos(\beta x) + C_2\sin(\beta x)) \].Applying this to our roots, we get:\[ y(x) = e^{-x}(C_1\cos(2x) + C_2\sin(2x)) \].This formula combines exponential decay with oscillatory motion.

Initial conditions are given values for the function and/or its derivatives at a specific point, often used to determine the particular solution. In our case: y(0) = 1, y'(0) = 0.First, substitute these initial conditions into the general solution to find constants \(C_1 \) and \( C_2 \). Starting with y(0) = 1, we get: \( e^{0}(C_1\cos(0) + C_2\sin(0)) = C_1 = 1 \).Next, differentiate the general solution:\[ y'(x) = e^{-x}(C_1(-\cos(2x) - 2\sin(2x)) + C_2(-\sin(2x) + 2\cos(2x)) ) \].Substituting the initial condition y'(0) = 0, we find:\[ y'(0) = -C_1 + 2C_2 = 0 \], leading to:\( C_2 = \frac{1}{2} \).These constants, \( C_1 = 1 \) and \( C_2 = \frac{1}{2} \), give the particular solution:\[ y(x) = e^{-x}(\cos(2x) + \frac{1}{2}sin(2x)) \].This solution satisfies both the differential equation and the initial conditions.