91Ó°ÊÓ

A third-order differential equation involves the third derivative of a function. The general form is given by: \(a_3 y''' + a_2 y'' + a_1 y' + a_0 y = 0\). Here, \(a_3, a_2, a_1,\) and \(a_0\) are constants.In this exercise, we worked with the equation: \(y''' - 10y'' + 25y' = 0\). Notice it doesn’t have a \(y\) term. This type of differential equation is called 'homogeneous' because it equals zero.Homogeneous differential equations are simpler to solve. The method involves assuming a solution of the form \(y = e^{rt}\), leading us to the characteristic equation. This assumption helps us convert a difficult differential equation into an algebraic equation that's easier to solve.Working with third-order equations introduces complexity beyond second-order ones. These require finding three roots. These roots will directly influence the form of the solution.

The characteristic equation plays a key role in solving linear homogeneous differential equations. We start by assuming a solution of the form \(y = e^{rt}\).When we substitute \(y = e^{rt}\) into the differential equation, the derivatives transform it into an algebraic equation. For our example: \(y''' - 10y'' + 25y' = 0\), replacing \(y\) with \(e^{rt}\) yields: \(r^{3}e^{rt} - 10r^{2}e^{rt} + 25re^{rt} = 0\). Factoring out \(e^{rt}\) (which is never zero), we obtain: \(r^{3} - 10r^{2} + 25r = 0\).Now, we solve for \(r\) by finding the roots of the characteristic equation. This involves factoring the polynomial. In our case, we factor as follows:\(r(r^{2} - 10r + 25) = 0\). Further simplifies to \(r(r - 5)^{2} = 0\). Thus, \(r = 0\) and \(r = 5\) (with a multiplicity of 2).These roots are essential in determining the form of our general solution.

The general solution of a differential equation is a combination of terms involving the roots of the characteristic equation. For a homogeneous differential equation, if our characteristic equation has distinct roots, the solution typically involves exponential functions.In our example, the roots were \(r = 0\) and \(r = 5\) (with multiplicity 2). This means we have:- One distinct root: \(r = 0\) leads to a term \(e^{0t} = 1\). - One repeated root: \(r = 5\) (with multiplicity 2) gives terms \(e^{5t}\) and \(te^{5t}\).Combining these, we get the general solution: \(y(t) = C_{1} + C_{2}e^{5t} + C_{3}te^{5t}\).Here, \(C_1, C_2,\) and \(C_3\) are constants determined by initial conditions of the problem.This form illustrates how different types of roots (real, repeated, complex) contribute unique terms in the overall solution of the differential equation.