Question

Find the general solution of the given differential equation. In Problems 11 and \(12 \mathrm{g}\) is an arbitrary continuous function. $$ y^{\prime \prime}-5 y^{\prime}+6 y=g(t) $$

Short answer

Question: Find the general solution of the given inhomogeneous second-order linear differential equation: $$y'' - 5y' + 6y = g(t)$$ where g(t) is an arbitrary continuous function. Answer: The general solution of the given inhomogeneous equation is: $$y(t) = C_1e^{2t} + C_2e^{3t} + y_p(t)$$ where C1 and C2 are arbitrary constants and yp(t) is a particular solution corresponding to the given function g(t).

Step-by-step solution

Homogeneous Equation and Complementary Function

First, we need to find the complementary function by solving the homogeneous version of the given equation (when g(t) = 0): $$ y'' - 5y' + 6y = 0 $$ To solve this second-order linear homogeneous differential equation, we assume a trial solution of the form: $$ y = e^{rt} $$ Differentiating this trial solution twice with respect to t, we get: $$ y'=re^{rt}, \quad y'' = r^2e^{rt} $$ Substituting these derivatives and the trial solution into the given homogeneous equation, we obtain the characteristic equation for r: $$ r^2e^{rt} - 5re^{rt} + 6e^{rt} = 0 $$

Characteristic Equation and Roots

Solving the characteristic equation, we first factor out the common term e^(rt): $$ e^{rt}(r^2 -5r + 6) = 0 $$ Since e^(rt) is never zero, we can safely ignore it and focus on finding the roots of the polynomial equation: $$ r^2 - 5r + 6 = 0 $$ Factoring this, we get: $$ (r - 3)(r - 2) = 0 $$ This yields two distinct real roots, r1 = 2 and r2 = 3.

Complementary Function

Given the roots r1 and r2, we can now write down the complementary function yc(t) as: $$ y_c(t) = C_1e^{2t} + C_2e^{3t} $$ where C1 and C2 are arbitrary constants.

Particular Solution

To obtain a particular solution for the inhomogeneous equation, we use the method of undetermined coefficients. However, as g(t) is an arbitrary continuous function, we cannot find a specific form of yp(t). That being said, when you are provided with a particular g(t) later in the course or future problems, you can use the method of undetermined coefficients, the method of variation of parameters, or other suitable methods to find a particular solution.

General Solution

Finally, the general solution of the given inhomogeneous equation is the sum of the complementary function (yc(t)) and a particular solution (yp(t)): $$ y(t) = y_c(t) + y_p(t) = C_1e^{2t} + C_2e^{3t} + y_p(t) $$ where C1 and C2 are arbitrary constants and yp(t) is a particular solution corresponding to the given function g(t).

Key concepts

Homogeneous Differential Equation

A homogeneous differential equation is a type of differential equation in which the function and its derivatives are set to zero. In simple terms, it means there are no external forces or influences acting on the system. To illustrate this concept, consider the equation given in the problem:
\[ y'' - 5y' + 6y = 0 \] The left-hand side consists solely of the derivatives of the function \(y\). Since the equation is set to zero, indicating no forcing term \(g(t)\), it is referred to as homogeneous. Solving this form of the equation helps us identify the complementary function which forms part of the overall solution.

Characteristic Equation

The characteristic equation is derived from a homogeneous linear differential equation. It's a crucial step in solving second-order differential equations. It simplifies the problem by converting differential operations into algebraic forms. We achieve this by assuming a solution of the form \(y = e^{rt}\), where \(r\) is a constant, which is substituted back into the homogeneous equation.
For example:\[ r^2 e^{rt} - 5r e^{rt} + 6e^{rt} = 0\] Factoring out \(e^{rt}\) (which is never zero), gives:\[ r^2 - 5r + 6 = 0\]Solving this polynomial equation gives us the characteristic equation's roots, which are essential in developing the complementary function.

Complementary Function

The complementary function, often denoted as \(y_c(t)\), forms part of the general solution to differential equations. After finding the roots of the characteristic equation, these roots are used to construct \(y_c(t)\). Each root corresponds to one part of the solution. If the roots are distinct and real, as in our problem where the roots were 2 and 3, the complementary function is:\[ y_c(t) = C_1 e^{2t} + C_2 e^{3t}\]Here, \(C_1\) and \(C_2\) are arbitrary constants. These constants are determined based on initial or boundary conditions, which might be provided in further contexts.

Undetermined Coefficients

The method of undetermined coefficients is a technique used to find a particular solution to non-homogeneous differential equations. Unfortunately, in this exercise, a specific form of \(g(t)\) is not given, so the method does not provide a direct solution for a particular solution here. However, the approach typically involves guessing a form for \(y_p(t)\), the particular solution, based on \(g(t)\), and determining the coefficients through substitution.When you have \(g(t)\), assume a form of \(y_p(t)\) similar in form to \(g(t)\), differing in terms only if it clashes with the complementary function. This method allows you to pinpoint a specific solution for unique external inputs specified by \(g(t)\).

General Solution

The general solution of a differential equation combines the complementary function, \(y_c(t)\), with a particular solution, \(y_p(t)\). This integration covers all possible solutions of a differential equation.
In the context of the given problem, the general solution is:\[ y(t) = y_c(t) + y_p(t) = C_1 e^{2t} + C_2 e^{3t} + y_p(t)\]Here, \(C_1\) and \(C_2\) are arbitrary constants, and \(y_p(t)\) represents the particular solution which caters to \(g(t)\). Although \(y_p(t)\) remains unspecified without \(g(t)\), once \(g(t)\) is known, the precise form of the general solution can be determined.