Question

Determine the values of \(r\) for which the given differential equation has solutions of the form \(y=e^{t} .\) $$ y^{\prime \prime \prime}-3 y^{\prime \prime}+2 y^{\prime}=0 $$

Short answer

The values of r for which there exists a solution of the form y = e^(rt) for the given third-order homogeneous linear differential equation are r = 0, r = 1, and r = 2.

Step-by-step solution

Substitute the given solution form into the differential equation

Given the differential equation: $$ y^{\prime \prime \prime}-3 y^{\prime \prime}+2 y^{\prime}=0 $$ Let the solution be in the form $$ y=e^{r t}. $$ Now find the first, second, and third derivatives of this solution form. First derivative: $$ y' = r e^{r t} $$ Second derivative: $$ y'' = r^2 e^{r t} $$ Third derivative: $$ y''' = r^3 e^{r t} $$ Now plug these derivatives into the given differential equation: $$ (r^3 e^{rt}) - 3(r^2 e^{rt}) + 2(re^{rt}) = 0 $$

Simplify and obtain a polynomial equation in r

Since \(e^{rt}\) is always non-zero, we can divide both sides of the equation by \(e^{rt}\) to simplify it: $$ r^3 - 3r^2 + 2r = 0 $$ Now we are left with a polynomial equation in r.

Factor and find the roots of the polynomial

Factor the polynomial equation in r: $$ r(r^2 - 3r + 2) = 0 $$ Now factor the quadratic inside the parenthesis: $$ r(r - 1)(r - 2) = 0 $$ The roots of this equation are \(r = 0\), \(r = 1\), and \(r = 2\).

Conclusion

The values of \(r\) for which the given differential equation has solutions of the form \(y = e^{t}\) are \(r = 0\), \(r = 1\), and \(r = 2\).

Key concepts

Ordinary Differential Equations

An ordinary differential equation (ODE) is an equation that involves functions and their derivatives. In these equations, the function is defined for one independent variable, generally time. Solving an ODE means finding a function that satisfies the equation when the derivatives are replaced by that function's derivatives.

For example, the given ODE in the exercise, \[ y''' - 3y'' + 2y' = 0 \], is a third-order linear homogeneous ODE with constant coefficients. It's linear because each of its terms is either a derivative of \( y \) or \( y \) itself, all to the first power, and it's homogeneous because there's no term that's just a function of time without \( y \). A significant aspect of ODEs is their initial conditions or boundary values that can provide unique solutions under suitable circumstances.

Characteristic Equation

The characteristic equation is a key tool for finding the solutions to linear homogeneous differential equations with constant coefficients, such as the one presented in the exercise. It is obtained by substituting the trial solution, usually in the form of an exponential function \( e^{rt} \), into the differential equation and setting the result as a polynomial equal to zero.

Using this method transforms the ODE into an algebraic equation without derivatives, which is much easier to solve. For our given ODE, the characteristic equation is \( r^3 - 3r^2 + 2r = 0 \). Solving this algebraic equation provides the values of \( r \) which we refer to as the roots. These roots are crucial because they define the exponential solutions to the differential equation.

Exponential Solutions

Exponential solutions to differential equations arise naturally when dealing with constant coefficient linear ODEs. If an ODE has the form \( y''' - ay'' + by' = 0 \), it suggests the solution might be an exponential function, which, when differentiated, remains a scaled version of itself. In the exercise, we seek solutions of the form \( y = e^{rt} \).

Upon finding the solutions to the characteristic equation, we get specific values for \( r \): \( r = 0 \), \( r = 1 \), and \( r = 2 \). Each of these values provides an exponential solution to the original ODE: \( e^{0t} \) (which simplifies to 1, representing a constant solution), \( e^{t} \), and \( e^{2t} \). These solutions are critical as they form the general solution of the ODE, which can be combined linearly to fit specific boundary conditions or initial values.