Question

Obtain the general solution of \(y^{\prime \prime \prime}-5 y^{\prime \prime}+6 y^{\prime}=0\). [Hint: Make the change of dependent variable \(u(t)=y^{\prime}(t)\), determine \(u(t)\), and then antidifferentiate to obtain \(y(t) .\) ]

Short answer

Question: Determine the general solution of the following third-order linear homogeneous ordinary differential equation: \(y^{\prime \prime \prime}(t)-5y^{\prime \prime}(t)+6y^{\prime}(t)=0\). (Hint: Define a new dependent variable \(u(t)=y^{\prime}(t)\), and solve the resulting second-order linear ODE for \(u(t)\). Then, antidifferentiate to find the general solution for \(y(t)\).) Answer: The general solution of the given third-order linear homogeneous ordinary differential equation is: $$y(t)= \frac{C_1}{2} e^{2t}+\frac{C_2}{3} e^{3t} + C_3$$ where \(C_1\), \(C_2\), and \(C_3\) are constants of integration.

Step-by-step solution

Define the new dependent variable

The hint suggests using a new dependent variable \(u(t)=y^{\prime}(t)\). Therefore, we can also rewrite \(y^{\prime \prime}(t)\) and \(y^{\prime \prime \prime}(t)\) in terms of \(u(t)\): $$y^{\prime \prime}(t)=u^{\prime}(t)$$ $$y^{\prime \prime \prime}(t)=u^{\prime \prime}(t)$$ Now let's substitute these into the original equation.

Substitute the new variable into the equation

Equation for substitution: $$y^{\prime \prime \prime}-5 y^{\prime \prime}+6 y^{\prime}=0$$ Replacing the new dependent variable \(u(t)\), we get: $$u^{\prime \prime}(t)-5u^{\prime}(t)+6 u(t)=0$$

Solve the second-order linear ordinary differential equation for u(t)

Now we have a simpler second-order linear ordinary differential equation for \(u(t)\): $$u^{\prime \prime}(t)-5u^{\prime}(t)+6 u(t)=0$$ The characteristic equation for this ODE is : $$r^2-5r+6=0$$ Upon factoring this equation, we get \((r-2)(r-3)=0\). Therefore, \(r=2\) and \(r=3\) are the roots of the characteristic equation, and the general solution for \(u(t)\) is given by: $$u(t)=C_1 e^{2t}+C_2 e^{3t}$$

Antidifferentiate to obtain y(t)

Recall that \(u(t)=y^{\prime}(t)\). Thus, to find the general solution for \(y(t)\), we need to integrate \(u(t)\) with respect to \(t\). $$y^{\prime}(t)= C_1 e^{2t}+C_2 e^{3t}$$ Integrating once, we get: $$y(t)= \frac{C_1}{2} e^{2t}+\frac{C_2}{3} e^{3t} + C_3$$ Here, \(C_1\), \(C_2\), and \(C_3\) are constants of integration that could be determined by any initial conditions if they were given in the problem. Given that we're asked to find the general solution of the equation, this is our final answer.

Key concepts

Characteristic Equation

When solving linear ordinary differential equations (ODEs), we often encounter a step where we need to find the roots of a polynomial known as the 'characteristic equation.' This approach is particularly useful for linear differential equations with constant coefficients.

Consider the earlier example from the exercise where the ODE is simplified to a second-order differential equation in terms of a new variable, leading to the formation of a characteristic equation \(r^2-5r+6=0\). By solving this algebraic equation, we obtain the values of \(r\), which represent the roots and are essential in formulating the general solution for the ODE.

The characteristic equation helps in understanding the behavior of the solution without having to solve the ODE completely. For instance, the roots can be real and distinct, real and repeated, or complex conjugates, and each scenario affects the form of the general solution differently.

After determining the roots, the general solution of the second-order differential equation would be \(u(t)=C_1 e^{rt}+ C_2 e^{rt}\) for each distinct root \(r\), where \(C_1\) and \(C_2\) are constants.

Linear Ordinary Differential Equation

Linear ordinary differential equations are a class of differential equations where the unknown function and its derivatives appear linearly, meaning they are not multiplied together, raised to a power, or composed with another function.

In the original problem, we are given a third-order linear ODE, \(y^{\prime \prime \prime}-5y^{\prime \prime}+6y^\prime=0\). By applying the hint provided in the exercise to use a change of variable, we reduce the order of the ODE, simplifying the problem to a more manageable second-order linear ODE

Understanding how to transform and tackle linear ODEs is a vital skill in mathematics. It allows us to handle complex real-world problems that can be modeled by these types of equations, from natural phenomena in physics to rate-of-change problems in economics.

The crux of solving linear ODEs often lies in finding the characteristic equation and using its roots to construct the general solution.

Integration of Functions

Integration is a fundamental concept in calculus, often considered the inverse operation to differentiation. When solving differential equations, integration helps us to recover the original function from its derivative.

In our exercise, once we have determined the general solution for the new variable \(u(t)=y^\prime(t)\), we need to integrate to find the original function \(y(t)\). The process of integrating \(u(t)\) with respect to \(t\) allows us to ascertain \(y(t)\), the general solution of the original ODE.

For the given exercise, integrating \(C_1 e^{2t}+ C_2 e^{3t}\) yields \(\frac{C_1}{2} e^{2t}+\frac{C_2}{3} e^{3t}+C_3\), where \(C_3\) is an additional constant of integration. This final form represents the solution to the original third-order differential equation, showcasing how integration serves to tie together the overall process of solving differential equations.

Understanding the techniques of integration, including the recognition of standard integral forms, is critical when attempting to solve ODEs, as it allows for the reconstruction of solutions from their derivatives.