Question

Find the solution of the given initial value problem and plot its graph. How does the solution behave as \(t \rightarrow \infty ?\) $$ 6 y^{\prime \prime \prime}+5 y^{\prime \prime}+y^{\prime}=0 ; \quad y(0)=-2, \quad y^{\prime}(0)=2, \quad y^{\prime \prime}(0)=0 $$

Short answer

Answer: The general solution is \(y(t) = C_1 e^{\frac{-1}{2}t} + C_2 e^{-t} + C_3 t e^{-t}\). Question: What are the values of the constants C_1, C_2, and C_3 after applying the initial conditions? Answer: The values of the constants are \(C_1 = -2\), \(C_2 = 0\), and \(C_3 = 1\). Question: What is the particular solution of the ODE given the initial conditions? Answer: The particular solution is \(y(t) = -2 e^{\frac{-1}{2}t} + t e^{-t}\). Question: What is the behavior of the solution as \(t \rightarrow \infty\)? Answer: As \(t \rightarrow \infty\), the solution tends to zero.

Step-by-step solution

Find the general solution for the ODE

To find the general solution, let's first consider the characteristic equation that corresponds with this ODE: $$ 6m^3 + 5m^2 + m = 0 $$ We can simplify it by factoring out m: $$ m(6m^2 + 5m + 1) = 0 $$ Now we need to solve the quadratic equation to determine the possible values of m.

Solve the quadratic equation

Apply the quadratic formula to the remaining equation, where \(a=6\), \(b=5\), and \(c=1\): $$ m = \frac{-b\pm\sqrt{b^2-4ac}}{2a} = \frac{-5\pm\sqrt{5^2 - 4 \cdot 6 \cdot 1}}{12} $$ Simplify the expression to get the two possible values for m: $$ m_1 = \frac{-1}{2} \text{ and } m_2 = -1 $$

Write the general solution

With the characteristic roots found, we can now write the general solution for the third-order ODE as follows: $$ y(t) = C_1 e^{m_1 t} + C_2 e^{m_2 t} + C_3 t e^{m_2 t} $$ which simplifies to: $$ y(t) = C_1 e^{\frac{-1}{2}t} + C_2 e^{-t} + C_3 t e^{-t} $$

Apply the initial conditions to find the particular solution

We have the initial conditions: \(y(0) = -2\), \(y^{\prime}(0) = 2\), and \(y^{\prime\prime}(0) = 0\). We need to differentiate y(t) twice and then use these conditions to find the values of \(C_1\), \(C_2\), and \(C_3\). $$ y'(t) = -\frac{1}{2} C_1 e^{\frac{-1}{2}t} - C_2 e^{-t} - C_3 e^{-t} + C_3 t e^{-t} $$ $$ y''(t) = \frac{1}{4} C_1 e^{\frac{-1}{2}t} + C_2 e^{-t} + C_3 e^{-t} - C_3 t e^{-t} $$ Use the initial conditions: $$y(0) = C_1 + C_2 = -2$$ $$y'(0) = -\frac{1}{2}C_1 - C_2 - C_3 = 2$$ $$y''(0) = \frac{1}{4}C_1 + C_2 + C_3 = 0$$

Solve for constants C_1, C_2, and C_3

To solve for the constants, use matrices or any other method you prefer. The solution will be: $$ C_1 = -2, \quad C_2 = 0, \quad C_3 = 1 $$ Now, substitute these values back into the general equation: $$ y(t) = -2 e^{\frac{-1}{2}t} + t e^{-t} $$

Examine behavior as t approaches infinity and plot the graph

As \(t \rightarrow \infty\), the terms involving exponentials will decay to zero since their exponents are negative. Therefore, the solution will tend to zero. To plot the graph, use a graphing calculator or software and enter the function for the particular solution: $$ y(t) = -2 e^{\frac{-1}{2}t} + t e^{-t} $$

Key concepts

Differential Equations

Differential equations play a fundamental role in mathematics, physics, engineering, and many other disciplines, as they describe the relationship between functions and their derivatives. Specifically, they allow us to model the change of a system over time or space.

For the initial value problem in question, we're dealing with a third-order ordinary differential equation (ODE). The order of an ODE is determined by the highest derivative present, which in this case is the third derivative of the function y with respect to the variable t. Inferring from the context of the problem, this could model a physical system where acceleration (the second derivative of position with respect to time) itself is changing, such as in advanced classical mechanics scenarios involving jerk, which is the rate of change of acceleration.

The inclusion of initial conditions, such as the specific values for the function and its first two derivatives at t=0, allows us to find a unique solution tailored to a specific scenario. This is crucial for real-world applications, where the general behavior of a system must conform to certain conditions at a starting point in time.

Characteristic Equation

When solving linear differential equations with constant coefficients, a characteristic equation is a powerful tool. It transforms the problem of solving a differential equation into one of solving a polynomial equation.

The roots of the characteristic equation, which are the solutions to this polynomial, provide us with critical information about the behavior of the differential equation's solutions. For instance, in our exercise, the characteristic equation is a cubic polynomial, and its roots determine the terms of the general solution of the ODE. This is because each root represents an eigenvalue related to an exponential solution of the system.

If the characteristic equation yields real and distinct roots, the corresponding solutions to the ODE will be real exponential functions. On the other hand, complex roots would indicate oscillatory behavior in the solutions, such as sine and cosine functions multiplied by exponentials. By understanding the characteristic equation, we gain insight into the type of functions that will form the general solution.

Laplace Transform

The Laplace transform is a versatile and powerful integral transform used frequently for analyzing linear time-invariant systems. It is especially useful for simplifying the process of solving differential equations.

While not directly used in our step-by-step solution, understanding the Laplace transform can provide additional methods to tackle differential equations. It works by transforming a function of time, t, into a function of a complex variable, s. This transformation changes differential equations into algebraic equations, which are often easier to solve.

The Laplace transform also provides a straightforward way to apply initial conditions, which is a considerable advantage when solving initial value problems. After solving for the function in the Laplace domain, the inverse Laplace transform is used to revert back to the time domain, giving us the solution to the original problem. The Laplace transform is particularly effective when dealing with systems subject to Heaviside (step) functions or Dirac delta (impulse) functions.

Graph of Solutions

Visualizing the solutions of differential equations is particularly insightful as it allows us to see the behavior of the system the equation is describing. The graph illustrates how the value of the function changes with respect to time or any other independent variable.

In our exercise, plotting the graph of the particular solution to the ODE gives us a clearer picture of how the system behaves over time. Given the exponential terms in the solution, we can predict that as time progresses, these terms will diminish due to their negative exponents, leading the solution to approach zero, as confirmed by our analysis for large t.

Graphs serve not only as visual confirmations of analytical conclusions but can also reveal aspects of the solutions that are not immediately apparent. For example, the presence of transient behavior, stability, and long-term trends are all easily recognizable on a well-crafted graph. In educational settings, encouraging students to convert their mathematical findings into visual forms fosters a deeper understanding and appreciation for the material.