Question

For the given differential equation, (a) Determine the roots of the characteristic equation. (b) Obtain the general solution as a linear combination of real-valued solutions. (c) Impose the initial conditions and solve the initial value problem. $$ y^{\prime \prime \prime}+4 \pi^{2} y=0, \quad y(1)=2, \quad y^{\prime}(1)=1 $$

Short answer

In conclusion, the general solution to the given differential equation is: $$ y_p(t) = c_1 e^{r_1t} + c_2 \cos(r_2t) + c_3 \sin(r_2t) $$ with the roots \(r_1\), \(r_2\), and \(r_3\), given by: $$ r_1 = \sqrt[3]{-4\pi^2}, \quad r_2 = \sqrt[3]{-4\pi^2}\cdot e^{\frac{2\pi}{3}i}, \quad r_3 = \sqrt[3]{-4\pi^2}\cdot e^{\frac{4\pi}{3}i} $$ We were not able to determine unique values for the constants \(c_1\), \(c_2\), and \(c_3\) using the given initial conditions. Additional constraints or information would be required to find a unique particular solution.

Step-by-step solution

Write down the given differential equation and initial conditions

The given differential equation is: $$ y^{\prime \prime \prime} + 4\pi^2 y = 0 $$ with initial conditions: $$ y(1) = 2, \quad y^{\prime}(1) = 1 $$

Determine the roots of the characteristic equation

To find the characteristic equation for the given differential equation, replace y with \(e^{rt}\), where r is the roots of the equation. We get: $$ r^3 + 4\pi^2 = 0 $$ The characteristic equation is a cubic equation, and we can factor it as: $$ r^3 = - 4\pi^2 $$ The roots are the cube roots of the left side of the equation: $$ r = \sqrt[3]{-4\pi^2} $$ Let's find the three roots: $$ r_1 = \sqrt[3]{-4\pi^2} \omega^0 = \sqrt[3]{-4\pi^2} $$ $$ r_2 = \sqrt[3]{-4\pi^2} \omega^1 = \sqrt[3]{-4\pi^2}\cdot e^{\frac{2\pi}{3}i} $$ $$ r_3 = \sqrt[3]{-4\pi^2} \omega^2 = \sqrt[3]{-4\pi^2}\cdot e^{\frac{4\pi}{3}i} $$ where \(\omega = e^{2\pi i /3}\) is the cube root of unity.

Obtain the general solution as a linear combination of real-valued solutions

Using the roots \(r_1\), \(r_2\), and \(r_3\), we can write the general solution as a linear combination of real-valued solutions: $$ y_p(t) = c_1 e^{r_1t} + c_2 e^{r_2t} + c_3 e^{r_3t} $$ Since \(r_2\) and \(r_3\) are complex conjugates, the general solution can be written as a sum of a real part and an imaginary part. Using Euler's formula, we can rewrite the general solution as: $$ y_p(t) = c_1 e^{r_1t} + c_2 \cos(r_2t) + c_3 \sin(r_2t) $$

Impose the initial conditions and solve the initial value problem

Now we will use the initial conditions to find the particular solution. First, apply the initial condition \(y(1) = 2\): $$ 2 = c_1 e^{r_1(1)} + c_2 \cos(r_2(1)) + c_3 \sin(r_2(1)) $$ Next, apply the initial condition \(y^{\prime}(1) = 1\): $$ 1 = c_1 r_1 e^{r_1(1)} - c_2 r_2 \sin(r_2(1)) + c_3 r_2 \cos(r_2(1)) $$ Now, we must solve the above system of two linear equations with three variables \(c_1\), \(c_2\), and \(c_3\). Without losing the generality, it is impossible to determine unique values for \(c_1\), \(c_2\), and \(c_3\) from this equation system. However, the general solution is presented and can be adapted to suit additional information or additional constraints.

Key concepts

Characteristic Equation

Understanding the characteristic equation is a fundamental step in solving differential equations, especially when dealing with linear homogeneous differential equations with constant coefficients. The characteristic equation emerges when we make a pivotal assumption: we assume that solutions take the form of an exponential function, typically denoted by \( e^{rt} \), where \( r \) is an unknown that we wish to solve for.

For example, if we have a differential equation like \( y''' + 4\pi^2 y = 0 \), we would substitute \( y \) with \( e^{rt} \) and solve for \( r \). This substitution transforms our differential equation into a polynomial, where the roots of the polynomial \( r \) hold the key to constructing the general solution. It's an elegant way to turn the problem of solving differential equations into an algebraic one.

After obtaining the characteristic equation, which in our case is \( r^3 + 4\pi^2 = 0 \), we solve for the root \( r \). Calculating the roots tells us how our solution behaves, whether it oscillates, exponentially grows, or decays. In practice, the characteristic equation lets us peek into the soul of the differential equation, laying out its fundamental solutions.

Initial Value Problem

An initial value problem is a type of differential equation accompanied by specific conditions that dictate the value of the unknown function and, possibly, some of its derivatives at a given point. These conditions aid in determining a unique solution from the family of all possible solutions that a differential equation might have.

To illustrate, consider the previous differential equation with defined initial conditions: \( y(1) = 2 \) and \( y'(1) = 1 \). These conditions are what make our problem an 'initial value problem.' They serve as the anchor for our solution, ensuring we don't just have a general idea of the solution's shape but have a precise function that specifically satisfies the conditions at a given point in time.

Solving an initial value problem involves integrating the general solution with the initial conditions. In our example, by using these conditions, we can find the exact values for the constants \( c_1, c_2, \) and \( c_3 \). Often, in the real world, these initial conditions represent physical realities, like the initial speed of an object or the initial amount of a substance, shaping the specifics of the differential equation we're investigating.

Method of Variation of Parameters

When dealing with nonhomogeneous differential equations, the method of variation of parameters is a powerful technique that allows us to find a particular solution that complements the complementary (homogenous) solution. The beauty of this method lies in its capability to tackle equations where the nonhomogeneous term does not fit neatly into methods such as the method of undetermined coefficients.

In essence, this method involves determining functions to replace the arbitrary constants in the complementary solution. It's like we're saying, 'instead of having constant multipliers, let's have these multipliers vary in a particular way to satisfy the nonhomogeneous part of the equation.'

The execution of this method includes finding a set of particular functions \( u_1(t) \) and \( u_2(t) \), which, when replacing the constants in the complementary solution, will provide the specific shape of the curve we're looking to describe, managing to fulfill the 'particular' role needed to solve the non-homogeneous equation.

General Solution of Differential Equation

The general solution of a differential equation represents a family of functions that encapsulates all possible solutions to the equation. It includes arbitrary constants which are later determined by the initial or boundary conditions. The general solution combines the homogeneous and particular solutions, especially in the context of linear differential equations with constant coefficients.

To practically derive a general solution, one might first calculate the complementary (homogeneous) solution which solves the equation when the nonhomogeneous part is set to zero. In our exercise, the complementary solution \( y_c(t) = c_1 y_1(t) + c_2 y_2(t) \) is formed from the general solutions of the homogeneous part of the differential equation.

Next, the particular solution is found, this time accounting for the nonhomogeneous term (like a source or force acting on the system), and can often be the trickiest part. In the method of variation of parameters, we determine specific forms for the parameters to get this solution. Merging these, we achieve the general solution which is malleable enough to meet various initial conditions and versatile in covering all scenarios dictated by the differential equation.