Problem 18 - Calculus: Early Transcendentals

Question

Suppose that \(y(t)\) is a solution to \(a y^{\prime \prime}+b y^{\prime}+c y=0, y\left(t_{0}\right)=0, y^{\prime}\left(t_{0}\right)=0 .\) Show that \(y(t)=0\).

Step-by-Step Solution

Answer

The solution is \( y(t) = 0 \) for all \( t \).

Step 1: Understanding Initial Conditions

We are given that a function \( y(t) \) satisfies a second-order linear homogeneous differential equation \( a y'' + b y' + c y = 0 \). The initial conditions are \( y(t_0) = 0 \) and \( y'(t_0) = 0 \). These indicate that at the point \( t_0 \), both the function and its first derivative are zero.

Step 2: Analyzing the Differential Equation

The differential equation is homogeneous, which means it has the form \( a y'' + b y' + c y = 0 \). Such an equation typically has solutions that are exponential functions or related forms depending on the characteristic equation. However, specific solutions depend on the initial conditions provided.

Step 3: Applying Initial Conditions to General Solution

The general solution for a homogeneous second-order differential equation is usually expressed as \( y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} \), where \( r_1 \) and \( r_2 \) are roots of the characteristic equation derived from \( a y'' + b y' + c y = 0 \). When applying initial conditions \( y(t_0) = 0 \) and \( y'(t_0) = 0 \), we set \( C_1 e^{r_1 t_0} + C_2 e^{r_2 t_0} = 0 \) and \( C_1 r_1 e^{r_1 t_0} + C_2 r_2 e^{r_2 t_0} = 0 \).

Step 4: Solving System of Equations

The system of equations from the initial conditions is:1. \( C_1 e^{r_1 t_0} + C_2 e^{r_2 t_0} = 0 \)2. \( C_1 r_1 e^{r_1 t_0} + C_2 r_2 e^{r_2 t_0} = 0 \)These equations indicate that both the solution and its derivative at \( t_0 \) are zero regardless of the coefficients \( C_1 \) and \( C_2 \). However, since the system results in both coefficients being zero, this implies the entire function \( y(t) \) can only be zero everywhere to satisfy these conditions.

Step 5: Conclusion Using Initial Conditions

From Step 4, we concluded that the coefficients \( C_1 \) and \( C_2 \) must both be zero for the equation and its derivative to satisfy the initial conditions, hence \( y(t) = 0 \) for all \( t \). This is consistent with the mathematical requirement that for a solution of a homogeneous differential equation with zero initial conditions, the zero solution is the only possible solution.

Key Concepts

Homogeneous differential equation

A homogeneous differential equation is a special type of differential equation where all terms involve the dependent variable or its derivatives, and there are no constant or forcing terms. In the case of a second-order homogeneous differential equation, we typically see it represented as \(a y'' + b y' + c y = 0\). Here:

• \(a, b,\) and \(c\) are constants.
• \(y''\) is the second derivative of \(y\) with respect to \(t\).
• \(y'\) is the first derivative of \(y\) with respect to \(t\).
• \(y\) is the function of \(t\).

This equation is termed "homogeneous" because the right-hand side equals zero. The solutions to these equations typically involve exponential functions or sine and cosine functions, which correspond to different types of systems such as mechanical or electrical systems. In this exercise, the equation's solutions must also respect any given initial conditions, leading to unique or trivial solutions like \(y(t) = 0\) when such conditions imply zero coefficients for all elements in the general solution.

Characteristic equation

The characteristic equation is a crucial concept when solving homogeneous linear differential equations. This equation is derived from the differential equation \(a y'' + b y' + c y = 0\) by replacing the derivatives with algebraic terms involving a characteristic root \(r\). To find this equation, we assume a solution of the form \(y(t) = e^{rt}\), where \(r\) is a constant. By differentiating, we obtain:

• \(y'(t) = r e^{rt}\)
• \(y''(t) = r^2 e^{rt}\)

Substituting these into the original differential equation leads to \(a r^2 e^{rt} + b r e^{rt} + c e^{rt} = 0\). Factoring out \(e^{rt}\) gives us the characteristic equation: \[ ar^2 + br + c = 0 \]The roots of this quadratic equation, \(r_1\) and \(r_2\), determine the form of the general solution. If the roots are real and distinct, the solution is a combination of exponentials. If they are complex, it involves sines and cosines. Special cases arise when there are repeated roots, leading to solutions involving \(t\) times the exponential functions to account for the multiplicity.

Initial conditions

Initial conditions are values specified for both the dependent variable and its derivatives at a particular point. In differential equations, they are crucial for finding a unique solution to the problem. In this exercise, we have the initial conditions:

• \(y(t_0) = 0\)
• \(y'(t_0) = 0\)

These conditions mean that at time \(t_0\), both the function and its first derivative must be zero. Applying these conditions involves substituting \(t_0\) into the general solution of the differential equation. For example, if the general solution is \(y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t}\), the initial conditions lead to a system of equations:

• \(C_1 e^{r_1 t_0} + C_2 e^{r_2 t_0} = 0\)
• \(C_1 r_1 e^{r_1 t_0} + C_2 r_2 e^{r_2 t_0} = 0\)

The solution to this system typically results in \(C_1\) and \(C_2\) being zero when the only solution compatible with the zero initial conditions is the zero function itself: \(y(t) = 0\) for all \(t\). This reflects that the general solution must be consistent with the specifics of the given initial conditions for meaningful physical or theoretical outcomes.