Question

find the fundamental set of solutions specified by Theorem 3.2.5 for the given differential equation and initial point. $$ y^{\prime \prime}+4 y^{\prime}+3 y=0, \quad t_{0}=1 $$

Short answer

Based on the given differential equation and the step-by-step solution provided, the fundamental set of solutions for the differential equation is: $$ \{e^{-3t}, e^{-t}\} $$

Step-by-step solution

Write down the characteristic equation

The characteristic equation of the given differential equation is: $$ r^2 + 4r + 3 = 0 $$

Find the roots of the characteristic equation

We can solve the quadratic equation by factoring or using the quadratic formula, but in this case, the characteristic equation can be factored as: $$ (r + 3)(r + 1) = 0 $$ Hence, the roots are \(r_1 = -3\) and \(r_2 = -1\).

Write down the general solution

Since we have found two distinct real roots, the general solution of the differential equation is given by: $$ y(t) = C_1 e^{-3t} + C_2 e^{-t} $$ Now, we have the general solution, and we can find the fundamental set of solutions by applying the initial condition.

Apply the initial condition

The given initial point is \((t_0, y_0) = (1, 0)\). Substitute \(t_0=1\) into the general solution: $$ 0 = C_1 e^{-3(1)} + C_2 e^{-1(1)} $$ However, finding the constants \(C_1\) and \(C_2\) in this case is not necessary, because we only need the fundamental set of solutions. Notice that we do not have a value for \(y_0\) which would allow us to find a unique solution by solving for \(C_1\) and \(C_2\).

Write down the fundamental set of solutions

Since we have a second-order homogeneous linear differential equation, the fundamental set of solutions consists of the two linearly independent solutions given by the general solution. Hence, the fundamental set of solutions can be written as: $$ \{e^{-3t}, e^{-t}\} $$

Key concepts

Second-Order Differential Equation

A second-order differential equation involves the second derivative of a function. The equation we are examining is in the form:\[y'' + 4y' + 3y = 0\]Here, \(y\) is a function of \(t\), \(y'\) is the first derivative, and \(y''\) is the second derivative.

• "Second-order" indicates that the highest derivative is the second derivative \(y''\).
• Such equations often describe systems where the rate of change of the rate of change (like acceleration) is considered.

These equations are crucial in expressing real-world phenomena such as the motion of mechanical systems, electrical circuits, and more.
The second-order linear differential equation can additionally be categorized based on other properties, such as homogeneity and linearity.

Linear Homogeneous Differential Equations

A linear homogeneous differential equation can be recognized by its structure and the condition that all terms involve either the function or its derivatives.
In the equation \(y'' + 4y' + 3y = 0\), neither standalone constants nor functions of \(t\) appear.

• "Linear" here means that the function \(y(t)\) and its derivatives appear to the power of 1.
• "Homogeneous" indicates that the entire equation equals zero, without any external function term on the right-hand side.

Linear homogeneous equations are significant in various studies because they allow for the superposition principle. This means that if \(y_1(t)\) and \(y_2(t)\) are solutions, then any linear combination \(C_1y_1(t) + C_2y_2(t)\) is also a solution. This makes finding solutions systematic and applicable to complex scenarios like vibrations and waves.

Characteristic Equation

The characteristic equation is vital to solving second-order linear differential equations. It helps find the solution by converting a differential equation into an algebraic one.
To derive it, replace the derivatives in the differential equation with powers of a variable (commonly \(r\)). For \(y'' + 4y' + 3y = 0\), the characteristic equation becomes:\[r^2 + 4r + 3 = 0\]

• Solving this quadratic equation provides the roots \(r_1\) and \(r_2\).
• The roots inform us about the general solution of the differential equation.

In this context, the equation allows us to factor it into: \[(r + 3)(r + 1) = 0\]yielding \(r_1 = -3\) and \(r_2 = -1\). These roots describe the exponents in the general solution's terms, guiding us to the fundamental set of solutions.

Initial Value Problem

An initial value problem (IVP) involves not only finding a general solution to a differential equation but also determining a specific solution that meets given conditions.
For the given differential equation, the initial condition is typically specified at a particular point \((t_0, y_0)\). Even though in our problem, \((t_0 = 1, y_0)\) lacks enough information to determine specific constants for a unique solution, it provides direction in certain contexts.

• The condition \(y(t_0) = y_0\) helps identify particular solutions in other scenarios.
• Solving an IVP ensures that solutions accurately fit physical or real-life initial conditions.

While the fundamental set of solutions \( \{e^{-3t}, e^{-t}\} \) provides the basis, in many cases, specific initial conditions are necessary to pin down unique solutions.