Question

In Problems \(1-4\), solve the given differential equation subject to the given condition. Note that \(y(a)\) denotes the value of \(y\) at \(t=a\). $$ \frac{d y}{d t}=-6 y, y(0)=4 $$

Short answer

The solution is \( y(t) = 4e^{-6t} \).

Step-by-step solution

Recognize the Type of Differential Equation

The given differential equation \( \frac{d y}{d t} = -6y \) is a first-order linear ordinary differential equation with constant coefficients.

Write the General Solution

For a differential equation of the form \( \frac{d y}{d t} = ky \) where \( k \) is a constant, the general solution is \( y(t) = Ce^{kt} \). Here, \( k = -6 \), so the general solution is \( y(t) = Ce^{-6t} \).

Apply the Initial Condition to Find Specific Solution

Substitute the initial condition \( y(0) = 4 \) into the general solution to find \( C \). We have \( 4 = Ce^{-6 \times 0} = C \). Thus, \( C = 4 \).

Write the Particular Solution

Substitute \( C = 4 \) back into the general solution to get the particular solution: \( y(t) = 4e^{-6t} \).

Key concepts

Ordinary Differential Equations

An ordinary differential equation (ODE) is a mathematical expression involving a function and its derivatives. It describes how a function's values change and is often used to model real-world phenomena such as motion, growth, and decay. In an ODE, the function depends on only one independent variable. For example, in our exercise, the independent variable is time \( t \), and the function is \( y(t) \).

ODEs can vary greatly in complexity, but first-order ODEs, like the one given, involve only the first derivative of the function \( y \). The goal is to find the unknown function that satisfies the equation under given conditions. By solving an ODE, we determine how a process evolves over time. In our exercise, the ODE \( \frac{dy}{dt} = -6y \) represents exponential decay, a common scenario in physical sciences.

• They are "ordinary" because they contain no partial derivatives.
• Used for modeling various dynamic systems.
• Commonly classified by their order, linearity, and homogeneity.

Constant Coefficients

In the realm of differential equations, constant coefficients are an important concept. Essentially, a constant coefficient is a coefficient that doesn't change—it remains fixed as the variables in the equation change. This creates a simpler form of differential equations, which are much easier to solve.

In our specific case, the differential equation \( \frac{dy}{dt} = -6y \) has a constant coefficient of \(-6\). This means that no matter what substitution or manipulations we apply to solve the equation, \(-6\) remains unchanged. Constant coefficients allow for the direct method of applying exponential functions as solutions because the rate of change represented by the derivative is proportional to the value of the function, leading to solutions of the form \( Ce^{kt} \).

• Simplifies the solving process compared to variable coefficients.
• Enables use of exponential solutions for first-order linear ODEs.
• In our problem, a negative value indicates decay.

Particular Solution

A particular solution to a differential equation is a specific solution that satisfies the differential equation as well as any initial conditions or boundary conditions provided. For any linear differential equation, many functions may satisfy the equation itself. However, the particular solution is one that fits the additional conditions.

In the exercise, after finding the general solution \( y(t) = Ce^{-6t} \), we apply the initial condition \( y(0) = 4 \). By substituting \( t = 0 \) and \( y(0) = 4 \), we solve for \( C \), giving us \( C = 4 \). Therefore, the particular solution is \( y(t) = 4e^{-6t} \). The initial condition directs us to the one precise function amongst potentially infinite general solutions.

• Particular solution fits both the differential equation and the initial condition.
• Essential for real-world applications where initial values are known.
• In our example, represents a specific mode of exponential decay with \( y(0) = 4 \).

Initial Conditions

Initial conditions in a differential equation provide the values of the function (or its derivatives) at a specific point in the domain, often used to find a unique solution. In essence, they "anchor" the general solution, determining which version fits a particular situation.

Our exercise presents the initial condition \( y(0) = 4 \). This condition tells us the exact value of the function \( y \) when \( t = 0 \). Applying this condition helps us find the particular constant \( C \) in the general solution \( y(t) = Ce^{-6t} \). As a result, \( C = 4 \) and we end up with the particular solution \( y(t) = 4e^{-6t} \).

• Define the starting point of the function.
• Turn general solutions into particular ones.
• Important in scenarios involving prior states or specific times.