Problem 29 - Calculus

Question

Solve \(y^{\prime}=3 y+1\) with \(y_{0}=0\) by assuming \(y=A e^{3 t}+B\) and determining \(A\) and \(B\).

Step-by-Step Solution

Answer

The solution is \(y = \frac{1}{3}e^{3t} - \frac{1}{3}\).

Step 1: Base Solution Assumption

We are given the differential equation \(y' = 3y + 1\). Assuming \(y = Ae^{3t} + B\) as a solution, we have to find \(A\) and \(B\). Take the derivative of \(y\) based on the assumed solution.

Step 2: Differentiate the Assumed Solution

Differentiating \(y = Ae^{3t} + B\) with respect to \(t\), we get \(y' = 3Ae^{3t}\).

Step 3: Substitute into the Differential Equation

Now substitute \(y\) and \(y'\) into the differential equation \(y' = 3y + 1\). This gives us \(3Ae^{3t} = 3(Ae^{3t} + B) + 1\).

Step 4: Simplify the Equation

Simplifying the equation \(3Ae^{3t} = 3Ae^{3t} + 3B + 1\), we can cancel \(3Ae^{3t}\) from both sides, yielding \(0 = 3B + 1\).

Step 5: Solve for Constant B

From the equation \(3B + 1 = 0\), solving for \(B\) gives \(B = -\frac{1}{3}\).

Step 6: Initial Condition Application

Use the initial condition \(y(0) = 0\) indicating when \(t = 0\), \(y = 0\). Substitute \(t = 0\) and \(y = 0\) into \(y = Ae^{3t} + B\).

Step 7: Solve for Constant A

Applying the initial condition \(0 = Ae^{0} + (-\frac{1}{3})\), we have \(A - \frac{1}{3} = 0\). Solving for \(A\) gives \(A = \frac{1}{3}\).

Step 8: Final Solution

With both constants determined, the solution is \(y = \frac{1}{3}e^{3t} - \frac{1}{3}\).

Key Concepts

Initial Value Problems

Initial value problems are a type of differential equation problem where you're not only asked to find the general solution, but also a specific solution that satisfies an initial condition. These initial conditions are values that the function and its derivatives must fulfill at the beginning of the problem.
For example, consider the differential equation we had: \(y' = 3y + 1\). The initial condition given was \(y(0) = 0\). This means, when \(t = 0\), the value of \(y\) must be 0. The initial condition is crucial because it enables us to determine specific constants in our solution, leading us to a unique answer.
Solving an initial value problem usually involves:

• Identifying the differential equation and the initial condition.
• Finding a general solution to the differential equation.
• Applying the initial condition to find the constants specific to the problem.

By applying this step-by-step approach, you resolve not just any solution but the one that exactly meets the criteria given.

Exact Solutions

The term 'exact solutions' refers to solving differential equations analytically through methods that yield a precise and complete expression for the solution. In contrast to numerical or approximate solutions, exact solutions provide a specific formula that describes the behavior of the system for all permissible inputs.
In our exercise, we assumed a solution form \(y = Ae^{3t} + B\). This assumption aids in solving the equation by substitution, enabling us to adjust \(A\) and \(B\) to satisfy both the differential equation and initial conditions. Exact solutions often employ methods such as separation of variables, integrating factors, or characteristic equations, depending on what's suitable for the given differential equation.
This process is like solving a puzzle:

• Assume a potential form of the solution.
• Use algebraic manipulation to substitute and simplify.
• Solve for constants using additional conditions like initial values.

Exact solutions give us a definitive guide to how a dynamic system behaves over time.

Exponential Functions

Exponential functions are mathematical expressions in the form \(f(t) = a e^{bt}\), where \(e\) is a constant (approximately equal to 2.718) known as Euler’s number. These functions grow or decay at rates proportional to their current value and are common in the solutions of differential equations due to their unique properties.
Our differential equation solution assumed the form \(y = A e^{3t} + B\). Here, the term \(e^{3t}\) represents exponential growth due to the positive coefficient 3 multiplying \(t\). This forms a natural solution for our differential equation, \(y' = 3y + 1\), as exponential terms often appear when solving linear differential equations with constant coefficients.
Key features of exponential functions in differential equations include:

• Predictable growth or decay patterns.
• The derivative of an exponential function is proportional to the function itself, which neatly fits many dynamic systems.
• Flexibility in modeling diverse phenomena, from population growth to radioactive decay.

Understanding how to use exponential functions effectively is paramount in solving and interpreting differential equations.