Question

Solve the following initial value problems. $$y^{\prime}(t)-3 y=12, y(1)=4$$

Short answer

Answer: The particular solution for the given initial value problem is \(y(t) = -4 + \frac{8}{e^3}e^{3t}\).

Step-by-step solution

Identify the type of ODE

We are given a first-order linear ODE in the form of: $$y'(t) + p(t)y = q(t)$$ Here, \(p(t) = -3\) and \(q(t) = 12\). Since this equation meets the criteria of a first-order linear ODE, we can proceed with solving it.

Find the integrating factor

An integrating factor is a function that is used to simplify the given ODE so that it becomes easier to integrate. For the given first-order linear ODE, the integrating factor (\(I(t)\)) is calculated as: $$I(t) = e^{\int p(t) dt}$$ Here, \(p(t) = -3\). So, $$I(t) = e^{\int -3 dt} = e^{-3t}$$

Multiply the ODE by the integrating factor

To simplify the ODE, we will multiply the entire equation by the integrating factor we determined in step 2. $$e^{-3t} (y'(t) - 3y) = 12e^{-3t}$$

Integrate both sides of the equation

The left side of the equation can be rewritten as a product rule derivative, which simplifies the ODE further: $$\frac{d}{dt}(y(t)e^{-3t}) = 12e^{-3t}$$ Now, we will integrate both sides with respect to t: $$\int \frac{d}{dt}(y(t)e^{-3t}) dt = \int 12e^{-3t} dt$$ This results in: $$y(t)e^{-3t} = -4e^{-3t} + C$$

Solve for y(t)

We now isolate y(t) from the equation to find the general solution: $$y(t) = -4 + Ce^{3t}$$

Apply the initial condition to find the particular solution

We will use the initial condition provided: \(y(1) = 4\). Plug the initial condition into the general solution: $$4 = -4 + Ce^{3}$$ Solving for C, we obtain: $$C = \frac{8}{e^3}$$

Write the final solution

Now, we have found the value of C and can write the particular solution to the given initial value problem: $$y(t) = -4 + \frac{8}{e^3}e^{3t}$$

Key concepts

Integrating Factor

When dealing with first-order linear ordinary differential equations (ODEs), one useful method of solving them is through the integrating factor. This technique involves finding a suitable function that simplifies the original equation.
In general, an ODE in the form \( y'(t) + p(t)y = q(t) \) can be challenging to solve directly, so we introduce an integrating factor, \( I(t) = e^{\int p(t) \, dt} \). This factor transforms the equation into something that resembles the result of the product rule of calculus.

• Multiply the entire ODE by the integrating factor.
• This creates a scenario where the left side of the equation becomes the derivative of the product \( y(t)I(t) \), making it easier to integrate both sides.

This clever step sets the stage for more straightforward integration, ultimately aiding in finding the general solution.

Initial Value Problem

Initial Value Problems (IVPs) are specific types of differential equations that come with an initial condition, such as a known value of the function at a certain point. Solving an IVP means not only finding the general behavior of the function explaining the system but also a specific function that meets the initial condition.
For example, with the equation \( y'(t) - 3y = 12 \), the initial condition \( y(1) = 4 \) asks us to identify the function \( y(t) \) that satisfies both the equation and this specific condition.
This additional piece of data reduces the infinite set of possible solutions (general solutions) to one particular solution. This solution is uniquely suited to model what actually happens based on the given initial information.

General Solution

The general solution to a differential equation represents a family of all possible solutions, depending on arbitrary constants. For a first-order differential equation, this process typically involves integrating the equation and expressing the result in terms of a general constant, here denoted as \( C \).
In the problem with the equation \( y'(t) - 3y = 12 \), the solution obtained is \( y(t) = -4 + Ce^{3t} \).

• This general solution outlines how solutions behave and contains different possibilities based on the value of \( C \).
• By altering \( C \), you can model different initial conditions, thus demonstrating various potential system behaviors.

The power of the general solution is in its ability to provide a flexible framework, allowing adjustments to match specific scenarios through initial conditions.

Particular Solution

From the general solution, we derive the particular solution by incorporating the initial conditions provided in the problem. This solution is an exact function that fits the differential equation as well as the initial condition.
Using \( y(1) = 4 \), substitute into the general solution \( y(t) = -4 + Ce^{3t} \) to determine \( C \):

• Calculate by substituting the condition: \( 4 = -4 + Ce^{3} \)
• Solving for \( C \) gives: \( C = \frac{8}{e^3} \)

After determining \( C \), we can write the specific equation \( y(t) = -4 + \frac{8}{e^3}e^{3t} \).
This process demonstrates how specific conditions filter the general solution down to one precise answer relevant to the original problem's context.