Question

Solve the following initial value problems. $$y^{\prime}(t)=1+e^{t}, y(0)=4$$

Short answer

Question: Solve the initial value problem $$y^{\prime}(t) = 1 + e^{t}$$ with the initial condition $$y(0) = 4$$, and find the function y(t) that satisfies both the given ODE and the initial condition. Answer: The function y(t) that satisfies the given ODE and the initial condition is $$y(t) = t + e^{t} + 3$$.

Step-by-step solution

Identify the given ODE and initial condition

We are given the following first-order ODE: $$y^{\prime}(t)=1+e^{t}$$ and the initial condition: $$y(0)=4$$

Solve the ODE using direct integration

Since the ODE is just a first-order ODE with no y(t) term, we can directly integrate both sides with respect to t: $$\int y^{\prime}(t) dt = \int (1+e^{t}) dt$$ Integrating each term separately, we get: $$y(t) = t + \int e^{t} dt$$ Now, integrate the exponential term: $$y(t) = t + e^{t} + C$$ where C is the integration constant.

Use the initial condition to solve for the integration constant

We are given the initial condition $$y(0) = 4$$. Substitute this into the equation for y(t) that we found above: $$4 = 0 + e^{0} + C$$ $$4 = 1 + C$$ Solving for C, we get: $$C = 3$$

Write the final solution

Now that we have the value of C, we can write down the final solution: $$y(t) = t + e^{t} + 3$$ This is the function y(t) that satisfies both the given ODE and the initial condition.

Key concepts

Initial Value Problems

Initial value problems are a type of differential equation that comes with a specific condition. This condition, known as the initial condition, specifies the value of the solution at a particular point. In simpler terms, to solve an initial value problem, you need both:

• A differential equation that describes how y changes with t.
• A point or value, called the initial condition, that the solution must satisfy at a specific time, like y(0) = 4 in our problem.

Such conditions help ensure we find the exact solution from potentially many possible solutions.
By applying the initial condition provided, we can calculate unknown constants and reach a unique solution.

First-Order Ordinary Differential Equations

First-order ordinary differential equations (ODEs) are the simplest kind of differential equations, where the highest derivative is the first derivative. The general form is given by:

• \[ y'(t) = f(t, y) \]

In our example, however, the ODE simplifies to:

• \[ y^{\prime}(t)=1+e^t \]

Here, there is no y(t) term on the right-hand side, which allows us to integrate directly. This straightforward structure makes these ODEs generally easier to tackle first.
First-order ODEs often model various real-world situations where rates of change matter, such as population growth, radioactive decay, or cooling/heating processes.

Integration

Integration is a key mathematical process for solving differential equations. It is essentially the reverse of differentiation. When you integrate, you find the original function whose derivative you have.
For the ODE in the exercise, we integrated both sides to find y(t) as:

• \[ \int y'(t) dt = \int (1 + e^t) dt \]

By integrating, we find:

• \[ y(t) = t + e^t + C \]

It's important to remember the constant of integration, C. This constant represents any constant value which would disappear when differentiated. Later on, you solve for C using the initial condition.

Exponential Functions

Exponential functions are mathematical expressions where a constant base is raised to a variable exponent, often the natural number e. These functions model exponential growth or decay and often appear in differential equations.
In our problem:

• \[ y^{\prime}(t)=1+e^{t} \]
• e^t is an exponential term.

When addressing these in solutions, their unique properties allow exponential functions to stay largely unchanged when differentiated or integrated. Upon integrating e^t, it remains as e^t, simplifying calculations significantly.
Such features make exponential functions particularly useful in modeling real-life scenarios such as compound interest, population dynamics, and radioactive decay. Understanding these can immensely help in recognizing patterns and behaviors in differential equations.