Question

A special class of first-order linear equations have the form \(a(t) y^{\prime}(t)+a^{\prime}(t) y(t)=f(t),\) where \(a\) and \(f\) are given functions of \(t.\) Notice that the left side of this equation can be written as the derivative of a product, so the equation has the form $$a(t) y^{\prime}(t)+a^{\prime}(t) y(t)=\frac{d}{d t}(a(t) y(t))=f(t)$$ Therefore, the equation can be solved by integrating both sides with respect to \(t.\) Use this idea to solve the following initial value problems. $$t y^{\prime}(t)+y=1+t, y(1)=4$$

Short answer

Answer: The solution to the initial value problem is $$y(t) = 1 + \frac{1}{2}t + \frac{3}{t}$$.

Step-by-step solution

Identify the functions a(t) and f(t)

In our given problem, $$t y^{\prime}(t)+y(t)=1+t$$, we can identify the functions as: a(t) = t f(t) = 1+t Our goal is to rewrite the equation in the form $$\frac{d}{d t}(a(t) y(t))=f(t)$$

Rewrite the equation in the required form

We know that the equation can be written as $$\frac{d}{d t}(t y(t)) = 1 + t$$

Integrate both sides

Now, integrate both sides with respect to t: $$\int \frac{d}{d t}(t y(t)) dt = \int (1 + t) dt$$ On the left side, the integral and the derivative cancel out, leaving: $$t y(t) = \int (1 + t) dt$$ Now, we can integrate the right side: $$t y(t) = t + \frac{1}{2}t^2 + C$$

Solve for y(t)

Divide both sides by t to get y(t): $$y(t) = 1 + \frac{1}{2}t + \frac{C}{t}$$

Apply the initial condition

We are given the initial condition y(1) = 4, so we can plug this in to find C: $$4 = 1 + \frac{1}{2}(1) + \frac{C}{1}$$ Solving for C, we get: $$C = 3$$

Write the particular solution

With the value of C, we can now write the particular solution: $$y(t) = 1 + \frac{1}{2}t + \frac{3}{t}$$ This is the solution to the initial value problem.

Key concepts

Integration

Integration is a fundamental concept in calculus. It's the process of finding a function when you know its derivative. In other words, integration is about reversing differentiation. When you're given a differential equation, like the one from the exercise, integrating both sides helps you find the original function. To integrate, follow these steps:

• Identify what you're integrating. It can be a simple polynomial or a more complex expression.
• Use the correct integration techniques. Polynomials, trigonometric and exponential functions each have their own methods.
• Add a constant of integration, often denoted as "C". This constant is vital because when you differentiate, any constant loses its identity as it results in zero derivative.

In our problem, integrating is key to solving the equation. The integral of the left side effectively reverses the derivative, while the integral of the right side must be calculated explicitly. This approach simplifies the equation and provides a direct path to solving for the function initially unknown.

Initial Value Problems

Initial value problems (IVPs) are differential equations paired with additional information called initial conditions. These conditions specify the value of the unknown function at a particular point, ensuring the solution fits not just the differential equation but also matches real-world or initial situation parameters.For example, in our exercise, after finding the general solution through integration, we are given the initial condition \( y(1) = 4 \). This piece of information is crucial because:

• It helps determine the unique constant "C" in our integrated solution.
• It ensures the solution curve not only solves the equation, but passes through the given point, anchoring the function accurately on a graph.

Solving IVPs requires you to substitute the initial condition into the general solution. Typically, this results in an algebraic equation where you solve for "C". Once determined, this value finalizes the particular solution, satisfying both the differential equation and the initial condition.

Identifying Functions

Identifying functions within a differential equation is a vital first step in solving it. It involves discerning the correct roles of each function or term, which guides the method of solution to be adopted.In first-order linear differential equations like the one in the exercise, functions are often categorized as follows:

• Identify \( a(t) \) - the function multiplied by the derivative of the unknown such as \( y' \).
• Identify \( f(t) \) - the function on the opposite side of the equal sign when the equation is rearranged to the specific form.

In our example, recognizing that \( a(t) = t \) and \( f(t) = 1 + t \) allows us to rewrite the equation in an integrable form. This reformulation leverages a property known as the derivative of a product, critically simplifying the process to find the solution. Identifying these functions correctly ensures the process flows smoothly from start to finish.