Question

Consider the general first-order linear equation \(y^{\prime}(t)+a(t) y(t)=f(t) .\) This equation can be solved, in principle, by defining the integrating factor \(p(t)=\exp \left(\int a(t) d t\right) .\) Here is how the integrating factor works. Multiply both sides of the equation by \(p\) (which is always positive) and show that the left side becomes an exact derivative. Therefore, the equation becomes $$p(t)\left(y^{\prime}(t)+a(t) y(t)\right)=\frac{d}{d t}(p(t) y(t))=p(t) f(t)$$ Now integrate both sides of the equation with respect to t to obtain the solution. Use this method to solve the following initial value problems. Begin by computing the required integrating factor. $$y^{\prime}(t)+\frac{1}{t} y(t)=0, y(1)=6$$

Short answer

Question: Solve the first-order linear differential equation using the integrating factor method and the given initial condition: \(y^{\prime}(t)+\frac{1}{t} y(t)=0\), \(y(1) = 6\). Answer: \(y(t) = \frac{6}{t}\)

Step-by-step solution

Find the integrating factor \(p(t)\)

The function \(a(t)\) is given by \(\frac{1}{t}\). We need to find the integrating factor by computing the integral of \(a(t)\): $$p(t) = \exp\left(\int a(t) dt\right) = \exp\left(\int \frac{1}{t} dt\right)$$ Computing the integral, $$p(t) = \exp(\ln(t)) = t$$

Multiply both sides of the equation by the integrating factor

The original equation is given by: $$y^{\prime}(t)+\frac{1}{t} y(t)=0$$ Multiply both sides by the integrating factor \(p(t) = t\): $$t\left(y^{\prime}(t)+\frac{1}{t} y(t)\right)= t\cdot0$$ This simplifies to: $$ty^{\prime}(t) + y(t) = 0$$

Show that the left side becomes an exact derivative

We want to show that the left side of the equation becomes an exact derivative: $$\frac{d}{dt}(ty(t)) = ty^{\prime}(t) + y(t) = 0$$

Integrate both sides of the equation with respect to \(t\)

Now we integrate both sides of the equation: $$\int \frac{d}{dt}(ty(t)) dt = \int 0 dt$$ This gives us: $$ty(t) = C$$ where \(C\) is the constant of integration.

Solve for \(y(t)\) using the initial condition

To find the constant \(C\), we use the initial condition \(y(1) = 6\): $$y(t) = \frac{C}{t} \Rightarrow y(1) = \frac{C}{1} \Rightarrow C = 6$$ Now, we can write the solution for \(y(t)\): $$y(t) = \frac{6}{t}$$

Key concepts

Integrating Factor

When solving first-order linear differential equations, an integrating factor is a powerful tool that simplifies the process. The main idea is to multiply both sides of the differential equation by a function (the integrating factor) that is specifically chosen to make the equation easier to solve. This function, denoted as p(t), is usually an exponential function of the integral of a(t), which is the coefficient function in the original differential equation.

• The integrating factor is defined as: \[ p(t) = \exp\left(\int a(t) \, dt\right) \]
• It is chosen to turn the left-hand side of the differential equation into an exact derivative, which can be easily integrated.
• In the example given, we calculated the integrating factor to be: \[ p(t) = \exp(\ln(t)) = t \]

In essence, applying the integrating factor is a preparatory step that modifies the differential equation into a form where both sides can be directly integrated to find the solution. It is an essential technique for students to master when tackling initial value problems involving first-order linear differential equations.

Initial Value Problems

Initial value problems (IVPs) are a category of differential equations that come with an additional piece of information: the value of the unknown function (commonly y(t)) at a specific point, referred to as the initial condition. In our example, the initial condition is y(1) = 6. This value is crucial because it determines the particular solution to the differential equation out of the infinite number of general solutions.

• The initial condition provides a specific point through which the solution curve must pass.
• After finding the general solution to the differential equation, we use the initial condition to solve for the constant of integration.
• In the final step of our solution, using y(1) = 6 allowed us to determine that the constant C was 6, leading to the particular solution: \[ y(t) = \frac{6}{t} \]

Students should appreciate the significance of initial conditions as they transform the general solution of a differential equation into one that is relevant and applicable to the problem at hand.

Exact Derivative

An exact derivative is a term used to describe a derivative that represents the rate of change of one function with respect to another, where the functions are related in such a way that they can be combined into a single differentiated function. This concept is central to the method involving integrating factors.

• In our step-by-step solution, we've shown that multiplying the given differential equation by the integrating factor transforms the left side into the exact derivative \[ \frac{d}{dt}(ty(t)) \].
• Recognizing an exact derivative allows us to integrate both sides with respect to t, simplifying the process of finding the solution.
• The ability to identify and work with exact derivatives is critical for students in order to effectively solve differential equations.

Working with exact derivatives can sometimes require practice to identify and apply correctly, but once mastered, it simplifies the process of integration and is a key step in solving differential equations.

Integration

Integration is a fundamental concept in calculus, indispensable in solving differential equations. It is the process of finding the integral of a function, which is essentially the 'inverse' of differentiation.

• In our problem, once we had the exact derivative, we performed integration on both sides to find the general solution ty(t) = C.
• Integration allows us to calculate the area under a curve, the accumulated quantity, or, in the case of differential equations, the original function given its rate of change.
• Even after integrating, don't forget the importance of applying the initial conditions to find the exact value of the integration constants as they are essential for completing the problem.

Students often find integration challenging because it not only involves understanding the theoretical aspects but also requires practice to develop the skills needed to integrate functions efficiently. Nevertheless, grasping the concept of integration is vital to solving many types of problems in mathematics, physics, engineering, and beyond.