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. $$e^{-t} y^{\prime}(t)-e^{-t} y=e^{2 t}, y(0)=4$$

Short answer

Question: Determine the solution of the initial value problem $$e^{-t} y^{\prime}(t)-e^{-t} y=e^{2 t}, y(0)=4$$. Answer: The solution of the initial value problem is given by $$y(t) = e^t\left(\frac{1}{2}e^{2t} + \frac{7}{2}\right)$$

Step-by-step solution

Identify a(t), a'(t), and f(t)

In the given equation \(e^{-t} y^{\prime}(t)-e^{-t} y=e^{2 t}\), we have \(a(t) = e^{-t}\), \(a'(t) = -e^{-t}\), and \(f(t) = e^{2t}\).

Verify that the left side of the equation can be written as the derivative of a product

Using the identified functions, the left side of the equation is \(a(t) y^{\prime}(t) + a^{\prime}(t) y(t) = e^{-t} y^{\prime}(t) - e^{-t} y(t)\). We want to verify that this expression can be written as the derivative of a product. In fact, we can write it as \(\frac{d}{dt}(a(t)y(t))\), which in this case is \(\frac{d}{dt}(e^{-t}y(t))\).

Integrate both sides of the equation with respect to t

Now that we have verified the left side of the equation can be written as the derivative of a product, we will integrate both sides of the equation with respect to \(t\). $$\int \frac{d}{dt}(e^{-t}y(t)) dt = \int e^{2t} dt$$ Integrating the left side, we get \(e^{-t}y(t)\), and integrating the right side, we get \(\frac{1}{2}e^{2t} + C\), where \(C\) is the constant of integration. Therefore, we have: $$e^{-t}y(t) = \frac{1}{2}e^{2t} + C$$

Apply the initial condition to find the value of C

We are given the initial condition \(y(0) = 4\). To find the value of \(C\), we substitute \(t=0\) and \(y(0)=4\) into our equation: $$e^{0}y(0) = \frac{1}{2}e^{2(0)} + C$$ Simplifying and solving for \(C\), we get: $$4 = \frac{1}{2} + C$$ So, \(C = \frac{7}{2}\)

Solve for y(t)

Now that we have found the value of \(C\), we can solve for \(y(t)\) by isolating it in the equation: $$e^{-t}y(t) = \frac{1}{2}e^{2t} + \frac{7}{2}$$ Divide both sides by \(e^{-t}\): $$y(t) = e^t\left(\frac{1}{2}e^{2t} + \frac{7}{2}\right)$$ The solution of the initial value problem is: $$y(t) = e^t\left(\frac{1}{2}e^{2t} + \frac{7}{2}\right)$$

Key concepts

Product Rule for Derivatives

The product rule for derivatives helps us find the derivative of a product of two functions. It's a vital tool in calculus and here in this problem, we see it used in a first-order linear differential equation. The rule states that if you have a product of two functions, say \( u(t) \) and \( v(t) \), then the derivative of their product with respect to \( t \) is given by:

• \( \frac{d}{dt}(u(t) \cdot v(t)) = u'(t)v(t) + u(t)v'(t) \)

In our given equation, the left side can be interpreted as such a derivative of a product. Specifically, it's written as \( \frac{d}{dt}(e^{-t}y(t)) \). By recognizing this, we are then able to simplify and proceed with solving the differential equation by integration.
This step is crucial as it converts the complex-looking left side of the equation into a more manageable form, which we can then integrate. It’s all about recognizing patterns in expressions.

Initial Value Problems

An initial value problem is a type of differential equation that comes with a specified value, referred to as the initial condition. This value is given for the unknown function at a specific point, typically \( t = 0 \). In our example, we have \( y(0) = 4 \).
Initial conditions are crucial as they allow us to find the particular solution of a differential equation. Without them, we can only find a general solution that includes an arbitrary constant value of integration. By plugging in the initial condition, we can solve for this constant and thus identify the unique specific solution that not only satisfies the equation but also respects the initial condition.

• First, solve the differential equation generally.
• Then, substitute the initial value to solve for the arbitrary constant.

This ensures that the solution works for the specific scenario we're considering.

Integration Techniques

Integration is a core technique used in calculus to find the area under a curve or to solve differential equations. In this exercise, we're utilizing integration to solve the equation after recognizing the product rule's role.

• The left-hand side of the equation can be simplified to \( \frac{d}{dt}(e^{-t}y(t)) \), which then allows us to equate it to the right-hand side \( e^{2t} \) and integrate both sides.
• Integrating the left side results in recovering the original function from its derivative: \( e^{-t}y(t) \).
• The right side's integration involves finding the antiderivative of \( e^{2t} \), which results in \( \frac{1}{2}e^{2t} + C \).
• Always remember to add a constant of integration \( C \), which represents any constant term that could have been lost during differentiation.

This method of integrating both sides is standard when dealing with linear differential equations and helps you proceed towards finding the specific solution, especially when combined with the initial value conditions.