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: Solve the given initial value problem: \(e^{-t} y^{\prime}(t)-e^{-t} y=e^{2 t}\) with the initial condition, \(y(0) = 4\). Answer: The solution to the given initial value problem is \(y(t) = \frac{1}{2} e^{3t} + \frac{7}{2} e^{t}\).

Step-by-step solution

Identify the given elements

In our given equation \(e^{-t} y^{\prime}(t)-e^{-t} y=e^{2 t}\), we can identify the elements as: - \(a(t) = e^{-t}\) - \(a^{\prime}(t) = -e^{-t}\) - \(f(t) = e^{2t}\)

Write the left side as derivative of a product

As we can see, the left side of our equation can be written as the derivative of the product of \(a(t)\) and \(y(t)\). So, we rewrite our equation as: \(\frac{d}{dT}(a(t) y(t))=f(t)\) Substitute the expressions for \(a(t)\) and \(f(t)\), we get: \(\frac{d}{dT}(e^{-t} y(t))=e^{2t}\)

Integrate both sides

Now we will integrate both sides of the equation with respect to \(t\): \(\int \frac{d}{dT}(e^{-t} y(t)) dt = \int e^{2t} dt\) Using the Fundamental Theorem of Calculus, we have: \(e^{-t} y(t) = \frac{1}{2} e^{2t} + C\)

Solve for the integration constant

Now we will use the initial value, \(y(0) = 4\), to solve for the integration constant \(C\): \(e^{0} (4) = \frac{1}{2} e^{0} + C\) \(4 = \frac{1}{2} + C\) Therefore, \(C = \frac{7}{2}\)

Find the function y(t)

Finally, we solve for the function \(y(t)\): \(e^{-t} y(t) = \frac{1}{2} e^{2t} + \frac{7}{2}\) Multiply both sides by \(e^{t}\): \(y(t) = \frac{1}{2} e^{3t} + \frac{7}{2} e^{t}\) So, the solution to the given initial value problem is: \(y(t) = \frac{1}{2} e^{3t} + \frac{7}{2} e^{t}\)

Key concepts

Initial Value Problem

An initial value problem is a type of differential equation that comes with a specific condition known as an 'initial value'. This value specifies the state of the function at a particular point, often when time, denoted by \(t\), is zero. This helps us find a unique solution to the differential equation instead of an entire family of solutions.
This is crucial because differential equations often yield general solutions with arbitrary constants. The initial value condition allows us to determine the exact value of these constants, providing a specific solution that starts out by fulfilling the requirement set, such as \(y(0) = 4\) in our example.

• The given initial value helps simplify and solve the equation.
• Determines the constant of integration precisely.
• Makes the problem applicable to real-world scenarios by specifying initial states.

Understanding the initial value problem is essential for solving specific situations described by differential equations.

Integrating Both Sides

When we face a differential equation, one powerful technique to solve it is by integrating both sides. This approach is particularly useful when the equation is expressed as the derivative of a product, just like in our given problem.
By integrating the entire equation with respect to the variable \(t\), we effectively 'reverse' the differentiation, aiming to find the original function or its particular solution. This step requires us to apply rules of integration which might turn the problem into a simpler algebraic form.

• Helps arrive at the integrated form of the equation.
• Transforms a differential equation into a manageable mathematical expression.
• Requires careful substitution and solving for constants from initial conditions.

Take care, though, to integrate both sides correctly, considering all parts involved to ensure an accurate solution to the original problem.

Fundamental Theorem of Calculus

At the heart of calculus lies the Fundamental Theorem of Calculus, which bridges the concept of differentiation with integration. It essentially states that integration and differentiation are inverse processes.
In our example, we used this theorem implicitly when integrating both sides of the differential equation. The left side's integration yields the antiderivative of the product, aligning perfectly with what this theorem explains.
The theorem also provides a way to evaluate definite integrals efficiently, but in a first-order linear differential equation setup, it helps simplify the understanding of how integrated results match back to their differentiated form.

• Shows the relationship between differentiation and integration.
• Gives a clear procedure to solve definite integrals.
• Essential for solving and understanding differential equations.

Mastery of this theorem is vital as it guarantees that your integrated solution indeed represents the antiderivative of the derivative you started with, making it a cornerstone in solving differential equations.