Problem 2 - Applied Calculus

Question

If the solution to \(y^{\prime}=x+4 y\) passes through the point \((3,1),\) what is the slope of the solution at that point?

Step-by-Step Solution

Answer

The slope at (3,1) is 7.

Step 1: Understanding the Differential Equation

The given differential equation is linear with the form \(y' = x + 4y\). This is a first-order linear differential equation.

Step 2: Finding the Solution Function

To find the solution, we must solve the differential equation. This can be approached by using an integrating factor \( \mu(x) = e^{\int 4 \, dx} = e^{4x} \). Multiply the entire differential equation by this integrating factor.

Step 3: Applying the Integrating Factor

Using the integrating factor, we transform the equation to find a function \( y(x) \). This is \(e^{4x} y' = x e^{4x} + 4 e^{4x} y\), leading to \( (e^{4x} y)' = x e^{4x} \).

Step 4: Integrating Both Sides

Integrate both sides with respect to \(x\). The right side becomes \( \int x e^{4x} \, dx \), which can be solved using integration by parts. This gives the particular integral. Calculate to find \( y(x) = e^{-4x} \left(C + \int x e^{4x} \, dx \right) \).

Step 5: Solving the Integral

Perform integration by parts on \( \int x e^{4x} \, dx \), where \( u = x \) and \( dv = e^{4x} \, dx \). Thus, \( du = dx \) and \( v = \frac{1}{4} e^{4x} \). The integral is \(\frac{1}{4} x e^{4x} - \frac{1}{16} e^{4x} + C' \).

Step 6: Substituting the Initial Condition

We know the solution passes through \((3,1)\). Substitute \(x = 3\) and \(y = 1\) into the general solution to find the particular constant \(C\) in the solution.

Step 7: Finding the Derivative at the Point

To find the slope at \((3,1)\), use the derivative \(y' = x + 4y\). Substitute \(x = 3\) and \(y = 1\) into the equation to calculate \(y'\).

Key Concepts

First-Order Linear Differential Equation

A first-order linear differential equation is a type of differential equation that involves the first derivative of a function and the function itself. It is called "first-order" because it involves the first derivative, and "linear" because the function and its derivative appear to the power of one.
This kind of differential equation generally has the form: \[ y' + P(x)y = Q(x) \]

• \( y' \) represents the derivative of the dependent variable \( y \) with respect to the independent variable \( x \).
• \( P(x) \) and \( Q(x) \) are known functions of the independent variable.

These equations are common in physics and engineering, often modeling systems that change at rates proportional to their current state.

Integrating Factor

The integrating factor is a specially designed function that simplifies the process of solving a linear differential equation. By multiplying the entire equation by this factor, we transform it into a form that can be easily integrated.
For our differential equation \( y' = x + 4y \), an integrating factor \( \mu(x) \) is calculated as: \[ \mu(x) = e^{\int P(x)\,dx} = e^{\int 4\,dx} = e^{4x} \]

• Purpose: The integrating factor allows us to rewrite the differential equation in a way that makes the left side a total derivative.
• Application: Multiply the original equation by \( \mu(x) \) so it becomes easier to integrate.

This approach is extremely useful when dealing with linear differential equations as it systematically reduces complexity.

Initial Condition

An initial condition is information provided that allows us to find a unique solution to a differential equation. It typically involves the value of the function at a specific point.
This condition ensures we know not only the general behavior of the solution but also the exact path it takes. In the problem we explored:

• The initial condition is presented as the solution passes through the point \((3,1)\).
• We set up the equation solution such that when \(x = 3\), \(y = 1\).

Applying initial conditions secures one unique answer from the infinitely many solutions a differential equation might suggest.

Integration by Parts

Integration by parts is a technique used to find the integral of a product of two functions. It is derived from the product rule for differentiation and is pivotal in solving certain integrals that appear complex at first glance.
For the integral of the form \( \int u \, dv \), integration by parts formula is:\[ \int u \, dv = uv - \int v \, du \]

• Purpose: This technique helps to transform an integral into a simpler form.
• Example: In the step \( \int x e^{4x} \, dx \), we choose \( u = x \) and \( dv = e^{4x} \, dx \). Then, \( du = dx \) and \( v = \frac{1}{4} e^{4x} \).

It simplifies integrals in solving differential equations, making it easier to find the function solution.