Problem 9
Question
Solve each differential equation. $$4 x-3 y^{\prime}=5$$
Step-by-Step Solution
Verified Answer
The general solution of the differential equation \(4x - 3y' = 5\) is \(y = \frac{2}{3}x^2 - \frac{5}{3}x + C\).
1Step 1: Identify the Type of Differential Equation
First, observe the equation to determine its type. The given differential equation, \(4 x - 3 y' = 5\), is a first-order linear differential equation with the derivative of \(y\) (denoted as \(y'\)) appearing without any powers or functions applied on it.
2Step 2: Rearrange to Solve for the Derivative \(y'\)
To find the derivative \(y'\), solve for \(y'\) by isolating it on one side. You will get \(y' = \frac{4x - 5}{3}\).
3Step 3: Integrate Both Sides with Respect to \(x\)
Now, integrate both sides of the equation with respect to \(x\) to find \(y\). The integration on the right side becomes \(\frac{1}{3} \int (4x - 5) dx\).
4Step 4: Perform the Integration
Carry out the integral \(\frac{1}{3} \int (4x - 5) dx\) to find \(y\). The integral of \((4x - 5)\) with respect to \(x\) is \(2x^2 - 5x\), so multiplying it by \(\frac{1}{3}\) gives \(\frac{2}{3}x^2 - \frac{5}{3}x\).
5Step 5: Add the Constant of Integration
After integrating, add the constant of integration \((C)\) to the result because the integral of a derivative gives a family of functions. This gives you the general solution \(y = \frac{2}{3}x^2 - \frac{5}{3}x + C\).
Key Concepts
First-Order Linear Differential EquationsIntegrating Differential EquationsConstant of Integration
First-Order Linear Differential Equations
When dealing with differential equations, identifying the type of equation is crucial for determining the most effective approach to solve it. A first-order linear differential equation is a starting point for many students as it is the simplest form of differential equations that involve derivatives. In essence, it can be represented in the form \(y' + p(x)y = q(x)\), where \(y\) is the unknown function of \(x\), \(y'\) denotes the derivative of \(y\) with respect to \(x\), and \(p(x)\) and \(q(x)\) are functions of \(x\) only. The differential equation given in our exercise, \(4 x - 3 y' = 5\), fits this category after a bit of rearrangement.
One of the key properties of a first-order linear differential equation is its linearity, meaning that both \(y\) and its derivative \(y'\) are to the first power and not multiplied by each other. The linearity makes these types of equations manageable because they can be solved using straightforward integration methods after rearranging terms to isolate the derivative \(y'\).
In certain contexts, especially when the equation involves only \(x\) as in our exercise, it becomes even simpler. The absence of \(y\) itself in the equation means you do not have to deal with expressions containing both \(x\) and \(y\), simplifying the integration process. This type is also known as the separable first-order linear differential equation.
One of the key properties of a first-order linear differential equation is its linearity, meaning that both \(y\) and its derivative \(y'\) are to the first power and not multiplied by each other. The linearity makes these types of equations manageable because they can be solved using straightforward integration methods after rearranging terms to isolate the derivative \(y'\).
In certain contexts, especially when the equation involves only \(x\) as in our exercise, it becomes even simpler. The absence of \(y\) itself in the equation means you do not have to deal with expressions containing both \(x\) and \(y\), simplifying the integration process. This type is also known as the separable first-order linear differential equation.
Integrating Differential Equations
Integrating a differential equation is the process of finding a function that satisfies the relationship described by the derivative in the equation. In practical terms, this means we are trying to reverse the differentiation process to find the original function from which its rate of change is derived.
The beauty of integrating differential equations comes from understanding that it is a process to recover the 'whole' from the 'rate of change'. In the exercise provided, step 3 and step 4 demonstrate this beautifully where the integral of \(4x - 5\) with respect to \(x\) reclaims the original function \(y\). It's like piecing together a torn-up letter by matching the edges of each tear—it requires patience and a discerning eye for the underlying pattern.
When integrating, it’s also important to pay attention to the limits of integration if they are provided. In indefinite integrals (like in our example), we are looking for a general solution and hence include a constant of integration to account for all possible answers.
The beauty of integrating differential equations comes from understanding that it is a process to recover the 'whole' from the 'rate of change'. In the exercise provided, step 3 and step 4 demonstrate this beautifully where the integral of \(4x - 5\) with respect to \(x\) reclaims the original function \(y\). It's like piecing together a torn-up letter by matching the edges of each tear—it requires patience and a discerning eye for the underlying pattern.
Standard Form and Integration Technique
For a smooth integration process, we reorganize the differential equation into a standard and manageable form. Integration techniques such as u-substitution, integration by parts, or, as in this case, straightforward polynomial integration are then employed. Recognizing the type of integrals you encounter and the tools at your disposal, such as integral tables or rules like the power rule, is therefore paramount.When integrating, it’s also important to pay attention to the limits of integration if they are provided. In indefinite integrals (like in our example), we are looking for a general solution and hence include a constant of integration to account for all possible answers.
Constant of Integration
The constant of integration, often denoted as \(C\), is a fundamental component of integrating differential equations. It represents an infinite number of possible functions that all share the same differentiated form. This stems from the fact that indefinite integration (without specific bounds or limits) can only tell us the general form of a function, but not the precise one we started with.
One can think of this constant as the 'C' in a 'Choose Your Own Adventure' book. Different choices (or values for \(C\)) lead to different paths (or functions), yet all these paths are capable of fitting the criteria set by the initial differential equation. That's why, after integration, we always include \(C\) to address this multitude of possibilities.
If you're given additional information, like an initial condition, you can solve for \(C\) to find the specific solution tailored to that condition. Until then, \(C\) remains a representation of the rich diversity of solutions possible from a single differential equation, embodying the principle that there is often more than one way to fulfill conditions or solve a problem in mathematics.
One can think of this constant as the 'C' in a 'Choose Your Own Adventure' book. Different choices (or values for \(C\)) lead to different paths (or functions), yet all these paths are capable of fitting the criteria set by the initial differential equation. That's why, after integration, we always include \(C\) to address this multitude of possibilities.
Importance in Solutions
In Step 5 of the solution for our provided exercise, the constant of integration \(C\) is added to the term \(\frac{2}{3}x^2 - \frac{5}{3}x\) after integration. This is indicative of all the potential specific solutions. Without \(C\), we would be wrongly stating that there is only one unique solution.If you're given additional information, like an initial condition, you can solve for \(C\) to find the specific solution tailored to that condition. Until then, \(C\) remains a representation of the rich diversity of solutions possible from a single differential equation, embodying the principle that there is often more than one way to fulfill conditions or solve a problem in mathematics.
Other exercises in this chapter
Problem 9
Using the given boundary condition, find the particular solution. $$\left(x^{3}+y^{3}\right) d x-x y^{2} d y=0, x=1, y=0$$
View solution Problem 9
Solve each differential equation and find the approximate value of \(y\) requested. Start at the given boundary value and use a slope field or Euler's graphical
View solution Problem 10
Find the general solution of each differential equation. Try some by calculator. $$\left(1-2 x^{2} y\right) \frac{d y}{d x}=2 x y^{2}$$
View solution Problem 10
Find the general solution to each differential equation. Try some by calculator. $$y^{\prime}=x^{2} e^{-3 y}$$
View solution