Chapter 3: Problem 75
Find the solution to the initial value problem. $$ y^{\prime}=8 x-\ln x-3 x^{4}, y(1)=5 $$
Short Answer
Expert verified
The solution to the IVP is
\[ y = 4x^2 - x \ln x + x - \frac{3}{5}x^5 + \frac{3}{5} \].
Step by step solution
01
Identify and Set Up the Integral
The given initial value problem (IVP) involves a first-order differential equation: \[y' = 8x - \ln x - 3x^4\] To find the solution, integrate both sides with respect to \(x\). This yields a general solution expression for \(y\) in terms of \(x\).
02
Perform the Integration
Integrate each term separately: - The integral of \(8x\) is \(4x^2\).- The integral of \(-\ln x\) is \(-x \ln x + x\), using integration by parts.- The integral of \(-3x^4\) is \(-\frac{3}{5}x^5\).Thus, the integral becomes:\[ y = 4x^2 - (x \ln x - x) - \frac{3}{5}x^5 + C\]
03
Simplify the Expression
Simplify the expression obtained from integration:\[ y = 4x^2 - x \ln x + x - \frac{3}{5}x^5 + C\]
04
Apply Initial Condition to Solve for C
Use the given initial condition \(y(1) = 5\) to solve for \(C\).Substitute \(x = 1\) and \(y = 5\) in the simplified equation:\[ 5 = 4(1)^2 - 1 \ln 1 + 1 - \frac{3}{5}(1)^5 + C\]Simplify:\[ 5 = 4 + 1 - \frac{3}{5} + C\]This simplifies further to solve for \(C\):\[ 5 = \frac{20}{5} + \frac{5}{5} - \frac{3}{5} + C\]\[ 5 = \frac{22}{5} + C\]\[ C = 5 - \frac{22}{5}\]\[ C = \frac{25}{5} - \frac{22}{5}\]\[ C = \frac{3}{5}\]
05
Write the Final Solution
Substitute the found value of \(C\) back into the general solution:\[ y = 4x^2 - x \ln x + x - \frac{3}{5}x^5 + \frac{3}{5}\] This is the solution to the initial value problem.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First-Order Differential Equation
A first-order differential equation involves derivatives that are first-order, meaning the highest derivative is taken once. In this context, the equation given is represented as \( y' = 8x - \ln x - 3x^4 \). This means that the rate of change of \( y \), with respect to \( x \), is dependent on a combination of a linear term \( 8x \), a logarithmic term \(- \ln x \), and a polynomial term \(-3x^4\).
First-order differential equations are among the simplest differential equations you will encounter, yet they introduce important methods for solving, such as integration. To solve such equations, we seek to determine a function \( y(x) \) that satisfies both the differential equation and any given initial conditions when plugged into the equation.
First-order differential equations are among the simplest differential equations you will encounter, yet they introduce important methods for solving, such as integration. To solve such equations, we seek to determine a function \( y(x) \) that satisfies both the differential equation and any given initial conditions when plugged into the equation.
Integration by Parts
Integration by parts is a fundamental technique used to solve certain integrals where direct integration is not straightforward. This method applies the formula: \[ \int u \, dv = uv - \int v \, du \] where \( u \) and \( dv \) are parts of the integrand you choose strategically.
In the context of the problem, integration by parts is used to solve \( \int - \ln x \, dx \). Here, if you let \( u = \ln x \) and \( dv = dx \), you find \( du = \frac{1}{x}dx \) and \( v = x \). Applying the integration by parts formula, the integral becomes:
\[ -\int \ln x \, dx = -\left[ x \ln x - \int x \cdot \frac{1}{x} \, dx \right] \]
This results in \(-x \ln x + x\), which is utilized in solving the differential equation. Knowing when and how to apply this technique can make solving integrals much easier.
In the context of the problem, integration by parts is used to solve \( \int - \ln x \, dx \). Here, if you let \( u = \ln x \) and \( dv = dx \), you find \( du = \frac{1}{x}dx \) and \( v = x \). Applying the integration by parts formula, the integral becomes:
\[ -\int \ln x \, dx = -\left[ x \ln x - \int x \cdot \frac{1}{x} \, dx \right] \]
This results in \(-x \ln x + x\), which is utilized in solving the differential equation. Knowing when and how to apply this technique can make solving integrals much easier.
Constant of Integration
In the process of solving differential equations, especially when integrating, a constant of integration \( C \) arises. This constant represents the family of all possible solutions, as integration is essentially about summing up infinitesimally small changes, leaving an arbitrary constant.
Once we perform integration, we obtain a general solution that includes \( C \). To find the specific value of \( C \), we use initial conditions provided in the problem. Here, the condition \( y(1) = 5 \) helps us find the particular value of \( C \) that makes the solution not just general but specific to our problem.
Using this initial condition allows us to solve for \( C \) and tailor the general solution to meet the exact requirements of a given scenario. Keeping track of \( C \) is crucial for determining the unique solution.
Once we perform integration, we obtain a general solution that includes \( C \). To find the specific value of \( C \), we use initial conditions provided in the problem. Here, the condition \( y(1) = 5 \) helps us find the particular value of \( C \) that makes the solution not just general but specific to our problem.
Using this initial condition allows us to solve for \( C \) and tailor the general solution to meet the exact requirements of a given scenario. Keeping track of \( C \) is crucial for determining the unique solution.
General Solution
The general solution is the form we arrive at after integrating a differential equation, and it represents a family of functions. In this problem, after integrating \( y' = 8x - \ln x - 3x^4 \), we obtain: \[ y = 4x^2 - x \ln x + x - \frac{3}{5}x^5 + C \] where \( C \) is the constant of integration.
The general solution helps in conceptualizing all possible solutions that the differential equation can have, dependent on different values of \( C \). By substituting different values for \( C \), we can generate different solutions, each distinct but all consistent with the original differential equation. However, to find a specific solution, an initial condition or additional requirement is used to determine the precise value of \( C \), ultimately leading to the exact solution of the problem.
This way, the general solution serves both as a theoretical exploration of possibilities and a practical step in determining the exact answer.
The general solution helps in conceptualizing all possible solutions that the differential equation can have, dependent on different values of \( C \). By substituting different values for \( C \), we can generate different solutions, each distinct but all consistent with the original differential equation. However, to find a specific solution, an initial condition or additional requirement is used to determine the precise value of \( C \), ultimately leading to the exact solution of the problem.
This way, the general solution serves both as a theoretical exploration of possibilities and a practical step in determining the exact answer.
