Problem 15
Question
In Exercises \(11-20,\) solve the initial value problem explicitly. $$\frac{d y}{d x}=-\frac{1}{x^{2}}-\frac{3}{x^{4}}+12\( and \)y=3\( when \)x=1$$
Step-by-Step Solution
Verified Answer
The explicit solution of the initial value problem is \(y = x + \frac{1}{x} - 3x^2 + 12x - 8\).
1Step 1: Integrate the differential equation
The differential equation is \(\frac{dy}{dx} = - \frac{1}{x^2} - \frac{3}{x^4} + 12\). In order to obtain a general solution, this needs to be integrated with respect to \(x\). Doing so, it results in \(y = \int (- \frac{1}{x^2} - \frac{3}{x^4} + 12) dx\). Solving this integral leads to the general solution \(y = x + \frac{1}{x} - 3x^2 + 12x + C\), where \(C\) is the constant of integration.
2Step 2: Use the initial condition to find the constant
The initial condition given is \(y = 3\) when \(x = 1\). This condition is used to resolve for the constant of integration. Substituting into the general solution gives \(3 = 1 + 1 - 3(1)^2 + 12(1) + C\). Solving this equation for \(C\) yields \(C = -8\).
3Step 3: Insert the constant into the general solution
Now, replace \(C\) with \(-8\) in the general solution: \(y = x + \frac{1}{x} - 3x^2 + 12x - 8\). This is the explicit solution of the given initial value problem.
Key Concepts
Differential EquationIntegrationConstant of IntegrationInitial Condition
Differential Equation
A differential equation is a mathematical equation that involves derivatives, which describe how a function changes. In the context of the given problem, \( \frac{dy}{dx} = -\frac{1}{x^2} - \frac{3}{x^4} + 12 \)describes the rate of change of the function \(y\) with respect to \(x\). This particular equation is straightforward because it involves terms of \(x\) with powers and constant coefficients. By solving this equation, we derive a function \(y(x)\) that fulfills the stated relationship between \(y\) and \(x\).
Differential equations play a crucial role in modeling real-world systems where quantities change continuously. By employing appropriate methods, like integration, one can find these functions that describe such dynamic systems.
Differential equations play a crucial role in modeling real-world systems where quantities change continuously. By employing appropriate methods, like integration, one can find these functions that describe such dynamic systems.
Integration
Integration is the process of reversing differentiation, and in this exercise, it is used to solve the differential equation. To find the solution, we integrate the expression:\[ y = \int \left(- \frac{1}{x^2} - \frac{3}{x^4} + 12 \right) dx \]Finding the antiderivatives of each term separately involves:
Once integrated, this yields the general solution:\( y = - \frac{1}{x} + \frac{1}{x^3} + 12x + C \).
Integration is an essential mathematical tool that allows us to determine antiderivatives, which can describe accumulated quantities.
- \( \int -\frac{1}{x^2} \, dx = \frac{1}{x} \)
- \( \int -\frac{3}{x^4} \, dx = \frac{1}{x^3} \)
- \( \int 12 \, dx = 12x \)
Once integrated, this yields the general solution:\( y = - \frac{1}{x} + \frac{1}{x^3} + 12x + C \).
Integration is an essential mathematical tool that allows us to determine antiderivatives, which can describe accumulated quantities.
Constant of Integration
When integrating a function, one always includes a constant of integration, denoted by \(C\). This constant accounts for any constant values that could have existed before differentiation, which do not affect derivatives.
In this initial value problem:\( y = - \frac{1}{x} + \frac{1}{x^3} + 12x + C \),\(C\) represents the constant of integration. Since differentiation of a constant equals zero, it doesn’t appear in the original differential equation, but it can be determined using initial conditions.
This step is crucial because there can be infinitely many solutions to a differential equation without specifying the constant using extra information.
In this initial value problem:\( y = - \frac{1}{x} + \frac{1}{x^3} + 12x + C \),\(C\) represents the constant of integration. Since differentiation of a constant equals zero, it doesn’t appear in the original differential equation, but it can be determined using initial conditions.
This step is crucial because there can be infinitely many solutions to a differential equation without specifying the constant using extra information.
Initial Condition
An initial condition is a piece of additional information that helps determine the specific solution to a differential equation, notably by resolving the constant of integration. In this exercise, the initial condition is given as\(y = 3\) when \(x = 1\).This condition helps to find the value of \(C\) in the general solution.
By substituting \(x = 1\) and \(y = 3\) into the general solution:\( 3 = 1 + 1 - 3(1)^2 + 12(1) + C \),we find \(C = -8\).
Initial conditions are vital in solving initial value problems because they pin down the exact solution from the family of possible solutions provided by the general equation.
By substituting \(x = 1\) and \(y = 3\) into the general solution:\( 3 = 1 + 1 - 3(1)^2 + 12(1) + C \),we find \(C = -8\).
Initial conditions are vital in solving initial value problems because they pin down the exact solution from the family of possible solutions provided by the general equation.
Other exercises in this chapter
Problem 15
In Exercises \(11-16,\) solve the initial value problem. Confirm your answer by checking that it conforms to the slope field of the differential equation. $$\fr
View solution Problem 15
In Exercises \(15-18,\) solve the differential equation. $$\frac{d y}{d x}=\frac{2 x-6}{x^{2}-2 x}$$
View solution Problem 16
In Exercises \(13-16,\) verify that \(\int f(u) d u \neq \int f(u) d x\) $$f(u)=\sin u\( and \)u=4 x$$
View solution Problem 16
In Exercises \(11-16,\) solve the initial value problem. Confirm your answer by checking that it conforms to the slope field of the differential equation. $$\fr
View solution