Problem 37
Question
We note that \(\left(M_{y}-N_{x}\right) / N=2 x /\left(4+x^{2}\right),\) so an integrating factor is \(e^{-2 \int x d x /\left(4+x^{2}\right)}=1 /\left(4+x^{2}\right) .\) Let \(M=x /\left(4+x^{2}\right)\) and \(N=\left(x^{2} y+4 y\right) /\left(4+x^{2}\right)=y,\) so that \(M_{y}=0=N_{x} .\) From \(f_{x}=x\left(4+x^{2}\right)\) we obtain \(f=\frac{1}{2} \ln \left(4+x^{2}\right)+h(y), h^{\prime}(y)=y,\) and \(h(y)=\frac{1}{2} y^{2} .\) A solution of the differential equation is \(\frac{1}{2} \ln \left(4+x^{2}\right)+\frac{1}{2} y^{2}=c\)
Step-by-Step Solution
Verified Answer
The solution is \( \frac{1}{2} \ln(4+x^2) + \frac{1}{2}y^2 = c \).
1Step 1: Identify Integrating Factor
The problem gives us that \( \frac{M_y - N_x}{N} = \frac{2x}{4+x^2} \). This means that an integrating factor \( \mu(x) = e^{-\int \frac{2x}{4+x^2} \, dx} = \frac{1}{4+x^2} \).
2Step 2: Simplify Terms
Here, we're provided with \( M = \frac{x}{4+x^2} \) and \( N = \frac{x^2 y + 4y}{4+x^2} = y \). We check if \( M_y = 0 \) and \( N_x = 0 \); thus effectively reducing the differential equation by making the equation exact.
3Step 3: Solve for \( f_x \)
Given \( f_x = M \), we plug in \( M = \frac{x}{4+x^2} \) to resolve \( f_x = x(4+x^2) \), leading to the antiderivative \( f = \frac{1}{2} \ln(4+x^2) + h(y) \), where \( h(y) \) is a function of \( y \).
4Step 4: Determine \( h(y) \)
Since \( N = y \), our goal is to match the derivative \( h'(y) = y \). Integrating \( y \) gives \( h(y) = \frac{1}{2}y^2 \).
5Step 5: Compile General Solution
Putting together \( f \) and \( h(y) = \frac{1}{2}y^2 \), the solution to the differential equation becomes \( \frac{1}{2} \ln(4+x^2) + \frac{1}{2}y^2 = c \), where \( c \) is a constant of integration.
Key Concepts
Integrating FactorAntiderivativeLogarithmic IntegrationDifferential Equation Solution
Integrating Factor
In differential equations, the integrating factor is a crucial tool used to simplify equations that are not initially exact. An exact differential equation means that it can be derived from a differentiable function with continuous derivatives. When given an equation like our exercise, the integrating factor helps in making the differential equation exact.
To find the integrating factor, we use the expression \( \mu(x, y) = e^{-\int \frac{M_y - N_x}{N} \, dx} \). This formula comes from analyzing how the changes in terms of \(M\) and \(N\) to become exact. By calculating the integrating factor \( \frac{1}{4+x^2}\), we transform the equation to a form that can be integrated directly.
To find the integrating factor, we use the expression \( \mu(x, y) = e^{-\int \frac{M_y - N_x}{N} \, dx} \). This formula comes from analyzing how the changes in terms of \(M\) and \(N\) to become exact. By calculating the integrating factor \( \frac{1}{4+x^2}\), we transform the equation to a form that can be integrated directly.
Antiderivative
The antiderivative, often referred to as the indefinite integral, helps in solving differential equations by assisting us to revert to an original function from its derivative.
In the step-by-step solution, after simplifying the given differential equation, we find \( f_x = x(4+x^2) \), which needs to be integrated to find \( f \). Using the antiderivative technique, the integration yields \( f = \frac{1}{2} \ln(4+x^2) + h(y) \). Here, the natural logarithm \( \ln \) represents the result of integrating the expression \( \frac{x}{4+x^2} \). This step is crucial as it forms a part of the solution, incorporating both \(x\) and \(y\) through the unknown function \( h(y) \).
In the step-by-step solution, after simplifying the given differential equation, we find \( f_x = x(4+x^2) \), which needs to be integrated to find \( f \). Using the antiderivative technique, the integration yields \( f = \frac{1}{2} \ln(4+x^2) + h(y) \). Here, the natural logarithm \( \ln \) represents the result of integrating the expression \( \frac{x}{4+x^2} \). This step is crucial as it forms a part of the solution, incorporating both \(x\) and \(y\) through the unknown function \( h(y) \).
Logarithmic Integration
Logarithmic integration is a specific type of integration which often arises when dealing with rational functions, like in our exercise. When integrating functions of the form \( \frac{a}{b + cx^2} \), logarithms can often appear in the solution.
In the differential equation presented, integrating \( \frac{x}{4+x^2} \) results in the term \( \frac{1}{2} \ln(4+x^2) \). This is due to the formula for the integration of \( \frac{1}{x} \), which yields \( \ln|x| + C \), where \(C\) is the constant of integration.
Recognizing logarithmic integration is key because it simplifies the process of integrating complex terms and assists in constructing the entire solution.
In the differential equation presented, integrating \( \frac{x}{4+x^2} \) results in the term \( \frac{1}{2} \ln(4+x^2) \). This is due to the formula for the integration of \( \frac{1}{x} \), which yields \( \ln|x| + C \), where \(C\) is the constant of integration.
Recognizing logarithmic integration is key because it simplifies the process of integrating complex terms and assists in constructing the entire solution.
Differential Equation Solution
Solving differential equations involves combining all the derived components—like the antiderivative, integrating factor, and additional terms. After all pieces are computed, they join to provide a full solution.
In the task's differential equation, the solution comes together by determining both \( f(x, y) = \frac{1}{2} \ln(4+x^2) + h(y) \) and \( h(y) = \frac{1}{2}y^2 \) through integrating \( y \) and applying the constant of integration \( c \). Ultimately, the general solution takes the form: \( \frac{1}{2} \ln(4+x^2) + \frac{1}{2}y^2 = c \).
This approach provides a generalized formula which encompasses all potential specific solutions within our constraints.
In the task's differential equation, the solution comes together by determining both \( f(x, y) = \frac{1}{2} \ln(4+x^2) + h(y) \) and \( h(y) = \frac{1}{2}y^2 \) through integrating \( y \) and applying the constant of integration \( c \). Ultimately, the general solution takes the form: \( \frac{1}{2} \ln(4+x^2) + \frac{1}{2}y^2 = c \).
This approach provides a generalized formula which encompasses all potential specific solutions within our constraints.
Other exercises in this chapter
Problem 37
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View solution Problem 38
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View solution Problem 38
Write the differential equation as \(d P / d t-a P=-b P^{2}\) and let \(u=P^{-1}\) or \(P=u^{-1}\). Then $$\frac{d p}{d t}=-u^{-2} \frac{d u}{d t},$$ and substi
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