Chapter 30: Problem 29
Using the given boundary condition, find the particular solution to each differential equation. Try some by calculator. $$y^{2} y^{\prime}=x^{2}, x=0 \text { when } y=1$$
Short Answer
Expert verified
The particular solution to the differential equation given the initial condition is \(y = 2x + \frac{3}{x}\).
Step by step solution
01
Rewrite the differential equation
The given differential equation is in the form of a first-order linear differential equation. We can write it as: \( y' + \frac{1}{x}y = 4 \).
02
Identify the integrating factor
We use the formula for the integrating factor, \(\mu(x) = e^{\int P(x) \, dx}\), where \(P(x)\) is the coefficient of \(y\) in the differential equation. Here, \(P(x) = \frac{1}{x}\), so the integrating factor is \(\mu(x) = e^{\int \frac{1}{x} \, dx} = e^{\ln|x|} = x\).
03
Multiply through by the integrating factor
Multiply every term in the differential equation by the integrating factor to get \(x(y' + \frac{1}{x}y) = 4x\) which simplifies to \(xy' + y = 4x\).
04
Recognize the left side as a product rule
The left side of the equation \(xy' + y\) is the derivative of \(xy\) with respect to \(x\), indicated by the product rule \(\frac{d}{dx}(xy)\).
05
Integrate both sides
Integrating both sides of the equation with respect to \(x\), we get \(\int d(xy) = \int 4x dx\), which yields \(xy = 2x^2 + C\), where \(C\) is the constant of integration.
06
Apply the initial condition
Apply the initial condition, which is \(x=1\) when \(y=5\), to solve for \(C\). Substituting into the integrated equation gives \(1*5 = 2*1^2 + C\). Therefore, \(C=3\).
07
Write the particular solution
Using the value of \(C\), the particular solution to the differential equation is \(xy = 2x^2 + 3\) or \(y = 2x + \frac{3}{x}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integrating Factor
One of the fundamental methods for solving first-order linear differential equations involves the use of an integrating factor. This mathematical tool transforms a non-exact differential equation into an exact one, enabling us to integrate both sides and find a solution.
An integrating factor is typically a function, denoted as \(\mu(x)\), that depends solely on the independent variable, in this case, \(x\). To find \(\mu(x)\), we identify the coefficient \(P(x)\) of \(y\) in the differential equation \(y' + P(x)y = Q(x)\), where \(Q(x)\) represents the rest of the terms. The integrating factor is then calculated as \(\mu(x) = e^{\int P(x) \, dx}\).
Let's consider the equation \(y' + \frac{1}{x}y = 4\). Here, \(P(x) = \frac{1}{x}\), and thus the integrating factor is \(x\), derived from \(e^{\int \frac{1}{x} \, dx} = e^{\ln|x|} = x\) when \(x > 0\). We multiply the entire differential equation by \(x\), the integrating factor, to make the left side a perfect derivative of a product of two functions—this is crucial for the subsequent integration step to find the solution.
An integrating factor is typically a function, denoted as \(\mu(x)\), that depends solely on the independent variable, in this case, \(x\). To find \(\mu(x)\), we identify the coefficient \(P(x)\) of \(y\) in the differential equation \(y' + P(x)y = Q(x)\), where \(Q(x)\) represents the rest of the terms. The integrating factor is then calculated as \(\mu(x) = e^{\int P(x) \, dx}\).
Let's consider the equation \(y' + \frac{1}{x}y = 4\). Here, \(P(x) = \frac{1}{x}\), and thus the integrating factor is \(x\), derived from \(e^{\int \frac{1}{x} \, dx} = e^{\ln|x|} = x\) when \(x > 0\). We multiply the entire differential equation by \(x\), the integrating factor, to make the left side a perfect derivative of a product of two functions—this is crucial for the subsequent integration step to find the solution.
Initial Condition
In the process of finding solutions to differential equations, an initial condition allows us to determine a specific, or particular solution, out of the family of general solutions.
The general solution of a first-order linear differential equation includes an arbitrary constant \(C\). Without an initial condition, we cannot specify the value of \(C\), and therefore, the solution remains general and not unique. By applying the initial condition—plugging a known value of \(x\) and its corresponding \(y\) into the general solution—we can solve for \(C\).
Considering our exercise, where the initial condition given is \(x=1\) when \(y=5\), we substitute these values into the general solution \(xy = 2x^2 + C\) to find \(5 = 2(1)^2 + C\), which simplifies to \(C=3\). The initial condition has allowed us to move from a broad, generic description of possible solutions to one specific answer that fits the given criteria.
The general solution of a first-order linear differential equation includes an arbitrary constant \(C\). Without an initial condition, we cannot specify the value of \(C\), and therefore, the solution remains general and not unique. By applying the initial condition—plugging a known value of \(x\) and its corresponding \(y\) into the general solution—we can solve for \(C\).
Considering our exercise, where the initial condition given is \(x=1\) when \(y=5\), we substitute these values into the general solution \(xy = 2x^2 + C\) to find \(5 = 2(1)^2 + C\), which simplifies to \(C=3\). The initial condition has allowed us to move from a broad, generic description of possible solutions to one specific answer that fits the given criteria.
Particular Solution
A particular solution is a single, specific solution to a differential equation that satisfies both the equation itself and any given initial or boundary conditions. It stands in contrast with the general solution, which represents a family of possible solutions typically involving one or more arbitrary constants.
In our exercise, once we have utilized the integrating factor to simplify the differential equation and applied the initial condition to find the value of the arbitrary constant \(C\), we can then write down the particular solution. Here, the particular solution that meets the boundary condition \(x=1\) when \(y=5\) is \(y = 2x + \frac{3}{x}\). This equation will hold true for any \(x\) and \(y\) along the curve described by this specific solution. Importantly, while the general solution is expressive of the system's behavior, the particular solution is the one that actually complies with the physical, real-world constraints or conditions provided.
In our exercise, once we have utilized the integrating factor to simplify the differential equation and applied the initial condition to find the value of the arbitrary constant \(C\), we can then write down the particular solution. Here, the particular solution that meets the boundary condition \(x=1\) when \(y=5\) is \(y = 2x + \frac{3}{x}\). This equation will hold true for any \(x\) and \(y\) along the curve described by this specific solution. Importantly, while the general solution is expressive of the system's behavior, the particular solution is the one that actually complies with the physical, real-world constraints or conditions provided.
