Chapter 4: Problem 7
First find the general solution (involving a constant C) for the given differential equation. Then find the particular solution that satisfies the indicated condition. (See Example 2.) \(\frac{d y}{d x}=\frac{x}{y} ; y=1\) at \(x=1\)
Short Answer
Expert verified
The particular solution is \( y = x \).
Step by step solution
01
Separate Variables
Start by separating variables in the differential equation \( \frac{d y}{d x} = \frac{x}{y} \). This means rearranging the equation to get one variable on each side, resulting in \( y \, dy = x \, dx \).
02
Integrate Both Sides
Integrate both sides of the equation to find the general solution. Integrate \( \int y \, dy \) and \( \int x \, dx \). This gives \( \frac{y^2}{2} = \frac{x^2}{2} + C \), where \( C \) is the integration constant.
03
Solve for the General Solution
Multiply everything by 2 to simplify the equation: \( y^2 = x^2 + 2C \). This represents the general solution of the differential equation.
04
Apply Initial Condition
Use the initial condition \( y = 1 \) at \( x = 1 \) to find \( C \). Substitute \( y = 1 \) and \( x = 1 \) into \( 1^2 = 1^2 + 2C \). This simplifies to \( 0 = 2C \), meaning \( C = 0 \).
05
Write the Particular Solution
Substitute \( C = 0 \) back into the general solution \( y^2 = x^2 + 2C \) to get \( y^2 = x^2 \). Solving this gives the particular solution \( y = x \) or \( y = -x \). Considering the positive initial condition \( y = 1 \) when \( x = 1 \), the appropriate solution is \( y = x \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
General Solution
The general solution of a differential equation is a representation of all possible solutions that a differential equation can have. It embodies a family of curves or functions with one or more arbitrary constants, typically denoted by a symbol like \( C \). For the given differential equation \( \frac{d y}{d x} = \frac{x}{y} \), the process to find the general solution involves separating the variables, as seen in the first step of the solution.
After separation, we combine all terms involving \( y \) on one side and \( x \) on the other, yielding \( y \, dy = x \, dx \).
By integrating both sides, we are able to obtain the expression \( \frac{y^2}{2} = \frac{x^2}{2} + C \). This is our general solution before we apply any specific conditions.
After separation, we combine all terms involving \( y \) on one side and \( x \) on the other, yielding \( y \, dy = x \, dx \).
By integrating both sides, we are able to obtain the expression \( \frac{y^2}{2} = \frac{x^2}{2} + C \). This is our general solution before we apply any specific conditions.
- Integration includes finding antiderivatives, which introduces the integration constant \( C \).
- Each value of \( C \) represents a different curve from the family of solutions.
Particular Solution
A particular solution of a differential equation results by substituting a specific condition into the general solution. It represents a single, unique solution curve, among the family of solutions described by the general solution.
In this exercise, the particular solution is based on the initial condition provided, which is \( y = 1 \) when \( x = 1 \).
In this exercise, the particular solution is based on the initial condition provided, which is \( y = 1 \) when \( x = 1 \).
- Substitute these values into the general solution \( y^2 = x^2 + 2C \) to find the specific \( C \).
- Doing this results in \( 0 = 2C \), leading to \( C = 0 \).
- Substituting \( C = 0 \) back into the general solution gives \( y^2 = x^2 \), which simplifies to two potential solutions, \( y = x \) or \( y = -x \).
Separation of Variables
Separation of variables is a fundamental technique for solving differential equations. It involves rearranging the equation so that all terms involving one variable (say \( y \)) are on one side while terms involving the other variable (\( x \)) are on the opposite side. This sets the stage for straightforward integration.
In our exercise, the original differential equation was \( \frac{d y}{d x} = \frac{x}{y} \). The separation technique led to \( y \, dy = x \, dx \).
In our exercise, the original differential equation was \( \frac{d y}{d x} = \frac{x}{y} \). The separation technique led to \( y \, dy = x \, dx \).
- Each side of the equation only contains one variable, allowing independent integration.
- This method simplifies the integration process, making it easier to solve the equation analytically.
Initial Condition
An initial condition in a differential equation problem is a specified value for the function and often its derivatives at a particular point. It is essential to find a particular solution from the general solution.
In this exercise, the initial condition given is \( y = 1 \) when \( x = 1 \). When we plug these values into the general solution equation \( y^2 = x^2 + 2C \), it immediately helps determine the constant \( C \).
In this exercise, the initial condition given is \( y = 1 \) when \( x = 1 \). When we plug these values into the general solution equation \( y^2 = x^2 + 2C \), it immediately helps determine the constant \( C \).
- Substituting the initial values gives \( 1^2 = 1^2 + 2C \), simplifying to \( 0 = 2C \), and hence \( C = 0 \).
- An initial condition ensures that the solution corresponds to the conditions in the actual problem context.
- Solving with this specific condition refines the general solution to describe a unique, specific curve, which in this case leads us to \( y = x \).