Chapter 8: Problem 9

Find the "general solution" (that is, a solution containing an arbitrary constant) of each of the following differential equations, by separation of variables. Then find a particular solution of each equation satisfying the given boundary conditions. \((1+y) y^{\prime}=y\) \(y=1\) when \(x=1\)

Short Answer

Expert verified

The particular solution is \(y + \ln|y| = x\).

Step by step solution

Rewrite the Differential Equation

Start by rewriting the given differential equation \((1 + y) y' = y\). Since \(y'\) denotes the derivative of \(y\) with respect to \(x\), rewrite it as \((1 + y) \frac{dy}{dx} = y\).

Separate the Variables

Rearrange the terms to get all \(y\)-terms on one side and \(x\)-terms on the other: \(\frac{1 + y}{y} dy = dx\). This can be further simplified to: \((1 + \frac{1}{y}) dy = dx\).

Integrate Both Sides

Integrate both sides of the equation with respect to their own variables: \[\int (1 + \frac{1}{y}) dy = \int dx\]. This results in: \[\int 1 dy + \int \frac{1}{y} dy = \int 1 dx\]. The integrals are: \[y + \ln|y| = x + C\], where \(C\) is the constant of integration.

Solve for the General Solution

Combine the terms to form the general solution: \[y + \ln|y| = x + C\].

Apply the Boundary Condition

Use the given boundary condition \(y(1) = 1\) to find the particular solution. Substitute \(x = 1\) and \(y = 1\) into the general solution: \[1 + \ln|1| = 1 + C\]. Since \(\ln|1| = 0\), the equation simplifies to: \[1 = 1 + C\], implying \(C = 0\).

Write the Particular Solution

With \(C = 0\), the particular solution is: \[y + \ln|y| = x\].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. They are used to model real-world systems and changes over time. Take the example \((1 + y) y' = y\). Here, the dependent variable is \(y\) and it changes with respect to the independent variable \(x\). The term \(y'\) (or \(\frac{dy}{dx}\)) represents the derivative of \(y\) with respect to \(x\), which gives the rate of change of \(y\) as \(x\) varies.
Solving differential equations can help us predict and understand the behavior of systems.
In this exercise, we will use a method called 'separation of variables' to solve the differential equation.

The Role of Boundary Conditions

Boundary conditions are additional constraints or specific values provided for the variables at certain points. They are used to find particular solutions from the general solution of a differential equation.
In our problem, the boundary condition given is \(y=1\) when \(x=1\). This helps us find the specific constant value for the integral solution.
Using the boundary condition, we can narrow down the range of possible solutions to one that fits the condition. For example, substituting \(x = 1\) and \(y = 1\) into our derived general solution \(y + \ln|y| = x + C\) allows us to solve for the constant \(C\). This is crucial in practical applications where exact conditions are known, such as in physics or engineering problems.

Integration in Solving Differential Equations

Integration is a mathematical process of finding the integral of a function, which is essentially the reverse operation of differentiation. In the context of differential equations, integration is used to find the antiderivatives of both sides of the the equation.
In our exercise, we needed to integrate both sides of the separated equation \((1 + \frac{1}{y}) dy = dx\):

\(\int 1 dy + \int \frac{1}{y} dy = \int 1 dx\)
This results in \(y + \ln|y| = x + C\)

The integration constant \(C\) is crucial but not specific until boundary conditions are applied. Using the integral form, we obtained the general solution, which later became specific by applying the boundary condition. This integration technique is key to turning differential equations into solvable algebraic expressions.