Question

For each of the following differential equations, separate variables and find a solution containing one arbitrary constant. Then find the value of the constant to give a particular solution satisfying the given boundary condition. Computer plot a slope field and some of the solution curves. \((1+y) y^{\prime}=y, \quad y=1\) when \(x=1\)

Short answer

The particular solution is \(y = \frac{e^{x-1}}{2-e^{x-1}}\).

Step-by-step solution

Separate the Variables

Start by rewriting the given differential equation \((1+y) y^{\prime} = y\). Recall that \(y^{\prime}\) represents \(\frac{dy}{dx}\). Substitute this in to get: \((1+y) \frac{dy}{dx} = y\). To separate the variables, divide both sides by \(y(1+y)\) to get: \(\frac{1}{y(1+y)} dy = dx\).

Integrate Both Sides

Integrate both sides of the equation. For the left side, use partial fraction decomposition: \[ \frac{1}{y(1+y)} = \frac{A}{y} + \frac{B}{1+y} \] Solving for \(A\) and \(B\): Set up the equation: \(1 = A(1+y) + By\). Compare coefficients to get: \(A = 1\) and \(B = -1\). So the integral becomes: \[ \int \left( \frac{1}{y} - \frac{1}{1+y} \right) dy = \int dx \] Evaluate the integrals to get: \[ \ln|y| - \ln|1+y| = x + C \]

Solve for y

Combine the logarithms on the left side: \[ \ln \left| \frac{y}{1+y} \right| = x + C \] Exponentiate both sides to eliminate the logarithm: \[ \left| \frac{y}{1+y} \right| = e^{x+C} = e^x e^C \] Let \(e^C = K\), a new constant: \[ \frac{y}{1+y} = Ke^x \]. Solving for \(y\) in terms of \(x\): \[ y = \frac{Ke^x}{1 - Ke^x} \]

Apply the Boundary Condition

Use the boundary condition \(y = 1\) when \(x = 1\) to find the specific constant \(K\). Plug in \(x = 1\) and \(y = 1\): \[ 1 = \frac{Ke}{1 - Ke} \] This simplifies to: \[ 1 - Ke = Ke \] Solving this for \(K\): \[ 1 = 2Ke \implies K = \frac{1}{2e} \]

Write the Particular Solution

Substitute the value of \(K\) back into the general solution: \[ y = \frac{\frac{e^x}{2e}}{1 - \frac{e^x}{2e}} = \frac{e^{x-1}}{2-e^{x-1}} \]

Plot the Slope Field and Solution Curves

To plot the slope field, first rewrite the original differential equation in terms of the slope: \(y' = \frac{y}{1+y}\). Use software tools or graphing calculators capable of plotting differential equations to visualize the slope field and the particular solution curve \(y = \frac{e^{x-1}}{2-e^{x-1}}\).

Key concepts

Variable Separation

Variable separation is a crucial technique for solving differential equations. It involves rearranging the equation to isolate variables on opposite sides. For example, let’s consider the equation \((1+y) y^{\text{'}} = y\). By substituting \(y^{\text{'}}\) with \(\frac{dy}{dx}\), we get: \((1+y) \frac{dy}{dx} = y\). To separate the variables, divide both sides by \(y(1+y)\) and multiply by \(dx\): \(\frac{1}{y(1+y)} dy = dx\). This step simplifies the integration process and brings us closer to solving the differential equation.

Boundary Conditions

Boundary conditions are given initial values that help determine the particular solution of a differential equation. They are essential in finding the specific constant in the general solution. For our example, we have the boundary condition \(y = 1\) when \(x = 1\). This means that when \(x\) is 1, the value of \(y\) must be 1. After solving the equation partially, we use this condition to find the constant \(K\). For instance, when we substitute \(x = 1\) and \(y = 1\) in the equation: \(1 = \frac{Ke}{1-Ke}\). By solving this, we determine \(K\).

Slope Fields

Slope fields provide a visual representation of a differential equation. They show the slope of the solution curves at different points, making the behavior of solutions easier to understand. To plot a slope field for the differential equation \((1+y) y^{\text{'}} = y\), you rewrite it as \(y' = \frac{y}{1+y}\). By plotting several points and drawing short line segments with slopes given by \(y'\) at each point, you build up the slope field. This helps visualize how solutions curve and evolve over time.

Particular Solution

A particular solution to a differential equation is a specific solution that satisfies the given boundary conditions. In our step-by-step solution, once we have integrated both sides, we arrive at a general solution containing an arbitrary constant \(C\): \[ \ln \left| \frac{y}{1+y} \right| = x + C \]. Using the boundary condition \(y = 1\) when \(x = 1\), we solve for \(C\) and get a specific value. This leads to the particular solution: \[ y = \frac{e^{x-1}}{2-e^{x-1}} \] which uniquely satisfies our boundary condition.

Integration Techniques

Integration is a fundamental part of solving differential equations, requiring accurate techniques. For our equation \(\frac{1}{y(1+y)} dy = dx\), we use partial fraction decomposition to integrate the left side. First, we decompose \(\frac{1}{y(1+y)}\) into \(\frac{A}{y} + \frac{B}{1+y}\), solving for \(A\) and \(B\): \(A = 1\) and \(B = -1\). This turns the integral into: \[ \int \left( \frac{1}{y} - \frac{1}{1+y} \right) dy = \int dx \]. After integrating, we get: \[ \ln|y| - \ln|1+y| = x + C \]. This shows how integrating step-by-step helps us find the solution progressively.