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. \(x y^{\prime}=y, \quad y=3\) when \(x=2\)

Short answer

The particular solution is \( y = \frac{3}{2}x \).

Step-by-step solution

Separate the Variables

Start by rewriting the given differential equation in a form that allows the variables to be separated. The equation is \( x y' = y \). Divide both sides by \( y \) and multiply by \( dx \) to obtain: \[ \frac{y'}{y} = \frac{1}{x} \rightarrow \frac{dy}{y} = \frac{dx}{x} \]

Integrate Both Sides

Integrate both sides of the equation: \[ \int \frac{dy}{y} = \int \frac{dx}{x} \] The integrals are: \[ \text{ln}|y| = \text{ln}|x| + C \] where \( C \) is the constant of integration.

Solve for y

Exponentiate both sides to solve for \( y \): \[ e^{\text{ln}|y|} = e^{\text{ln}|x| + C} \rightarrow |y| = e^{\text{ln}|x|} \times e^C \rightarrow |y| = x \times e^C \] Let \( A = e^C \). Therefore, our general solution is: \[ y = Ax \]

Apply the Boundary Condition

Use the boundary condition \( y = 3 \) when \( x = 2 \) to find \( A \): \[ 3 = A \times 2 \rightarrow A = \frac{3}{2} \] Thus, the particular solution is: \[ y = \frac{3}{2}x \]

Slope Field and Solution Curves

Plot the slope field and the solution curves for the differential equation. The slope at any point \((x, y)\) is given by \( y/x \). The solution curve satisfying the boundary condition is \( y = \frac{3}{2} x \).

Key concepts

Variable Separation

Variable separation is a fundamental technique for solving differential equations. This method works by rearranging the equation to isolate the variables on opposite sides. For our problem, the given differential equation is:

\[ x y' = y \]

By dividing both sides by \( y \) and multiplying by \( dx \), we separate the variables as follows:

\[ \frac{y'}{y} = \frac{1}{x} \rightarrow \frac{dy}{y} = \frac{dx}{x} \]

This new form is crucial because it enables us to integrate each side with respect to its own variable.

Integration

The next step after separating the variables is to integrate both sides of the equation. For the equation

\[ \frac{dy}{y} = \frac{dx}{x} \],

we integrate each side concerning its respective variable:

\[ \begin{aligned} \text{ln}|y| & = \text{ln}|x| + C \ \text{where } C & \text{is the constant of integration} \ \text{ln}|y| & = \text{ln}|x| + C \rightarrow e^{\text{ln}|x|} \times e^C \rightarrow |y| & = x \times e^C \rightarrow y = x \times e^C \rightarrow y = Ax \ \text{where} \ A &= e^C \end{aligned} \]

We see that integrating results in getting rid of the logarithms and simplifying our equation to a more manageable algebraic form. Letting \( A = e^C \), we get the general solution:

\[ y = Ax \].

Boundary Conditions

To find the particular solution, we use the given boundary condition \( y = 3 \) when \( x = 2 \). Plugging these values into our general solution \( y = Ax \), we determine the constant \( A \):

\[ y = Ax \rightarrow 3 = A \times 2 \rightarrow A = \frac{3}{2} \]

Therefore, our particular solution is:

\[ y = \frac{3}{2} x \].

Boundary conditions help us determine the specific function out of the family of solutions described by the general solution.

Slope Field

A slope field, or direction field, visually represents the solutions of a differential equation. Each point in the slope field has a small line segment with a slope equal to the value given by the differential equation at that point. For the differential equation \( x y' = y \), the slope at any point \((x, y)\) is given by:

\[ y' = \frac{y}{x} \].

Plotting the slope field for this equation helps us see the pattern of the solutions. By drawing some solution curves, including the one satisfying the boundary condition \( y = \frac{3}{2} x \), we can grasp the behavior of all potential solutions. This visualization aids in understanding the nature of the differential equation and the impact of different initial conditions.