Problem 11 - Differential Equations and Linear Algebra

Question

Verify that the given function (or relation) defines a solution to the given differential equation and sketch some of the solution curves. If an initial condition is given, label the solution curve corresponding to the resulting unique solution. (In these problems, \(c\) denotes an arbitrary constant.) $$y^{2}=c x, \quad 2 x d y-y d x=0, \quad y(1)=2$$

Step-by-Step Solution

Answer

The given function \(y^2 = cx\) is a solution to the given differential equation \(2x dy - y dx = 0\), as we have verified by finding the derivative \(\frac{dy}{dx} = \frac{c}{2y}\) and plugging it into the differential equation. For the initial condition \(y(1) = 2\), we found the unique solution to be \(y^2 = 4x\). Sketching the solution curves for different values of \(c\), we see a set of parabolas opening to the right for \(c > 0\), a horizontal line at \(y = 0\) for \(c = 0\), and no real solution for \(c < 0\). The unique solution corresponding to the initial condition is represented by the curve \(y^2 = 4x\).

Step 1: Find the derivative of the given function

First, we need to find the derivative of \(y^2 = cx\). In order to find the derivative, we can differentiate both sides of the equation with respect to \(x\). Differentiating \(y^2\) with respect to \(x\), we get: \[\frac{d(y^2)}{dx} = 2y \frac{dy}{dx}\]. Now, differentiating \(cx\) with respect to \(x\), we get: \[\frac{d(cx)}{dx} = c\] Equating the two expressions, we get: \[2y \frac{dy}{dx} = c\] Now we can solve for \(\frac{dy}{dx}\): \[\frac{dy}{dx} = \frac{c}{2y}\]

Step 2: Plug the derivative and the function into the differential equation

Next, we will substitute the obtained derivative \(\frac{dy}{dx}\) and the given function (\(y^2 = cx\)) into the differential equation \(2x dy - y dx = 0\), and check if it holds true. First, we will replace \(dy\) and \(dx\) with their corresponding expressions in terms of \(\frac{dy}{dx}\): \[2x dy = 2xy\frac{c}{2y}\rightarrow 2x \cdot \frac{c}{2y} = c\] Simplifying the equation, we get: \[2x \cdot \frac{c}{2y} = c\] As we can see, the left-hand side equals the right-hand side. This indicates that the given function satisfies the given differential equation.

Step 3: Determine the unique solution using the initial condition

Now we need to find the solution corresponding to the initial condition \(y(1) = 2\). To do this, we will substitute the initial condition into the given function \(y^2 = cx\), and solve for the arbitrary constant \(c\). Substituting the initial condition, we get: \[y^2 = c x=(2)^2 = c \cdot 1\] \[c = 4\] So for the unique solution corresponding to the initial condition, the equation is now: \[y^2 = 4x\]

Step 4: Sketch some solution curves

To sketch the solution curves, first, consider different values of \(c\). For instance, we can consider three cases: \(c > 0\), \(c = 0\), and \(c < 0\). 1. For \(c > 0\), the solution will be \(y^2 = cx\), which represents a set of parabolas opening to the right. The unique solution that corresponds to the initial condition \(y(1) = 2\) is, in this case, \(y^2 = 4x\). 2. For \(c = 0\), the solution will be \(y^2 = 0\), which represents a horizontal line at \(y = 0\). 3. For \(c < 0\), there will be no real solution, as \(y^2\) cannot be negative. Now you can sketch the given function for different values of \(c\) and label the unique solution corresponding to the initial condition (\(y^2 = 4x\)).

Key Concepts

Initial Condition

An *initial condition* in differential equations is a value that allows us to find a specific solution among a family of solutions. When we have a differential equation, there are usually multiple solutions depending on an arbitrary constant, often denoted as \(c\).
The initial condition helps to determine the value of this constant, making the solution unique.Let's consider our exercise: we have the differential equation \(2x dy - y dx = 0\) and a general solution \(y^2 = cx\). Here, \(c\) is an arbitrary constant. To find the specific solution that satisfies an initial condition, we use the information \(y(1) = 2\).
Inserting these values into the solution \(y^2 = cx\), we calculate:

• \((2)^2 = c \cdot 1\)
• \(c = 4\)

By solving the equation above, we have determined that for \(y(1) = 2\), \(c = 4\). Therefore, the specific solution is \(y^2 = 4x\), giving us the unique solution curve that meets the initial condition.

Solution Curves

*Solution curves* represent different possibilities for the behavior of solutions to a differential equation. They are often drawn graphically to provide a visual understanding of how the solutions change with different values of the constant \(c\) in the solution \(y^2 = cx\).
In our problem, the family of solution curves forms parabolas. By varying \(c\), we see different solution curves:

• For \(c > 0\), the solution curves are parabolas opening to the right.
• For \(c = 0\), the solution is \(y^2 = 0\) — a line at \(y = 0\).
• For \(c < 0\), there are no real solutions, as \(y^2\) cannot be negative.

In addition to illustrating solution behavior, solution curves help to identify which one corresponds to an initial condition. The curve \(y^2 = 4x\) is particularly important here, as it is the curve that meets our initial condition \(y(1) = 2\). By recognizing this specific parabolic curve, students can more accurately understand how unique solutions are determined in differential equations.

Parabolas

In the context of differential equations, *parabolas* are curves that we often encounter as solutions like in our equation \(y^2 = cx\).
Parabolas express a quadratic relation between two variables, here \(y\) and \(x\). Depending on the sign and magnitude of \(c\), these parabolas can open in different directions and vary in size.For our equation:

• If \(c > 0\), the parabola opens to the right. This means that as \(x\) increases, \(y\) can take on positive and negative values symmetrically around the x-axis.
• If \(c < 0\), theoretically, the parabola would open to the left, but as \(y^2\) cannot be negative, no real solutions exist.
• The question \(c = 0\) gives us a degenerate state where the parabola flattens to the horizontal line \(y = 0\).

Visualizing different parabolas resulting from varying \(c\) helps understand distinct solutions arising from differential equations.
Students can gain insights into the nature of equation solutions by sketching these curves and observing how factors like initial conditions shift the particular parabola we choose.