Problem 29 - Advanced Engineering Mathematics

Question

(a) Byinspection, find a one-parameter family of solutions of the differential equation \(x y^{\prime}=y .\) Verify that each member of the family is a solution of the initial-value problem \(x y^{\prime}=y, y(0)=0\) (b) Explain part (a) by determining a region \(R\) in the \(x y\) -plane for which the differential equation \(x y^{\prime}=y\) would have a unique solution through a point \(\left(x_{0}, y_{0}\right)\) in \(R\). (c) Verify that the picecwise-defined function $$ y=\left\\{\begin{array}{ll} 0, & x<0 \\ x, & x \geq 0 \end{array}\right. $$ satisfies the condition \(y(0)=0\). Determine whether this function is also a solution of the initial-value problem in part (a).

Step-by-Step Solution

Answer

The solution to the initial-value problem is a family of lines \( y = cx \) with \( x = 0 \) except in the region \( x \neq 0 \). The piecewise function also satisfies this problem.

Step 1: Find a One-Parameter Family of Solutions

The given differential equation is \( x y' = y \). To find a one-parameter family of solutions, assume the form \( y = cx \) where \( c \) is a parameter. Differentiate both sides with respect to \( x \), giving: \( y' = c \). Substituting the assumed solution in the differential equation gives:\[ x(c) = cx \]This equation holds true for all \( x \) and \( y \), so \( y = cx \) is indeed a solution.

Step 2: Verify Solution of the Initial-Value Problem

Set \( y = cx \) and substitute into the initial condition \( y(0)=0 \). This gives:\[ 0 = c \times 0 = 0 \]This equation is satisfied regardless of the value of \( c \), so every member of the family \( y = cx \) satisfies the initial-value problem condition \( y(0)=0 \).

Step 3: Determine Region R in xy-plane for Unique Solutions

For the differential equation \( x y' = y \), to have a unique solution at a point \((x_0, y_0)\), it must satisfy conditions for existence and uniqueness. The equation can be rewritten as \( y' = \frac{y}{x} \), which has a singularity at \( x=0 \). Therefore, unique solutions exist in regions where \( x eq 0 \). Hence, the region \( R \) in the \( xy \)-plane is \( R = \mathbb{R} \setminus \{ x=0 \} \).

Step 4: Check the Piecewise-Defined Function

Consider the piecewise function:\[ y = \begin{cases} 0, & x < 0 \ x, & x \geq 0 \end{cases} \] This function satisfies \( y(0) = 0 \). For \( x<0 \), \( y = 0 \) makes \( y' = 0 \), satisfying the equation \( x y' = y = 0 \). For \( x \geq 0 \), \( y = x \) gives \( y' = 1 \) and \( xy' = x \) also equals \( y = x \), satisfying the differential equation. Thus, it satisfies the initial-value problem.

Key Concepts

One-Parameter Family of Solutions

A one-parameter family of solutions refers to a collection of functions that can solve a given differential equation. These functions are parameterized by a constant, allowing various solutions depending on the parameter's value.
For the differential equation \(x y' = y\), the one-parameter family of solutions can be expressed as \(y = cx\), where \(c\) is an arbitrary constant (or parameter).
To confirm that this form is a solution, substitute \(y = cx\) into the differential equation:

• Differentiate \(y = cx\) to get \(y' = c\).
• Substitute \(y = cx\) and \(y' = c\) back into \(x y' = y\).
• This results in \(x(c) = cx\), which is valid for any value of \(c\).

Thus, \(y = cx\) covers all possible solutions governed by the differential equation's constraints.

Initial-Value Problems

An initial-value problem specifies the value of a solution at a particular point. It narrows down the possible solutions of a differential equation by adding another condition.
In our case, the initial-value problem is \(x y' = y\) combined with the condition \(y(0) = 0\). This additional constraint helps in selecting a member from the one-parameter family of solutions.
To solve for the initial condition:

• Substitute \(x = 0\) and \(y = 0\) into \(y = cx\).
• This results in \(0 = c \, * \, 0\), naturally satisfied by all values of \(c\).

Since the condition is inherently satisfied, every member of the one-parameter family \(y = cx\) can potentially solve the initial-value problem. However, it also suggests there may be more than one valid solution for the initial data.

Existence and Uniqueness Theorem

The Existence and Uniqueness Theorem provides necessary conditions for a differential equation's solutions to be unique. It assures us that under certain circumstances, a differential equation has exactly one solution passing through a specified point.
When dealing with \(x y' = y\), which can be rearranged to \(y' = \frac{y}{x}\), observe that there is a singularity at \(x = 0\).
The theorem implies:

• Solutions exist and are unique where the function and its derivative are continuous.
• Singuarity at \(x = 0\) prevents uniqueness here because dividing by zero is undefined.

Therefore, the region \(R\) of uniqueness excludes the line \(x = 0\), indicating unique solutions only exist where \(x eq 0\).

Piecewise Functions

Piecewise functions are those that have different expressions for different intervals of their domain. They are especially useful for capturing behavior that changes based on input values.
Consider the piecewise function:\[y = \begin{cases} 0, & x < 0 \ x, & x \geq 0 \end{cases} \]
This function satisfies the initial condition \(y(0) = 0\). Let's verify it as a solution:

• For \(x < 0\), \(y = 0\). Thus, \(y' = 0\) makes \(x y' = y = 0\) true.
• For \(x \geq 0\), \(y = x\) results in \(y' = 1\), thus \(x \, * \, 1 = x\). It matches with \(y = x\).

So, the piecewise function appropriately solves the differential equation across its whole defined range, respecting both segments of its condition.