Problem 34 - Advanced Engineering Mathematics

Question

For \\[ y^{\prime}+\frac{2 x}{1+x^{2}} y=\left\\{\begin{array}{ll} \frac{x}{1+x^{2}}, & 0 \leq x \leq 1 \\ \frac{-x}{1+x^{2}}, & x>1 \end{array}\right. \\] an integrating factor is \(1+x^{2}\) so that \\[ \left(1+x^{2}\right) y=\left\\{\begin{array}{ll} \frac{1}{2} x^{2}+c_{1}, & 0 \leq x \leq 1 \\ -\frac{1}{2} x^{2}+c_{2}, & x>1 \end{array}\right. \\] If \(y(0)=0\) then \(c_{1}=0\) and for continuity we must have \(c_{2}=1\) so that \\[ y=\left\\{\begin{array}{ll} \frac{1}{2}-\frac{1}{2\left(1+x^{2}\right)}, & 0 \leq x \leq 1 \\ \frac{3}{2\left(1+x^{2}\right)}-\frac{1}{2}, & x>1 \end{array}\right. \\]

Step-by-Step Solution

Answer

The solution is piecewise with \( y=\frac{1}{2}-\frac{1}{2(1+x^{2})} \) for \( 0 \leq x \leq 1 \) and \( y=\frac{3}{2(1+x^{2})}-\frac{1}{2} \) for \( x>1 \).

Step 1: Understanding the Differential Equation

The given differential equation is \( y^{\prime} + \frac{2x}{1+x^{2}} y = \left\{\begin{array}{ll} \frac{x}{1+x^{2}}, & 0 \leq x \leq 1 \ \frac{-x}{1+x^{2}}, & x>1 \end{array}\right. \). This indicates that we have two different expressions for the equation depending on whether \( x \) is between 0 and 1 or greater than 1.

Step 2: Identifying the Integrating Factor

An integrating factor for the linear differential equation is given as \( 1+x^{2} \). Multiplying the entire differential equation by this factor will simplify it into a format that allows integration to solve for \( y \).

Step 3: Applying the Integrating Factor

Multiply the entire differential equation \( y^{\prime} + \frac{2x}{1+x^{2}} y \) by \( 1+x^{2} \):\[ (1+x^{2})y^{\prime} + 2xy = \left\{ \begin{array}{ll} x, & 0 \leq x \leq 1 \ -x, & x > 1 \end{array} \right. \].This transforms it into a perfect derivative: \( \frac{d}{dx}((1+x^{2})y) \).

Step 4: Integrating on Both Sides

Integrate both sides of the equation:\[ \int \frac{d}{dx}((1+x^{2})y) \, dx = \int \left\{ \begin{array}{ll} x, & 0 \leq x \leq 1 \ -x, & x > 1 \end{array} \right. \, dx \]For \( 0 \leq x \leq 1 \), we find: \( (1+x^{2})y = \frac{1}{2}x^{2} + c_{1} \).For \( x > 1 \), we find: \( (1+x^{2})y = -\frac{1}{2}x^{2} + c_{2} \).

Step 5: Applying Initial Condition and Continuity

We use the initial condition \( y(0) = 0 \) to find \( c_{1} \). For \( x = 0 \), the equation becomes \( 1 \cdot 0 = \frac{1}{2} \cdot 0^{2} + c_{1} = 0 \). Hence \( c_{1} = 0 \).To ensure continuity at \( x = 1 \), set \( c_{2} = 1 \) to make both parts connect smoothly.

Step 6: Writing the Final Solution

The solution to the differential equation is:\[ y = \left\{\begin{array}{ll} \frac{1}{2} - \frac{1}{2(1+x^{2})}, & 0 \leq x \leq 1 \ \frac{3}{2(1+x^{2})} - \frac{1}{2}, & x > 1 \end{array}\right. \]This captures both segments of the solution, ensuring the function's continuity across the specified regions.

Key Concepts

Differential Equation

A differential equation is a mathematical equation that relates a function to its derivatives. In the given problem, the differential equation is expressed as:

• For \(0 \leq x \leq 1 \): \ \(y^{\prime} + \frac{2x}{1+x^{2}} y = \frac{x}{1+x^{2}}\)
• For \(x > 1 \): \ \(y^{\prime} + \frac{2x}{1+x^{2}} y = \frac{-x}{1+x^{2}}\)

This indicates two separate expressions based on the range of \(x\). Such equations are vital in modeling real-world systems that change at rates dependent on their current states, like population growth or thermal dynamics.
The challenge in solving these equations is usually finding an unknown function \(y(x)\) that satisfies the relationship between its derivative \(y'\) and other functions of \(x\). In this exercise, the differential equation has piecewise components, meaning it has different behaviors on different intervals.

Initial Condition

An initial condition in a differential equation problem provides a specific value for the unknown function at a particular point, often helping to determine the constant of integration. In our case, the initial condition is given as \(y(0) = 0\). This condition means:

• At \(x = 0\), the value of \(y\) must exactly be zero.
• This initial condition is applied to solve for the constant \(c_1\) when integrating to find \(y(x)\).

By substituting this condition into the integrated equation derived from applying the integrating factor, we effortlessly calculate \(c_1\). Initial conditions like this are crucial because they allow us to identify the specific solution that matches a specific scenario or state. They effectively reduce the number of potential solutions by specifying the behavior of the system at a precise starting point.

Continuity of Solutions

Continuity of solutions ensures that the solution of a differential equation does not have sudden jumps or discontinuities at certain points of the domain. In this exercise, it's essential to check the continuity at \(x = 1\), where our piecewise expressions for \(y\) switch.

• The condition for continuity means that both parts of the function \((0 \leq x \leq 1\) and \(x > 1\)) should have the same value at \(x = 1\).
• This requires finding the right constant \(c_2\) to ensure smooth continuation from one part to the other.

For this problem, we set \(c_2 = 1\) so the segments join seamlessly, ensuring that the behavior of \(y\) is predictable and consistent across the entire range of \(x\). Continuity in differential equations is often key in ensuring a solution not only mathematically solves the equation but also realistically represents physical phenomena.