Problem 31 - Advanced Engineering Mathematics

Question

For \(y^{\prime}+2 y=f(x)\) an integrating factor is \(e^{2 x}\) so that $$y e^{2 x}=\left\\{\begin{array}{ll} \frac{1}{2} e^{2 x}+c_{1}, & 0 \leq x \leq 3 \\ c_{2}, & x>3 \end{array}\right.$$ If \(y(0)=0\) then \(c_{1}=-1 / 2\) and for continuity we must have \(c_{2}=\frac{1}{2} e^{6}-\frac{1}{2}\) so that $$y=\left\\{\begin{array}{ll} \frac{1}{2}\left(1-e^{-2 x}\right), & 0 \leq x \leq 3 \\ \frac{1}{2}\left(e^{6}-1\right) e^{-2 x}, & x>3 \end{array}\right.$$

Step-by-Step Solution

Answer

The piecewise solution is correct for given initial and continuity conditions.

Step 1: Verify the Integrating Factor

We start with the differential equation \( y' + 2y = f(x) \) and are given the integrating factor \( e^{2x} \). Multiply the entire differential equation by this integrating factor to transform it into an exact equation.

Step 2: Multiply the Differential Equation

Multiply both sides of the equation by \( e^{2x} \): \( e^{2x}y' + 2e^{2x}y = e^{2x}f(x) \). This is equivalent to taking the derivative of \( y e^{2x} \), suggesting:\( \frac{d}{dx}(ye^{2x}) = e^{2x}f(x) \).

Step 3: Integrate the Equation

Integrate both sides with respect to \( x \):\( \int \frac{d}{dx}(ye^{2x}) \) equals \( \int e^{2x}f(x) \, dx \), leading to:\( ye^{2x} = \int e^{2x}f(x) \, dx + C \).

Step 4: Apply Initial Condition

Using the initial condition \( y(0) = 0 \), substitute \( x = 0 \) into the solution:\( ye^{2x} = \frac{1}{2}e^{2x} + c_1 \) becomes \( 0 = \frac{1}{2}e^0 + c_1 \), therefore \( c_1 = -\frac{1}{2} \).

Step 5: Determine Continuity at \(x=3\)

Address the continuity of the solution at \( x = 3 \):\( ye^{2x} = c_2 \) is continuous at \( x = 3 \), matching the piece corresponding to \( 0 \leq x \leq 3 \), \( \frac{1}{2}e^{6} + c_1 = c_2 \). With \( c_1 = -\frac{1}{2} \), it simplifies to \( c_2 = \frac{1}{2}e^{6} - \frac{1}{2} \).

Step 6: Write the Piecewise Solution

Substitute \( c_1 \, and \, c_2 \) into the general solutions for each interval:- For \( 0 \leq x \leq 3 \), solve \( ye^{2x} = \frac{1}{2}e^{2x} - \frac{1}{2} \), which simplifies to \( y = \frac{1}{2}(1 - e^{-2x}) \).- For \( x > 3 \), solve \( ye^{2x} = c_2 = \frac{1}{2}e^{6} - \frac{1}{2} \), which simplifies to \( y = \frac{1}{2}(e^{6} - 1)e^{-2x} \).

Step 7: Conclude with Final Piecewise Solution

Combine the results: \[ y=\begin{cases} \frac{1}{2}(1 - e^{-2x}), & 0 \leq x \leq 3 \\frac{1}{2}(e^{6} - 1)e^{-2x}, & x > 3 \end{cases} \]This represents the final piecewise function based on the determined values of \( c_1 \) and \( c_2 \).

Key Concepts

Integrating Factor

When faced with a linear first-order differential equation like \( y' + py = f(x) \), the integrating factor technique is a helpful method to simplify the problem. An integrating factor transforms the original, non-exact equation into an exact one, making it easier to solve.

In this exercise, the integrating factor given is \( e^{2x} \). By multiplying the entire differential equation by this factor, the left side becomes the derivative of the product of the unknown function \( y \) and the integrating factor. This transformation allows us to rewrite the differential equation as \( \frac{d}{dx}(ye^{2x}) = e^{2x}f(x) \). This transformed equation is now straightforward to integrate.

This method is powerful because it leverages the fact that the integrating factor is always of the form \( e^{\int p(x) \, dx} \), where \( p(x) \) is a function derived from the original differential equation. Once the equation is integrated, constants are determined using initial conditions, finalizing the solution.

Piecewise Functions

Piecewise functions are essential when the behavior of a function changes at particular points, often due to physical constraints or conditions in real-world problems. In this exercise, the solution to the differential equation is expressed as a piecewise function.

The function \( y \) is defined in such a way that it behaves differently over two intervals: when \( 0 \leq x \leq 3 \) and when \( x > 3 \). This detailed separation allows the solution to appropriately reflect different conditions or changes in the system dynamics. Because systems can have specific thresholds or breakpoints, piecewise functions provide a neat and clear way to represent such changes.

The importance of properly defining a piecewise function in this context lies in ensuring its continuity and correctness within each specified interval, especially at the boundaries. By doing so, we assure the mathematical integrity and physical realism of the solution.

Initial Conditions

Initial conditions are crucial to solving differential equations, as they provide specific values at the outset that help determine the "constants of integration" in the solution. In our problem, the condition \( y(0) = 0 \) acts as a pivotal tool for finding the correct value of the constant \( c_1 \).

After integrating the differential equation, an arbitrary constant is included in the solution. To find its exact value, substituting the initial condition into the solution is necessary. For this exercise, solving \( ye^{2x} = \frac{1}{2}e^{2x} + c_1 \) when \( x = 0 \) leads to \( 0 = \frac{1}{2}e^0 + c_1 \), thus determining \( c_1 = -\frac{1}{2} \).

Moreover, the initial conditions, paired with continuity requirements at specific points such as \( x = 3 \), help resolve any remaining constants, such as \( c_2 \). This ensures the final solution not only satisfies the differential equation but also meets the precise conditions at all given points, making the solution uniquely determined.