Problem 20 - Differential Equations and Linear Algebra

Question

Solve the given initial-value problem. \(y^{\prime}+y=f(x), \quad y(0)=3,\) where $$f(x)=\left\\{\begin{array}{ll} 1, & \text { if } x \leq 1 \\ 0, & \text { if } x > 1 \end{array}\right.$$

Step-by-Step Solution

Answer

The specific solution to the initial-value problem is: \[y(x)=\left\\{\begin{array}{ll} e^{-x}(e^x + 2), & \text { if } x \leq 1 \\\ 3e^{-x}, & \text { if } x > 1 \end{array}\right.\]

Step 1: Find the integrating factor

The integrating factor is given by the formula \(I(x)e^{\int P(x) dx}\), where P(x) is the coefficient of y in the given equation. In this case, P(x) = 1. Therefore, the integrating factor is: \[I(x)=e^{\int 1 dx} =e^x\]

Step 2: Multiply the given equation by the integrating factor

Multiply the given equation by the integrating factor, \(e^x\): \[e^x(dy/dx) + e^x y = e^x f(x)\]

Step 3: Integrate both sides

The left-hand side of the equation can be written as: \[\frac{d}{dx}(ye^x)\] Now integrate both sides of the equation with respect to x: \[\int\frac{d}{dx}(ye^x)dx = \int e^x f(x)dx\] Therefore, we have: \[ye^x = \int e^x f(x)dx + C\]

Step 4: Apply the given initial value to find the constant

Now using the initial condition, \(y(0) = 3\), we can find the constant C: \[3 = ye^0 = y(0)\] So, when x=0, y=3: \[3e^0 = 3\] From there, divide both sides by \(e^0\): \[y=3(\frac{e^x}{e^x})\] Now we have the equation: \[y=e^{-x}\left(\int e^x f(x)dx + 3\right)\]

Step 5: Determine the specific solution based on the given function

Substitute the given function f(x) into the equation: \[y=e^{-x}\left(\int e^x\left\\{\begin{array}{ll} 1, & \text { if } x \leq 1 \\\ 0, & \text { if } x > 1 \end{array}\right. dx + 3\right)\] Now, we have to separate the equation into two cases: Case 1: If \(x \leq 1\), then \(f(x) = 1\): \[y(x) = e^{-x}\left(\int e^x dx + 3\right) = e^{-x}(e^x + 3 - 1) = e^{-x}(e^x + 2)\] Case 2: If \(x > 1\), then \(f(x) = 0\): \[y(x) = e^{-x}\left(\int 0 dx + 3\right)=e^{-x}(C2)\] We need to find out what C2 is. To do this, we need to match the solutions at \(x=1\): \[3e^{-1} = e^{-1}(C2)\] Solve for C2: \[C2=3\] The specific solution is: \[y(x)=\left\\{\begin{array}{ll} e^{-x}(e^x + 2), & \text { if } x \leq 1 \\\ 3e^{-x}, & \text { if } x > 1 \end{array}\right.\]

Key Concepts

Integrating Factor

The integrating factor is a powerful tool used to solve first-order linear differential equations. It's particularly useful when you encounter an equation of the form \( y' + P(x)y = Q(x) \). The key to using an integrating factor lies in its ability to transform a non-exact equation into one that can easily be integrated.

To find the integrating factor, we use the formula \( I(x) = e^{\int P(x) dx} \), where \( P(x) \) is the coefficient of \( y \) in the equation. This formula allows us to multiply the entire differential equation by a carefully chosen expression that simplifies it, making it possible to express the left-hand side as the derivative of a product. In this way, the equation becomes \( \frac{d}{dx}(yI(x)) = I(x)Q(x) \).

In our specific example, \( P(x) = 1 \), so the integrating factor becomes \( e^x \). Multiplying the entire equation by \( e^x \) lets us rewrite the left-hand side as the derivative of \( ye^x \), thereby simplifying our integration process significantly. The role of the integrating factor is to facilitate finding a solution by making the equation integrable directly.

Initial-Value Problem

An initial-value problem is a type of differential equation that comes with extra information in the form of an initial condition. This additional piece of information specifies the value of the solution at a specific point, and helps us completely determine a unique solution.

When we say we have an initial-value problem, it usually means we are given a differential equation like \( y' + y = f(x) \) together with a condition such as \( y(0) = 3 \). This tells us that at \( x = 0 \), the value of \( y \) must be 3. The initial condition is critical because it helps us find the constant of integration \( C \) after integrating the differential equation.

• The differential equation itself provides general behavior.
• The initial condition tailors that general behavior to a specific instance.

By incorporating the initial condition into our solution process, we can solve for any unknown constants and determine the specific function that satisfies both the differential equation and the initial condition. In our example, the initial condition of \( y(0) = 3 \) was used to find the exact form of the constant in our solutions, leading to a unique function that fits the problem's requirements.

Piecewise Functions

Piecewise functions are interesting mathematical entities that are defined by different expressions depending on the value of the input variable. They are often used when the behavior of a system changes at specific values, requiring separate mathematical descriptions for different intervals.

In the context of differential equations, piecewise functions can be used to describe forcing functions like \( f(x) \) in our example: \[ f(x) = \begin{cases} 1, & \text{if } x \leq 1 \0, & \text{if } x > 1\end{cases} \]

This piecewise function indicates that \( f(x) \) engages active forcing with value 1 for \( x \leq 1 \) and ceases to influence \( y \) by turning to 0 when \( x > 1 \). Solving differential equations with piecewise functions requires addressing each segment separately, as each part can change the entire solution.

• Case analysis: evaluate the differential equation for each piece.
• Ensure continuity: match solutions at transition points (e.g., \( x = 1 \) in our problem).

This approach facilitates determining the overall solution by stitching together these separate pieces to form one cohesive function that behaves as required across its entire domain.