Problem 33 - Advanced Engineering Mathematics

Question

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

Step-by-Step Solution

Answer

The solution is piecewise: for \(0 \leq x \leq 1\), \(y = \frac{1}{2} + \frac{3}{2}e^{-x^2}\); for \(x > 1\), \(y = \left(\frac{1}{2}e + \frac{3}{2}\right)e^{-x^2}\).

Step 1: Identify the given ODE

The given differential equation is \( y' + 2xy = f(x) \). An integrating factor \( e^{x^2} \) is provided, but the equation does not explicitly specify \( f(x) \). Thus, we're solving for the solution using the integrating factor.

Step 2: Apply the integrating factor

Multiply the entire differential equation \( y' + 2xy = f(x) \) by the integrating factor \( e^{x^2} \). The equation becomes \( e^{x^2}y' + 2xe^{x^2}y = e^{x^2}f(x) \). This transforms into the derivative of the product: \( \frac{d}{dx}(ye^{x^2}) = e^{x^2}f(x) \).

Step 3: Determine the general solution

Integrating both sides with respect to \( x \), we have \( \int \frac{d}{dx}(ye^{x^2}) \, dx = \int e^{x^2}f(x) \, dx \), leading to \( ye^{x^2} = \int e^{x^2}f(x) \, dx + C \), where \( C \) is an arbitrary constant.

Step 4: Solve for parts with given solutions

The piecewise solution states \( ye^{x^2} = \begin{cases} \frac{1}{2}e^{x^2} + c_1, & 0 \le x \le 1 \ c_2, & x > 1 \end{cases} \). So, for \( 0 \le x \le 1 \), \( y = \frac{1}{2}e^{-x^2} + c_1e^{-x^2} \). For \( x > 1 \), \( y = c_2e^{-x^2} \).

Step 5: Use initial condition to find \( c_1 \)

We know that \( y(0) = 2 \). At \( x = 0 \), substituting into the equation gives \( 2 = \frac{1}{2} + c_1 \), leading to \( c_1 = \frac{3}{2} \).

Step 6: Ensure continuity at \( x = 1 \) to find \( c_2 \)

For continuity, set the expression for \( y \, e^{x^2} \) at \( x = 1 \) from both piecewise solutions equal. From \( 0 \le x \le 1 \), \( y = \frac{1}{2} + \frac{3}{2}e^{-1} \), and this must equal \( c_2 \) for \( x > 1 \). Therefore, \( c_2 = \frac{1}{2}e + \frac{3}{2} \).

Step 7: Write the complete solution

The final solution, maintaining continuity and considering piecewise conditions is: \[ y = \begin{cases} \frac{1}{2} + \frac{3}{2} e^{-x^2}, & 0 \leq x \leq 1 \ \left(\frac{1}{2} e + \frac{3}{2}\right) e^{-x^2}, & x > 1 \end{cases} \]

Key Concepts

Integrating Factor

In solving certain types of ordinary differential equations (ODEs), an integrating factor is a crucial tool. It's a specially chosen function that, when multiplied by the original equation, transforms it into something simpler. Essentially, it helps us to convert a non-exact equation into an exact one. This means that after applying the integrating factor, the left-hand side of the equation can be written as the derivative of a product.

For instance, consider the differential equation given in the exercise: \( y' + 2xy = f(x) \). The integrating factor provided is \( e^{x^2} \). When you multiply every term in the equation by this factor, the equation becomes:

• \( e^{x^2}y' + 2xe^{x^2}y = e^{x^2}f(x) \).

This equation now represents the derivative of a product:

• \( \frac{d}{dx}(ye^{x^2}) \).

The beauty of this transformation is that it simplifies the process of solving the differential equation, essentially allowing you to integrate both sides to find \( y \). The integrating factor method is especially useful in linear first-order differential equations, as demonstrated in this solution.

Piecewise Function

A piecewise function is a type of function that is defined by different expressions according to different intervals of the input variable. In this exercise, the solution for the differential equation is given as a piecewise function. This means the solution is different depending on the value of \( x \).

• For \( 0 \leq x \leq 1 \), the solution is \( ye^{x^2} = \frac{1}{2} e^{x^2} + c_1 \).
• For \( x > 1 \), the solution is \( ye^{x^2} = c_2 \).

The reason for having a piecewise solution often stems from discontinuities or conditions that change at certain points. It's essential to understand that each part of a piecewise function is independent in its specified range, but seamless transitions between these ranges are sometimes necessary to satisfy conditions like continuity.

Initial Condition

Initial conditions are essential in solving differential equations as they provide specific values that allow us to find unique solutions. These conditions are boundary values for the variable's function at a certain point, typically given as a value of \( y \) (or other dependent variable) at a particular \( x \). In our exercise, we've been provided with an initial condition \( y(0) = 2 \).

By substituting \( x = 0 \) and \( y = 2 \) into the piece of the piecewise function valid for \( 0 \leq x \leq 1 \), we can solve for the constant \( c_1 \). This specific initial condition translates the general solution to a particular one, making it unique and applicable to the problem we face. Initial conditions, as given, restrict the family of potential solutions down to one distinct path.

Continuity of Solutions

The concept of continuity in solutions involves ensuring that the solutions neither contain jumps nor disconnects at any point along their domain. For piecewise functions, this means that where different formulae define the solution across intervals, their outputs match at the thresholds.

In the exercise, ensuring continuity at \( x = 1 \) was critical. It necessitated setting the outcome of the function from the first part \( 0 \leq x \leq 1 \) equal to that of the part \( x > 1 \). This required that:

• \( \frac{1}{2} + \frac{3}{2}e^{-1} = c_2 \),

therefore finding \( c_2 \). Continuity is a crucial condition because many physical phenomena modeled by differential equations assume smooth changes without abrupt jumps. Ensuring solutions are continuous aligns the mathematical model closely with realistic scenarios.