Problem 30 - Calculus Early Transcendentals

Question

(a) Prove that solutions need not be unique for nonlinear initial-value problems by finding two solutions to $$ y \frac{d y}{d x}=x, \quad y(0)=0 $$ (b) Prove that solutions need not exist for nonlinear initial-value problems by showing that there is no solution for $$ y \frac{d y}{d x}=-x, \quad y(0)=0 $$

Step-by-Step Solution

Answer

(a) Solutions are \( y = x \) and \( y = -x \). (b) No solution exists.

Step 1: Rewrite the Differential Equation (a)

The given differential equation is \( y \frac{dy}{dx} = x \). We can rewrite it using separation of variables as \( y \, dy = x \, dx \).

Step 2: Integrate Both Sides (a)

Integrate both sides of the equation. The left side becomes \( \int y \, dy \), and the right side becomes \( \int x \, dx \). This results in \( \frac{y^2}{2} = \frac{x^2}{2} + C \).

Step 3: Solve for \( y \) by Applying Initial Conditions (a)

Substitute the initial condition \( y(0) = 0 \) into the integrated equation to determine the constant. Substituting gives \( \frac{0^2}{2} = \frac{0^2}{2} + C \), thus \( C = 0 \). The equation simplifies to \( y^2 = x^2 \).

Step 4: Find Two Possible Solutions (a)

Since \( y^2 = x^2 \), the solutions are \( y = x \) and \( y = -x \). Both are valid solutions that satisfy the initial condition \( y(0) = 0 \).

Step 5: Rewrite the Differential Equation (b)

The second differential equation is \( y \frac{dy}{dx} = -x \). Use separation of variables: \( y \, dy = -x \, dx \).

Step 6: Integrate Both Sides (b)

Integrate both sides of the equation. The left side becomes \( \int y \, dy \) and the right side becomes \( \int -x \, dx \). This yields \( \frac{y^2}{2} = -\frac{x^2}{2} + C \).

Step 7: Analyze the Integrated Equation (b)

The equation \( \frac{y^2}{2} = -\frac{x^2}{2} + C \) suggests that for any real \( x \), \( y^2 = -x^2 + 2C \). To satisfy the initial condition \( y(0) = 0 \), substitute \( x = 0 \), giving \( 0 = -0 + 2C \), so \( C = 0 \). Thus, \( y^2 = -x^2 \).

Step 8: Prove Nonexistence of Solution (b)

The equation \( y^2 = -x^2 \) implies \( y^2 \) is non-negative, but \(-x^2\) is always non-positive except when \( x = 0 \). Thus, these values cannot satisfy \( y^2 = -x^2 \) for non-zero \( x \), and no solution exists for \( x eq 0 \).

Key Concepts

Existence of Solution

The question of whether a solution exists for a nonlinear initial-value problem can be challenging. In mathematical terms, the existence of a solution is not always guaranteed. This is especially true for nonlinear differential equations, which can exhibit complex behaviors. Let's consider the function:\[y \frac{d y}{d x} = -x, \quad y(0) = 0\]At first glance, it may seem reasonable to think there should be a solution, but when we apply separation of variables and integrate both sides, we get:\[\frac{y^2}{2} = -\frac{x^2}{2} + C\]Substitute the initial condition \(y(0) = 0\), and you'll find:\[0 = -0 + 2C\]This leads to \(C = 0\), so the equation becomes \(y^2 = -x^2\). Here, things start to unravel, because \(y^2\) can never be negative for real numbers—a contradiction. Hence, no real solution exists when \(x eq 0\). By exploring these mathematics, we learn that not every equation with an initial condition has a corresponding solution, highlighting the complex nature of nonlinear problems.

Uniqueness of Solutions

In mathematics, uniqueness means there is only one solution to a problem under the given conditions. However, with nonlinear initial-value problems, uniqueness is not a given. A perfect example is: \(y \frac{d y}{d x} = x\) with \(y(0) = 0\).Through separation of variables and integration, we derive:\[\frac{y^2}{2} = \frac{x^2}{2} + C\]Substituting the initial condition \(y(0) = 0\), and solving for the constant, we find \(C = 0\). Therefore, the equation simplifies to \(y^2 = x^2\). This means:

• \(y = x\)
• \(y = -x\)

Both expressions satisfy the differential equation and the initial condition \(y(0) = 0\). This gives us two different solutions to the same problem, demonstrating that nonlinear initial-value problems can have multiple solutions. Thus, without additional constraints, uniqueness is not guaranteed.

Integration of Differential Equations

Integration is a fundamental process in solving differential equations, especially important in assessing initial-value problems. Integration helps us to find functions given their rate of change. Let's discuss its application to our given problems.For the differential equation:\[y \frac{d y}{d x} = x\]First, we perform separation of variables, setting up the integral of both sides:\[\int y \, dy = \int x \, dx\]Integrating gives us:\[\frac{y^2}{2} = \frac{x^2}{2} + C\]Through integration, we transform a differential equation into an algebraic expression. Using the initial condition \(y(0) = 0\), we can determine the constant of integration, which simplifies the equation further.However, the process and resulting functions can sometimes be unstable. As seen in the problem, once integrated, solutions can branch into more than one path or sometimes yield contradictions, as in the second problem \(y \frac{d y}{d x} = -x\). Recognizing these nuances is vital in understanding the behavior of solutions to differential equations.