Problem 5 - A First Course in Differential Equations with Modeling Applications

Question

Given that \(y=c_{1}+c_{2} x^{2}\) is a two-parameter family of solutions of \(x y^{\prime \prime}-y^{\prime}=0\) on the interval \((-\infty, \infty),\) show that constants \(c_{1}\) and \(c_{2}\) cannot be found so that a member of the family satisfies the initial conditions \(y(0)=0, y^{\prime}(0)=1 .\) Explain why this does not violate Theorem 4.1 .1.

Step-by-Step Solution

Answer

Constants \(c_1\) and \(c_2\) can't satisfy the initial conditions, not violating the theorem as no solution satisfies both conditions.

Step 1: Derive the First Derivative

Start by differentiating the function \( y = c_1 + c_2 x^2 \) with respect to \( x \). The derivative is \( y' = 2c_2 x \).

Step 2: Derive the Second Derivative

Differentiate \( y' = 2c_2 x \) with respect to \( x \) to get the second derivative: \( y'' = 2c_2 \).

Step 3: Substitute into Differential Equation

Substitute \( y \), \( y' \), and \( y'' \) into the differential equation \( x y'' - y' = 0 \). This yields \( x(2c_2) - (2c_2 x) = 0 \), which simplifies to \( 0 = 0 \). This means \( y = c_1 + c_2 x^2 \) indeed satisfies the differential equation.

Step 4: Apply Initial Condition for \( y \)

Apply the initial condition \( y(0) = 0 \). Substitute \( x = 0 \) into the function: \( c_1 + c_2(0)^2 = c_1 = 0 \). So, \( c_1 = 0 \).

Step 5: Apply Initial Condition for \( y' \)

Apply the initial condition \( y'(0) = 1 \). Substitute \( x = 0 \) into \( y' = 2c_2 x \), giving \( 2c_2(0) = 0 \). But \( y'(0) = 1 \), which cannot happen since 0 cannot equal 1. Thus, no solution satisfies both initial conditions.

Step 6: Explain Theorem 4.1.1

Theorem 4.1.1 typically states that there exists a unique solution for given initial conditions if the differential equation is linear and its coefficients satisfy continuity conditions. Our differential equation is satisfied by the family of solutions, but the initial conditions cannot be met simultaneously with this family. Thus, it does not violate the theorem since no solution from the family satisfies both conditions.

Key Concepts

Initial Value Problem

In mathematics, an Initial Value Problem (IVP) involves finding a function that satisfies a differential equation and meets specific initial conditions. These conditions typically define the value of the function, and potentially its derivatives, at a particular point. For instance, in our exercise, the search is for a solution to the differential equation with the following initial values:

• The value of the function at zero: \( y(0) = 0 \).
• The derivative of the function at zero: \( y'(0) = 1 \).

Solving an IVP involves integrating the differential equation and using the initial conditions to determine the constants in the solution. It is this process that distinguishes it from solving a general differential equation, offering a unique solution that fits the given conditions.

Second Order Differential Equation

Second order differential equations involve the second derivative of a function. Such equations are broader and include various forms, defining physical phenomena such as motion or waves. The equation in our exercise is:

• \( x y'' - y' = 0 \)

This equation relates the second derivative \( y'' \) and the first derivative \( y' \) of the function \( y \). The solution provided, \( y = c_{1} + c_{2} x^{2} \), is a two-parameter family. It means you can plug in different values for the constants \( c_{1} \) and \( c_{2} \) to obtain various solutions that satisfy the differential equation. Importantly, this relationship highlights the behavior and characteristics of the function by the way the derivatives connect with each other.

Theorem on Existence and Uniqueness

The Existence and Uniqueness Theorem gives us conditions under which we can expect to find a single, well-defined solution to a differential equation with given initial conditions. For a second order linear differential equation, the theorem requires:

• Continuity of the coefficients of the differential equation over an interval.
• Differentiability requirements are met, ensuring structural integrity of solutions.

In the provided exercise, although the equation \( x y'' - y' = 0 \) can be satisfied by a family of solutions, no combination of \( c_{1} \) and \( c_{2} \) fits both initial conditions (\( y(0) = 0 \) and \( y'(0) = 1 \)). This doesn't infringe on the theorem, as the theorem assures uniqueness only when it's possible to derive a solution; here, it was shown that no solution meets all conditions.

Solution Verification

Verifying a solution involves checking that a proposed function satisfies both the differential equation and the initial conditions. In this problem:

• First, derive the first and second derivatives from the solution \( y = c_1 + c_2 x^2 \).
• Insert these derivatives, and the function itself, into the differential equation.
• For it to be a solution, the equation must hold true for all values within the given domain.

For the initial conditions, substitute the values into the function and derivative to see if they meet \( y(0) = 0 \) and \( y'(0) = 1 \).Despite the function \( y = c_1 + c_2 x^2 \) verifying the differential equation, it failed to meet the initial conditions simultaneously. This process confirms whether a solution is viable, with the checks showing the conditions cannot simultaneously be met, hence no valid solution exists for the given conditions.