Problem 39 - A First Course in Differential Equations with Modeling Applications

Question

(a) Verify that \(y_{1}=x^{3}\) and \(y_{2}=|x|^{3}\) are linearly independent solutions of the differential equation \(x^{2} y^{\prime \prime}-4 x y^{\prime}+6 y=0\) on the interval \((-\infty, \infty)\). (b) For the functions \(y_{1}\) and \(y_{2}\) in part (a), show that \(W\left(y_{1}, y_{2}\right)=0\) for every real number \(x .\) Does this result violate Theorem 4.1.3? Explain. (c) Verify that \(Y_{1}=x^{3}\) and \(Y_{2}=x^{2}\) are also linearly independent solutions of the differential equation in part (a) on the interval \((-\infty, \infty)\). (d) Besides the functions \(y_{1}, y_{2}, Y_{1},\) and \(Y_{2}\) in parts (a) and (c), find a solution of the differential equation that satisfies \\[y(0)=0, y^{\prime}(0)=0\\] (e) By the superposition principle, Theorem \(4.1 .2,\) both linear combinations \(y=c_{1} y_{1}+c_{2} y_{2}\) and \(Y=c_{1} Y_{1}+c_{2} Y_{2}\) are solutions of the differential equation. Discuss whether one, both, or neither of the linear combinations is a general solution of the differential equation on the interval \((-\infty, \infty)\).

Step-by-Step Solution

Answer

The general solution is given by \(Y = c_1 x^3 + c_2 x^2\) on \((-\infty, \infty)\).

Step 1: Verify Linear Independence of Solutions from Part (a)

To verify linear independence, we check if a non-trivial combination \(c_1y_1 + c_2y_2 = 0\) implies \(c_1 = c_2 = 0\). Since \(y_1 = x^3\) and \(y_2 = |x|^3\) have different behaviors for negative \(x\) (\(x^3\) is negative, \(|x|^3\) is positive), these functions are linearly independent on each side of zero. Thus, they are linearly independent on \((-\infty, \infty)\).

Step 2: Show Wronskian for y1 and y2 is Zero

The Wronskian \(W(y_1, y_2)\) is computed as \[ W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \ y_1' & y_2' \end{vmatrix} = \begin{vmatrix} x^3 & |x|^3 \ 3x^2 & 3|x|^2 \text{sgn}(x)\end{vmatrix}\]This becomes zero because the second row is a scalar multiple of the first row for any \(x\), hence the Wronskian is zero for every real \(x\).

Step 3: Discuss Theorem 4.1.3 Violation

Theorem 4.1.3 states that linear independence is associated with a non-zero Wronskian on an interval. Since \(W(y_1, y_2)=0\) but they are linearly independent as shown in Step 1, this result suggests an exception due to the absolute value function \(|x|\) causing a point where derivatives are not analytically well-defined. Thus, the scenario doesn't completely violate the theorem as the functions behave differently across intervals.

Step 4: Verify Linear Independence for Part (c)

For \(Y_1 = x^3\) and \(Y_2 = x^2\), assume a linear combination \(c_1Y_1 + c_2Y_2 = 0\). This implies a polynomial \(c_1x^3 + c_2x^2 = 0\) which suggests \(c_1 = 0\) and \(c_2 = 0\) for all \(x\) since the polynomial must hold for any \(x\). Thus, \(Y_1\) and \(Y_2\) are linearly independent.

Step 5: Find a Solution for Given Conditions in Part (d)

The differential equation \(x^2 y'' - 4xy' + 6y = 0\) is homogeneous. Thus, we seek particular solutions combining \(Y_1\) and \(Y_2\) terms. For conditions \(y(0)=0\) and \(y'(0)=0\), solution can be attained using a different method since the standard basis fails at \(x=0\). Consider a generalized solution using a function like a polynomial satisfying initial conditions such as \(y = 0\) at these points, which could involve integral techniques or higher order terms.

Step 6: Discuss General Solution Using Superposition in Part (e)

Linear combinations given are \(y = c_1y_1 + c_2y_2\) and \(Y = c_1Y_1 + c_2Y_2\). Both satisfy the differential equation. However, \(y_1\) and \(y_2\) individually do not cover zero since they are indifferent at \(x=0\), thus, \(Y_1\) and \(Y_2\) cover this gap as they are expressed in standard polynomial terms. Therefore, using \(Y = c_1Y_1 + c_2Y_2\) can represent a general solution as it is more comprehensive over the interval \((-\infty, \infty)\).

Key Concepts

Linear Independence

When discussing linear independence in the context of differential equations, we refer to a set of functions that, when linearly combined, yield zero only if all the coefficient constants are zero. For example, if we have functions \(y_1 = x^3\) and \(y_2 = |x|^3\), they are considered linearly independent if the equation \(c_1 y_1 + c_2 y_2 = 0\) implies that \(c_1 = c_2 = 0\).

Linear independence is crucial because it ensures that the solutions to a differential equation cover the entire solution space available. This is important for forming the general solution to the differential equation. In the case of \(y_1=x^3\) and \(y_2=|x|^3\), despite their differing behaviors when \(x\) is less than zero, they are demonstrated to be linearly independent over the entire domain \((-\infty, \infty)\). This is because their behaviours are distinct enough, especially around zero, which allows them to remain linearly independent. However, note that linear independence doesn't directly imply that their Wronskian (a determinant used for checking independence) will be non-zero, especially if an alteration, like an absolute value, is involved.

Wronskian

The Wronskian is a determinant associated with a set of functions and their derivatives. It is used to determine linear independence of solutions to differential equations. For two functions \(y_1\) and \(y_2\), the Wronskian \(W(y_1, y_2)\) is given by:

• \[ W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \ y_1' & y_2' \end{vmatrix}\]

In the exercise, for the functions \(y_1 = x^3\) and \(y_2 = |x|^3\), their Wronskian matrix is:

• \[\begin{vmatrix} x^3 & |x|^3 \ 3x^2 & 3|x|^2 \text{sgn}(x)\end{vmatrix}\]
• Since the rows are scalar multiples, \(W(y_1, y_2) = 0\) for all \(x\)

A zero Wronskian might suggest linear dependence, but in this case, it arises from the non-analytic nature of \(|x|\) at zero. It's a unique scenario reflecting that caution is needed when the functions include non-analytic points.

Thus, the usual interpretation of the Wronskian doesn't hold because of the behavior of \(|x|\), and it does not fully violate the mathematical theorems governing these conditions due to this anomaly.

Superposition Principle

The Superposition Principle is foundational in solving linear differential equations. It states that if \(y_1\) and \(y_2\) are solutions of a linear homogeneous differential equation, any linear combination \(y = c_1 y_1 + c_2 y_2\) is also a solution. This principle is critical because it allows us to build the general solution from particular solutions.

Using this principle, both \(y = c_1y_1 + c_2y_2\) and \(Y = c_1Y_1 + c_2Y_2\) can solve the given differential equation. The value of this principle lies in creating solutions that can adopt coefficients reflecting initial or boundary conditions. It is particularly useful in cases where more than one solution exists for initial value problems.

In our exercise, while the Superposition Principle confirms that linear combinations of solutions satisfy the differential equation, particular attention is needed on when these combinations form a complete general solution. While \(y_1\) and \(y_2\) are solutions over most intervals, at \(x=0\) their mixed behavior implies that \(Y_1\) and \(Y_2\) are needed for a fully encompassing general solution over \((-\infty, \infty)\).

Homogeneous Differential Equation

A homogeneous differential equation has terms all dependent on the function or its derivatives, with no stand-alone constant term. Such an equation can typically be written in the form \(L[y] = 0\), where \(L\) is a linear differential operator.

In this exercise, the differential equation \(x^2 y'' - 4xy' + 6y = 0\) is homogeneous. This means any linear combination of its solutions produces another solution: an essential insight from the Superposition Principle. Since these equations are homogenized, their solutions often form a basis for vector spaces. Solve it by finding basis solutions and using linear combinations.

Homogeneous equations simplify the process of finding solutions, especially with initial conditions, as the equation structure naturally leads to solutions allowed by boundary conditions. Understanding the nature of the equation, as well as how to leverage properties like linear combination, is essential for applications in engineering and physics.