Problem 17 - Differential Equations and Linear Algebra

Question

Verify that the given function is a solution to the given differential equation \(\left(c_{1} \text { and } c_{2}\right.\) are arbitrary constants), and state the maximum interval over which the solution is valid. $$y(x)=c_{1} x^{1 / 2}+3 x^{2}, \quad 2 x^{2} y^{\prime \prime}-x y^{\prime}+y=9 x^{2}$$.

Step-by-Step Solution

Answer

The given function \(y(x) = c_1 x^{1/2} + 3x^2\) is a solution to the differential equation \(2x^2y'' - xy' + y = 9x^2\), and the maximum interval over which the solution is valid is \(x \ge 0\) or \([0, \infty)\).

Step 1: Calculate the first derivative of y(x)

Calculate the first derivative of the given function \(y(x) = c_1 x^{1/2} + 3x^2\): \(y'(x) = \frac{d}{dx}(c_1 x^{1/2} + 3x^2) = \frac{1}{2}c_1 x^{-1/2} + 6x\)

Step 2: Calculate the second derivative of y(x)

Now, we'll calculate the second derivative of our function \(y(x)\): \(y''(x) = \frac{d^2}{dx^2}(c_1 x^{1/2} + 3x^2) = -\frac{1}{4}c_1 x^{-3/2} + 6\)

Step 3: Substitute y(x), y'(x), and y''(x) into the differential equation

Substitute the original function \(y(x)\), first derivative \(y'(x)\), and second derivative \(y''(x)\) into the given differential equation and simplify: \(2x^2 (-\frac{1}{4}c_1 x^{-3/2} + 6) - x (\frac{1}{2}c_1 x^{-1/2} + 6x) + (c_1 x^{1/2} + 3x^2) = 9x^2\)

Step 4: Check if the substituted equation holds true

Simplify the equation and check if it holds true: \((-\frac{1}{2}c_1 x^{1/2} + 12x^2 - \frac{1}{2}c_1 x^{1/2} - 6x^2 + c_1 x^{1/2} + 3x^2) = 9x^2\) \((-c_1 x^{1/2} + 9x^2) = 9x^2\) Indeed, the given function is a solution to the given differential equation as both sides of the equation are equal.

Step 5: Determine the maximum interval over which the solution is valid

Since the given function \(y(x) = c_1 x^{1/2} + 3x^2\) and its derivatives are defined for all real numbers \(x\ge 0\), the maximum interval over which the solution is valid is \(x \ge 0\) or \([0, \infty)\).

Key Concepts

Second Derivative

The concept of a second derivative is a fundamental part of understanding the behavior of functions in calculus. In the context of differential equations, the second derivative represents the rate at which the function's rate of change is itself changing. Mathematically, if you have a function, say \(y(x)\), then the first derivative \(y'(x)\) measures the slope or rate of change of \(y\) with respect to \(x\). The second derivative \(y''(x)\) then measures how this slope changes as \(x\) itself varies.

For example, in our exercise, the function \(y(x) = c_1 x^{1/2} + 3x^2\) has its first derivative \(y'(x)\), which tells us how \(y\) is changing at any point. To gain deeper insight into the behavior of the function, such as concavity and points of inflection, we look to the second derivative \(y''(x)\). In this case, \(y''(x) = -\frac{1}{4}c_1 x^{-3/2} + 6\) is critical in verifying that \(y(x)\) adheres to the differential equation given.

The role of the second derivative is not limited to just this verification process. It can help explain the acceleration of an object in physics, the curvature of a graph, or the optimization of functions in economics by identifying maxima and minima. Therefore, understanding the concept of a second derivative can open doors to multiple areas of practical and theoretical applications.

Verification of Solutions

Verification of solutions is a necessary process in differential equations to confirm whether a proposed function is indeed a solution to a given differential equation. This involves checking if, when the function and its derivatives are substituted into the original equation, the equation holds true.

In the exercise, we are given the differential equation \(2 x^{2} y''-x y'+y = 9 x^{2}\) and a function \(y(x) = c_{1} x^{1 / 2}+3 x^{2}\) to verify. This requires calculating the first and second derivatives of \(y(x)\) and substituting them back into the differential equation. If after simplification, both sides of the equation are equal for all permitted values of \(x\), the function is a solution. In our case, after substituting and simplifying, we arrive at \((-c_1 x^{1/2} + 9x^2) = 9x^2\), indicating that the proposed function satisfies the differential equation.

Verification can also assist in understanding the function's behavior and predicting its future values, making it not just a mathematical exercise but a tool for practical problem-solving.

Maximum Interval of Validity

The maximum interval of validity is a critical consideration when working with solutions to differential equations. It refers to the range of the independent variable, usually denoted as \(x\), over which the solution is defined and satisfies the equation. The reason for identifying this interval is due to potential restrictions that can arise from the original function, its derivatives, or the equation itself.

For instance, in our exercise, we need to be cautious about where the function \(y(x) = c_1 x^{1/2} + 3x^2\) and its derivatives are defined. Because they involve a square root function and are defined for all real numbers \(x\textgreater0\), we state that the maximum interval of validity is \(x \textgreater 0\) or \([0, \textbackslash\text \textbackslashinfty)\). Outside of this interval, the function and its derivatives may not be real numbers, and the solution would not hold.

It's crucial to always consider this maximum interval when working with differential equations to ensure accurate predictions and understand the limitations of solutions in practical applications. It also helps avoid misinterpreting the results or applying the solutions outside the context where they are valid.