Problem 18 - 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. $$\begin{array}{l} y(x)=c_{1} x^{2}+c_{2} x^{3}-x^{2} \sin x. \\ x^{2} y^{\prime \prime}-4 x y^{\prime}+6 y=x^{4} \sin x. \end{array}$$

Step-by-Step Solution

Answer

The given function \(y(x) = c_1 x^2 + c_2 x^3 - x^2 \sin x\) is a solution to the differential equation \(x^2y''(x) - 4xy'(x) + 6y(x) = x^4 \sin x\). The solution is valid for the interval \((-\infty, \infty)\).

Step 1: Find the First Derivative of y(x)

Find the first derivative, \(y'(x)\): \(y'(x) = \frac{d}{dx} \left(c_1 x^2 + c_2 x^3 - x^2 \sin x\right)\) Using the power rule, chain rule and product rule, \(y'(x) = 2c_1 x + 3c_2 x^2 - 2x \sin x - x^2 \cos x\)

Step 2: Find the Second Derivative of y(x)

Now, find the second derivative, \(y''(x)\): \(y''(x) = \frac{d^2}{dx^2} \left(y'(x)\right)\) \(y''(x) = \frac{d}{dx} \left(2c_1 x + 3c_2 x^2 - 2x \sin x - x^2 \cos x\right)\) Using the power rule and product rule again, \(y''(x) = 2c_1 + 6c_2 x - 2 \sin x - 4x \cos x + 2x \cos x - x^2 \sin x\)

Step 3: Verify Given Function as a Solution to the Differential Equation

Now, substitute y(x), y'(x), and y''(x) into the given differential equation: \(x^2y''(x) - 4xy'(x) + 6y(x) = x^4 \sin x\) Plugging in the expressions for y(x), y'(x), and y''(x) and simplifying, \(x^2\left(2c_1 + 6c_2 x - 2 \sin x - 4x \cos x + 2x \cos x - x^2 \sin x\right) - 4x\left(2c_1 x + 3c_2 x^2 - 2x \sin x - x^2 \cos x\right) + 6\left(c_1 x^2 + c_2 x^3 - x^2 \sin x\right) = x^4 \sin x\) After simplification, we get \(x^4 \sin x = x^4 \sin x\) The equation stands true! Thus, given function y(x) is a solution to the given differential equation.

Step 4: Determine the Maximum Interval Over Which the Solution is Valid

Since the given function, y(x), is polynomial in nature and its only transcendental part is -x^2 * sin(x), the function will be valid for all x values. So, no maximum interval needs to be found in this case. Therefore, y(x) is a solution to the given differential equation for all x, and the solution is valid for the interval \((-\infty, \infty)\).

Key Concepts

First Derivative Calculation

Understanding how to calculate the first derivative is essential when dealing with differential equations. The first derivative, denoted as \(y'(x)\), represents the rate of change of a function at any given point. To find \(y'(x)\), we apply differentiation rules, such as the power rule, chain rule, and product rule.In our exercise, the function \(y(x) = c_{1} x^{2} + c_{2} x^{3} - x^{2} \text{sin} x\) is differentiated using these rules. The power rule tells us that the derivative of \(x^n\) is \(nx^{n-1}\). The chain rule is used for composite functions, and the product rule is applied when a function is a product of two other functions.

Second Derivative Calculation

Once the first derivative is found, the second derivative, denoted as \(y''(x)\), can be calculated to determine the curvature or the 'acceleration' of the original function. The second derivative helps describe the concavity of a graph and is crucial for identifying inflection points.In our given task, we further differentiate \(y'(x)\) to find \(y''(x)\). The process remains similar, using the power rule and product rule to perform the differentiation. It's key to perform each step carefully, ensuring each term is correctly derived to avoid any mistakes that could complicate the verification of the differential equation solutions.

Verification of Differential Equation Solutions

Verification is paramount to confirm that a proposed function is indeed a solution to a differential equation. To verify, we substitute the original function \(y(x)\), its first derivative \(y'(x)\), and its second derivative \(y''(x)\) into the differential equation and simplify. If the equation is satisfied—that is, both sides of the equation are equal—then the function is a valid solution.In this exercise, after substituting the calculated \(y(x)\), \(y'(x)\), and \(y''(x)\), we obtain \(x^{4} \text{sin} x = x^{4} \text{sin} x\), which confirms that the given function is indeed a correct solution to the differential equation provided.

Maximum Interval of Validity

The maximum interval of validity is the range of values over which the solution to a differential equation is defined and makes sense. Certain solutions may only be valid within a specific interval due to the presence of undefined expressions or physical constraints within the context of the problem.For polynomials or functions that are defined for all real numbers, as in the case of our exercise, the function and its derivatives do not impose any restrictions on the variable \(x\). Consequently, the solution \(y(x)\) is valid for all real numbers \(x\), and the maximum interval of validity is \((-\text{infinity}, \text{infinity})\). This means the equation will provide a meaningful solution for any value of \(x\) within the range of real numbers.