Problem 46 - Differential Equations and Linear Algebra

Question

Verify that the given function is a solution to the given differential equation. In these problems, \(c_{1}\) and \(c_{2}\) are arbitrary constants. $$\diamond y(x)=c_{1} x^{4}+c_{2} x^{-2}, x^{2} y^{\prime \prime}-x y^{\prime}-8 y=0, x>0$$.

Step-by-Step Solution

Answer

To verify the given function \(y(x) = c_{1}x^4 + c_{2}x^{-2}\) is a solution to the differential equation \(x^2y'' - xy' - 8y = 0\), we compute the first derivative as \(y'(x) = 4c_{1}x^3 - 2c_{2}x^{-3}\) and the second derivative as \(y''(x) = 12c_{1}x^2 + 6c_{2}x^{-4}\). Plugging these into the differential equation and simplifying, we obtain that the equation holds true, concluding that \(y(x)\) is a solution to the given differential equation.

Step 1: Compute the first derivative of \(y(x)\)

Using the power rule, the first derivative of the given function \(y(x) = c_{1} x^{4} + c_{2} x^{-2}\) is \[y'(x) = \frac{d}{dx}(c_{1}x^{4} + c_{2}x^{-2}) = 4c_{1}x^3 - 2c_{2}x^{-3}.\]

Step 2: Compute the second derivative of \(y(x)\)

Again using the power rule, we find the second derivative of \(y(x)\) as \[y''(x) = \frac{d^2}{dx^2}(c_{1}x^{4} + c_{2}x^{-2}) = \frac{d}{dx}(4c_{1}x^{3} - 2c_{2}x^{-3}) = 12c_{1}x^2 + 6c_{2}x^{-4}.\]

Step 3: Plug the function and its derivatives into the differential equation

Now, plug \(y(x)\), \(y'(x)\) and \(y''(x)\) into the given differential equation \(x^2y'' - xy' - 8y = 0\): \[x^2(12c_{1}x^2 + 6c_{2}x^{-4}) - x(4c_{1}x^3 - 2c_{2}x^{-3}) - 8(c_{1}x^4 + c_{2}x^{-2}) = 0.\]

Step 4: Simplify and check if the equation holds true

Multiplying through and simplifying, \begin{align*} &12c_{1}x^4 + 6c_{2}x^{-2} - 4c_{1}x^4 + 2c_{2}x^{-2} - 8c_{1}x^4 - 8c_{2}x^{-2} \\ =& (12c_{1}-4c_{1}-8c_{1})x^4 + (6c_{2} + 2c_{2} - 8c_{2})x^{-2} \\ =& 0x^4 + 0x^{-2} \\ =& 0. \end{align*} Since the given differential equation holds true for the given function \(y(x) = c_{1}x^4 + c_{2}x^{-2}\) and its derivatives, we can conclude that \(y(x)\) is indeed a solution to the given differential equation.

Key Concepts

Verifying Solutions of Differential Equations

When faced with a differential equation, one important skill is verifying whether a given function is a solution to it. To verify a solution, follow these steps:

• Find the derivatives of the proposed solution as required by the equation.
• Substitute the original function and its derivatives into the equation.
• Simplify the resulting expression to check whether it equals zero or the non-homogeneous part of the equation.

This method works because, if the expressions cancel out correctly, it means that the function behaves exactly as the differential equation describes throughout its domain. Our exercise required the verification of the function with arbitrary constants, which adds a general layer to the solution, representing a family of functions rather than a single function.

Power Rule for Differentiation

The power rule is an essential tool in calculus for finding the derivative of a function of the form \( f(x) = ax^n \), where \( a \) is a constant and \( n \) is a real number. According to the power rule, the derivative of \( f(x) \) with respect to \( x \) is \( f'(x) = anx^{n-1} \).

• Applying the power rule simplifies differentiation, as it provides a direct method to calculate the derivative without the limit definition.
• This rule is valuable in solving differential equations where powers of \( x \) are involved, as it helps to find derivatives quickly and accurately.

In our exercise, we applied the power rule for both positive and negative powers of \( x \), assisting in finding the first and second derivatives which were then used to verify the solution.

Second Derivative

Moving beyond the first derivative, the second derivative of a function, denoted as \( f''(x) \) or \( \frac{d^2}{dx^2}(f(x)) \), provides information about the curvature or concavity of the function's graph.

• The second derivative is simply the derivative of the first derivative.
• It tells us where a function is concave up or down and locates points of inflection where the concavity changes.

In differential equations, the second derivative appears often, particularly in second-order equations. To verify a solution, we must correctly compute the second derivative, as seen in the exercise, where we used the power rule to find \( y''(x) \) before substitifying into the differential equation.

Arbitrary Constants in Differential Equations

Differential equations often have solutions involving arbitrary constants, such as \( c_{1} \) or \( c_{2} \). These constants are featured when the solution to a differential equation includes an antiderivative.

• Arbitrary constants represent an infinite number of possible solutions, characterizing a family of curves.
• In initial value problems, these constants are determined by specific conditions provided in the problem.
• The general solution includes these constants and becomes particular once the initial conditions are applied.

As seen in our exercise, we verified a solution containing arbitrary constants \( c_{1} \) and \( c_{2} \). Because the equation held true irrespective of the values of these constants, it showed that the general solution was correctly formulated.