Problem 34 - A First Course in Differential Equations with Modeling Applications

Question

Verify that the given two-parameter family of functions is the general solution of the nonhomogeneous differential equation on the indicated interval. $$\begin{aligned}&2 x^{2} y^{\prime \prime}+5 x y^{\prime}+y=x^{2}-x\\\&y=c_{1} x^{-1 / 2}+c_{2} x^{-1}+\frac{1}{15} x^{2}-\frac{1}{6} x,(0, \infty)\end{aligned}$$

Step-by-Step Solution

Answer

The given function satisfies the differential equation as \( x^{2} - x \) is obtained after substitution and simplification.

Step 1: Differentiate the given solution

Start by differentiating the given function \( y = c_{1} x^{-1/2} + c_{2} x^{-1} + \frac{1}{15} x^{2} - \frac{1}{6} x \) to find \( y' \). Then differentiate \( y' \) to find \( y'' \). This will be used to substitute into the differential equation.

Step 2: First Derivative Calculation

Calculate the first derivative \( y' \):\[\begin{aligned}\&y' = \frac{d}{dx}(c_{1} x^{-1/2} + c_{2} x^{-1} + \frac{1}{15} x^{2} - \frac{1}{6} x)\&= -\frac{1}{2} c_{1} x^{-3/2} - c_{2} x^{-2} + \frac{2}{15} x - \frac{1}{6}.\end{aligned}\]

Step 3: Second Derivative Calculation

Now calculate the second derivative \( y'' \):\[\begin{aligned}\&y'' = \frac{d}{dx}(-\frac{1}{2} c_{1} x^{-3/2} - c_{2} x^{-2} + \frac{2}{15} x - \frac{1}{6})\&= \frac{3}{4} c_{1} x^{-5/2} + 2 c_{2} x^{-3} + \frac{2}{15}.\end{aligned}\]

Step 4: Substitute into Differential Equation

Substitute \( y \), \( y' \), and \( y'' \) into the differential equation \( 2x^{2} y'' + 5xy' + y = x^{2} - x \).

Step 5: Simplifying the Substitution

Plug the derivatives into the equation: \[2x^{2}\left(\frac{3}{4} c_{1} x^{-5/2} + 2 c_{2} x^{-3} + \frac{2}{15}\right) + 5x\left(-\frac{1}{2} c_{1} x^{-3/2} - c_{2} x^{-2} + \frac{2}{15} x - \frac{1}{6}\right) + c_{1} x^{-1/2} + c_{2} x^{-1} + \frac{1}{15} x^{2} - \frac{1}{6} x \]

Step 6: Verify the Identity

Combine and simplify the terms to verify the identity. Ensure that all terms involving \( c_{1} \) and \( c_{2} \) cancel out, leaving only \( x^{2} - x \) on the right, confirming that the function satisfies the differential equation.

Key Concepts

General Solution

The general solution of a differential equation is a formula that includes all possible solutions. It is key when dealing with differential equations, as it provides a comprehensive answer that considers all initial conditions. The general solution often involves arbitrary constants, which can be determined if specific initial conditions are given. In our case, the solution involves constants \(c_1\) and \(c_2\). These constants imply that the solution can vary based on different initial conditions or boundary values. The given general solution for the exercise is a two-parameter family of functions expressed as \( y = c_{1} x^{-1/2} + c_{2} x^{-1} + \frac{1}{15} x^{2} - \frac{1}{6} x \).
This solution captures all possible ways the function can behave according to the differential equation across the interval \((0, \infty)\). When a solution satisfies a differential equation for all values in a given interval, it confirms that it can describe every potential "shape" or behavior prescribed by the equation across that interval.

Derivatives

Derivatives represent the rate of change of a function concerning its variables. They are fundamental to understanding how functions behave and are indispensable in the process of solving differential equations.
To determine if a given function is a solution to a differential equation, we perform differentiation. In this exercise, we first found the first derivative \( y' \) of the given function. This involves differentiating the function term by term. The obtained first derivative is:

• \( y' = -\frac{1}{2} c_{1} x^{-3/2} - c_{2} x^{-2} + \frac{2}{15} x - \frac{1}{6} \)

Next, we calculate the second derivative \( y'' \), which helps in checking the solution against the given differential equation:

• \( y'' = \frac{3}{4} c_{1} x^{-5/2} + 2 c_{2} x^{-3} + \frac{2}{15} \)

These derivatives are plugged into the original differential equation to verify the solution.

Nonhomogeneous Differential Equation

A nonhomogeneous differential equation contains terms other than derivatives of the unknown function, meaning it does not equal zero on one side after simplification. It typically has the form: \[ P(x)y'' + Q(x)y' + R(x)y = g(x) \]where \( g(x) eq 0\).
In our case, the given differential equation is \( 2x^2 y'' + 5xy' + y = x^2 - x \), where \( g(x) = x^2 - x \) makes it nonhomogeneous.
Solving nonhomogeneous differential equations generally involves finding a complementary function (solution to the homogeneous part) and a particular solution (specific to \( g(x) \)). In our exercise, the given general solution includes both components, combined to satisfy the whole equation, ensuring each individual solution path is covered.

Two-Parameter Family of Functions

A two-parameter family of functions offers a broad range of potential solutions to a differential equation, represented by including two arbitrary constants. These parameters allow for flexibility, making it possible to adapt the solution to specific initial or boundary conditions you might face in real-world scenarios or specific tasks.
In our exercise, the function \( y = c_{1} x^{-1/2} + c_{2} x^{-1} + \frac{1}{15} x^{2} - \frac{1}{6} x \) includes parameters \( c_1 \) and \( c_2 \). These parameters essentially give us the freedom to shape our solution based on additional conditions that might be provided later on. When the solution verified by differentiating and substituting back into the equation still holds true, it confirms the robustness and utility of the two-parameter solution across numerous conditions.