Problem 8 - Calculus 3

Question

For each of the following problems, verify that the given function is a solution to the differential equation. Use a graphing utility to graph the particular solutions for several values of \(c_{1}\) and \(c_{2} .\) What do the solutions have in common? $$ [\mathrm{T}] x^{2} y^{\prime \prime}-2 y-3 x^{2}+1=0 $$ $$ y(x)=c_{1} x^{2}+c_{2} x^{-1}+x^{2} \ln (x)+\frac{1}{2} $$

Step-by-Step Solution

Answer

The given function is a solution to the differential equation as the substituted terms cancel out to zero.

Step 1: Differentiate the given function

First, we differentiate the given function \( y(x) = c_{1} x^{2} + c_{2} x^{-1} + x^{2} \ln(x) + \frac{1}{2} \). Calculate the first derivative \( y' \) and the second derivative \( y'' \).1. Find \( y' \): \[ y' = 2c_1x - c_2x^{-2} + 2x\ln(x) + x \]2. Find \( y'' \): \[ y'' = 2c_1 + 2c_2x^{-3} + 2\ln(x) + 3 \]

Step 2: Substitute into the differential equation

Substitute \( y \), \( y' \), and \( y'' \) into the differential equation \( x^{2} y^{ ext{''}} - 2y - 3x^{2} + 1 = 0 \).

Step 3: Simplify the expression

Simplify the equation by substituting the expressions from Step 2 into the differential equation:1. Insert \( y'' \): \[ x^2(2c_1 + 2c_2x^{-3} + 2\ln(x) + 3) \] This becomes: \[ 2c_1 x^2 + 2c_2 + 2x^2 \ln(x) + 3x^2 \]2. Insert \( -2y \) using \( y \): \[ -2(c_{1} x^{2} + c_{2} x^{-1} + x^{2} \ln(x) + \frac{1}{2}) \] This simplifies to: \[ -2c_{1} x^{2} - 2c_{2} x^{-1} -2x^{2} \ln(x) - 1 \]3. The rest of the equation: \( -3x^2 + 1 \)Combine everything,\[ 2c_1 x^2 + 2c_2 + 2x^2 \ln(x) + 3x^2 - (2c_{1} x^{2} + 2c_{2} x^{-1} + 2x^{2} \ln(x) + 1) -3x^2 + 1 = 0 \]

Step 4: Check for identity and verify equality to zero

Compare terms:- Notice that terms with \( x^2 \ln(x) \), \( c_1x^2 \), \( c_2x^{-1} \), and constants will cancel out due to subtraction of corresponding terms.\( 0 = 0 \) satisfies the differential equation, showing that \( y(x) \) is indeed a solution.

Step 5: Analyze solutions' graph

Graph solutions using different values for \( c_1 \) and \( c_2 \). Use a graphing utility to plot:- Keep \( x^2\ln(x) \), observe how the combination with \( c_1 \) and \( c_2 \) as linear combinations shifts the graph.Check common features:- The influence of \( c_1 \) shifts the parabola vertically.- The influence of \( c_2 \) adjusts the behavior near the origin due to the term \( x^{-1} \).

Key Concepts

Particular Solutions

Particular solutions in differential equations are those where specific constants, such as \( c_1 \) and \( c_2 \) in our function, are assigned specific values. These values allow us to find unique solutions within a family of general solutions created by the equation. In our exercise, the general solution is given, with constants \( c_1 \) and \( c_2 \) holding the potential to form numerous particular solutions.
When you assign different values to these constants, you generate particular solutions that confirm the function resolves the original differential equation. Here, by varying \( c_1 \) and \( c_2 \), each particular solution reflects a different curve on the graph, representing how these constants influence the specific shape of the solution curve within the general form.

Second Order Differential Equations

Second order differential equations involve the second derivative of a function, indicating a relation that includes \( y'' \). These equations are common in modeling physical phenomena, such as motion, where acceleration (second derivative of position) is considered.
The provided differential equation involves \( x^2y'' - 2y - 3x^2 + 1 = 0 \), a second order equation because of the \( y'' \) term. Solving it often involves finding two independent solutions, usually expressed with arbitrary constants (like \( c_1 \) and \( c_2 \)), to construct the general solution. This is why our function contains these constants in its terms. Understanding the role of the second derivative can uncover information such as concavity and inflection points, essential for graph interpretation.

Graphing Solutions

Graphing solutions of a differential equation helps visualize how solutions behave under different conditions imposed by varying constants. In our exercise, graphing the function \( y(x) = c_1x^2 + c_2x^{-1} + x^2\ln(x) + \frac{1}{2} \) for different values of \( c_1 \) and \( c_2 \) illuminates how these constants adjust the solution's graph.

• \( c_1 \) affects the solutions' parabolic components, shifting the graph vertically.
• \( c_2 \) affects the graph's behavior near the origin due to the \( x^{-1} \) term, potentially causing steep inclines or declines.

Graphing permits one to spot these shifts and precisely understand each component's influence, providing a comprehensive perspective on how the solution set behaves overall.

Integration Constant

In the context of differential equations, integration constants arise from the process of integration, which is often necessary to solve these equations. Constants like \( c_1 \) and \( c_2 \) emerge when finding antiderivatives, reflecting the infinite set of potential solutions that satisfy the equation.
Each integration of a differential equation yields a constant, reflecting that without specific initial conditions, multiple solutions exist. These constants enable the tailoring of a general solution to particular conditions, generating specific curves within the graph. When verifying a particular solution, these constants play a pivotal role, confirming that any value plugged in should still satisfy the original equation, indicating the flexibility and vastness of potential solution trends.