Question

Solve the initial value problem, given the fundamental set of solutions of the complementary equation. Where indicated by, graph the solution. $$ x^{3} y^{\prime \prime \prime}-5 x^{2} y^{\prime \prime}+14 x y^{\prime}-18 y=x^{3}, \quad y(1)=0, \quad y^{\prime}(1)=1, \quad y^{\prime \prime}(1)=7 ; \quad\left\\{x^{2}, x^{3}, x^{3} \ln x\right\\} $$

Short answer

Question: Verify if the given functions \(x^2\), \(x^3\), and \(x^3 \ln x\) are linearly independent. If linear independence is confirmed, find the particular solution of the linear ODE \(x^3y^{\prime \prime \prime} - 5x^2y^{\prime \prime} + 14xy^{\prime} - 18y = x^3\) with initial conditions \(y(1) = 0\), \(y^{\prime}(1) = 1\), and \(y^{\prime\prime}(1) = 7\). Provide the final solution and its graph. Answer: The given functions \(x^2\), \(x^3\), and \(x^3 \ln x\) are linearly independent. The particular solution to the ODE is the function \(y_p(x) = \frac{1}{25}x^3\). The final solution is \(y(x) = \frac{38}{25}x^2 - \frac{69}{25}x^3 + x^3\ln x + \frac{1}{25}x^3\). The solution can be graphed on the domain x > 0.

Step-by-step solution

Confirm Linear Independence of Given Solutions

We have a set of three solutions: \(\left\{x^2, x^3, x^3 \ln x\right\}\). To confirm if these are linearly independent, we can check their Wronskian. The Wronskian is given by the determinant of this matrix: $$ W(y_1, y_2, y_3) = \begin{vmatrix} x^2 & x^3 & x^3 \ln x \\ 2x & 3x^2 & x^3 (1 + \ln x) \\ 2 & 6x & x^2 (3 + \ln x) \\ \end{vmatrix} $$ We can compute the Wronskian: $$ W(y_1, y_2, y_3) = x^2(x^2(3+\ln x) - 0) - x^3(6x - (x^2(1+\ln x)) + x^3 \ln x(0) = x^4 (3+\ln x) $$ As the Wronskian is nonzero, we can conclude that the three given functions are linearly independent.

Find the Particular Solution using Undetermined Coefficients

The given ODE can be represented as: $$ L(y) = x^3y^{\prime \prime \prime} - 5x^2y^{\prime \prime} + 14xy^{\prime} - 18y = x^3 $$ We will try the particular solution of the form \(y_p = Ax^3\). Then, $$ y_p^{\prime} = 3Ax^2, \quad y_p^{\prime \prime \prime} = 6A $$ Substituting these in the ODE, we get: $$ x^3(6A) - 5x^2(6Ax) + 14x(3Ax^2) - 18(Ax^3) = x^3 $$ Which simplifies to: $$ Ax^3 - 18Ax^3 + 42Ax^3 = x^3 $$ Comparing the coefficients, we get: $$ A(1-18+42) = 1 $$ Solving for A, we have: $$ A = \frac{1}{25} $$ Thus, the particular solution is: $$ y_p(x) = \frac{1}{25}x^3 $$

Determine the Constants for the Given Solutions

Since we found the particular solution, the general solution can be written as: $$ y(x) = c_1x^2 + c_2x^3 + c_3x^3\ln x + \frac{1}{25}x^3 $$ Applying the initial conditions \(y(1) = 0\), \(y^{\prime}(1) = 1\), and \(y^{\prime\prime}(1) = 7\), we can find the constants \(c_1\), \(c_2\), and \(c_3\). Computing the derivatives, we have: $$ y^{\prime}(x) = 2c_1x + 3c_2x^2+ 3c_3x^2\ln x + c_3x^3(\frac{1}{x}) + \frac{3}{25}x^2 $$ And $$ y^{\prime\prime}(x) = 2c_1 + 6c_2x + 6c_3x\ln x + 3c_3x^2(\frac{1}{x})+\frac{6}{25}x $$ Now substituting the initial conditions we get: 1. \(y(1)=c_1+c_2+c_3+\frac{1}{25}=0\) 2. \(y^{\prime}(1)=2c_1+3c_2+3c_3+ \frac{3}{25}=1\) 3. \(y^{\prime\prime}(1)=2c_1+6c_2+6c_3 +\frac{6}{25}=7\) Solving this system of linear equations we get the following constants: $$ c_1 = \frac{38}{25}, \quad c_2 = -\frac{69}{25}, \quad c_3 = 1 $$

Write the Final Solution and Graph it

Now that we have all constants, we can write down the final solution: $$ y(x) = \frac{38}{25}x^2 - \frac{69}{25}x^3 + x^3\ln x + \frac{1}{25}x^3 $$ To graph the solution, plot this function along with the domain of interest. Note that this solution is valid for x > 0 as the functions of the fundamental set are defined only in this domain.

Key concepts

Wronskian

The Wronskian is a tool that helps us understand whether multiple functions are linearly independent or dependent. If we have several functions, like the ones given in our exercise \(\{x^2, x^3, x^3 \ln x\}\), the Wronskian tests if they can be expressed in terms of each other. To calculate it, we create a matrix. The rows come from the derivative of each function.
For example:

• The first row comes from the functions themselves.
• The second from their first derivatives.
• The third from their second derivatives.

Once the matrix is set up, we find the determinant. A glance at this part of the solution shows a non-zero determinant \( x^4 (3+\ln x) \) which means our functions are linearly independent.
Why is this important? Because linearly independent functions can construct the entire solution to a differential equation.

Linearly Independent Functions

Functions are linearly independent if none can be written as a combination of others. Think of them as unique building blocks in mathematics. When working with differential equations, these independent functions help form solutions that cover all possibilities.
In our example, we checked the independence using the Wronskian. When the Wronskian is not zero, then our set of functions has a unique role.

• This variety is crucial as it ensures that when combined with other possible solutions, like particular solutions, the complete solution truly meets the initial conditions.

Thus, recognizing and confirming linear independence lets us progress securely toward solving the differential equation.

Particular Solution

Finding a particular solution is like finding a specific path that satisfies the differential equation and its related conditions. Normally, differential equations demand solutions that fit a stipulated form. Here, we've tried a form \(y_p = Ax^3\) and observed what values might satisfy the equation.
Undetermined coefficients, in our case \(A\), help refine this path to exactly fit the scenario in the question. This offers us a piece of the puzzle that matches exactly with the established conditions.
Once we get the particular solution, \(y_p(x) = \frac{1}{25}x^3\), it leads toward the comprehensive solution adjusting to the initial values provided.

Differential Equations

Differential equations are equations involving derivatives of a function. They link a function with its derivatives, helping us model real-world situations. These equations arise when dealing with rates of change such as in physics or economics.
In this exercise, the differential equation involves third derivatives, making it a third-order differential equation. Solving it means identifying a function \(y(x)\) that satisfies the equation.

• To solve, we first ensure that the functions making up the solution set are linearly independent.
• Then, finding a particular solution fine-tunes the process to our specific initial conditions.

Through these steps, we construct a solution reflecting both the formula's structure and the initial value constraints.

Boundary Conditions

Boundary conditions help anchor our solutions of differential equations to specific starting values. They are integral along with initial value problems. For instance, the conditions \(y(1)=0\), \(y'(1)=1\), and \(y''(1)=7\) specify where the function or its derivatives start off.
These conditions let us find unknown constants in our general solution. By substituting initial conditions into the equation, we fix the constants \(c_1, c_2, c_3\) giving a unique solution.
Without boundary conditions, it would be impossible to confine the solution to a single path. These conditions dictate how the solution behaves and fits into real-world scenarios.