Question

Solve the initial value problem, given the fundamental set of solutions of the complementary equation. Where indicated by, graph the solution. $$ \begin{array}{l} x^{3} y^{\prime \prime \prime}+x^{2} y^{\prime \prime}-2 x y^{\prime}+2 y=x(x+1), \quad y(-1)=-6, \quad y^{\prime}(-1)=\frac{43}{6}, \quad y^{\prime \prime}(-1)=-\frac{5}{2} ; \\ \left\\{x, x^{2}, 1 / x\right\\} \end{array} $$

Short answer

Based on the given step by step solution, write a short answer. To solve the initial value problem (IVP) for the third-order linear differential equation with non-homogeneous terms, we first verified that the given fundamental set of solutions for the complementary equation is indeed correct. Then, we found a particular solution for the non-homogeneous equation and formed a linear combination of the fundamental solutions and the particular solution. Using the given initial conditions, we determined the constants in the general solution. The exact solution of the given IVP is: $$ y(x) = x - 5\frac{1}{x} $$

Step-by-step solution

Verify the fundamental set of solutions

The given fundamental set of solutions for the complementary equation is: $$ \{x, x^2, \frac{1}{x}\} $$ Before proceeding, let's quickly verify that these functions are solutions to the complementary equation: $$ x^3 y^{\prime \prime \prime} + x^2 y^{\prime \prime} - 2x y^{\prime} + 2y = 0 $$ By plugging in each function and checking if the equation holds.

Find the particular solution

Now let's find a particular solution for the non-homogeneous equation: $$ x^3 y^{\prime \prime \prime} + x^2 y^{\prime \prime} - 2x y^{\prime} + 2y = x(x+1) $$ Since the right side of the equation is a polynomial of degree 2, we will assume a solution of the form: $$ y_p = Ax^2 + Bx + C $$ Now, compute the derivatives and plug them into the non-homogeneous equation: First derivative: \(y_p' = 2Ax + B\) Second derivative: \(y_p'' = 2A\) Third derivative: \(y_p''' = 0\) Substituting these into the non-homogeneous equation yields: $$ (0) + x^2 (2A) - 2x(2Ax + B) + 2(Ax^2 + Bx + C) = x(x+1) $$ After simplifying and comparing coefficients, we have: $$ A = \frac{1}{2}, \quad B = 1, \quad C = 0 $$ Thus, the particular solution is: $$ y_p = \frac{1}{2}x^2 + x $$

Form a linear combination of the fundamental solutions and the particular solution

The general solution of the original non-homogeneous equation will be a combination of the complementary solutions and the particular solution found in previous steps: $$ y(x) = C_1x + C_2x^2 + C_3\frac{1}{x} + \frac{1}{2}x^2 + x $$

Determine the constants using the initial conditions

To find the exact solution, we will use the given initial conditions: $$ y(-1) = -6, \quad y'(-1) = \frac{43}{6}, \quad y''(-1) = -\frac{5}{2} $$ First, we find the derivatives of \(y(x)\): $$ y'(x) = C_1 + 2C_2x - C_3\frac{1}{x^2} + x $$ $$ y''(x) = 2C_2 + 2C_3\frac{1}{x^3} + 1 $$ Now, plug in the initial conditions and solve for the constants: $$ -6 = C_1 - C_2 - C_3 + \frac{1}{2} - 1 \implies C_1 - C_2 - C_3 = -\frac{9}{2} $$ $$ \frac{43}{6} = C_1 + 2C_2 - C_3 \implies C_1 + 2C_2 - C_3 = \frac{43}{6} $$ $$ -\frac{5}{2} = 2C_2 + 2C_3 + 1 \implies 2C_2 + 2C_3 = -\frac{7}{2} $$ Solving this system of equations, we find: $$ C_1 = 0, \quad C_2= -\frac{1}{2}, \quad C_3 = -5 $$

Write the exact solution and graph it if needed

The exact solution of the given initial value problem is: $$ y(x) = -\frac{1}{2}x^2 - 5\frac{1}{x} + \frac{1}{2}x^2 + x = x - 5\frac{1}{x} $$ If a graph of this solution is needed, plot the function \(y(x) = x - 5\frac{1}{x}\) over the desired interval.

Key concepts

Initial Value Problem

An Initial Value Problem (IVP) is a type of differential equation that not only seeks to find a general solution but also a specific solution that satisfies particular starting conditions. In this case, we are dealing with a third-order differential equation:\[x^{3} y^{\prime \prime \prime}+x^{2} y^{\prime \prime}-2 x y^{\prime}+2 y=x(x+1)\]The 'initial conditions' provided include specific values for the function and its derivatives at a certain point, here:

• \( y(-1) = -6 \)
• \( y'(-1) = \frac{43}{6} \)
• \( y''(-1) = -\frac{5}{2} \)

These conditions help to uniquely determine the constants of integration that appear in the solution of a differential equation, thus giving us a precise solution rather than a family of solutions.

Fundamental Set of Solutions

A Fundamental Set of Solutions is a group of functions that form the basis for the solutions of a homogeneous differential equation. For the given problem, the complementary equation (homogeneous version) is:\[x^{3} y^{\prime \prime \prime}+x^{2} y^{\prime \prime}-2 x y^{\prime}+2 y=0\]The fundamental set of solutions for this equation is:

• \( x \)
• \( x^2 \)
• \( \frac{1}{x} \)

These functions are linearly independent and any linear combination of them will solve the homogeneous equation. This concept is crucial because the general solution of a non-homogeneous differential equation combines these complementary solutions with a particular solution of the non-homogeneous equation.

Non-homogeneous Equation

A non-homogeneous equation involves a differential equation that equates to a non-zero function (often an external force or input), rather than zero. In the given exercise, our equation is non-homogeneous:\[x^{3} y^{\prime \prime \prime} + x^{2} y^{\prime \prime} - 2x y^{\prime} + 2y = x(x+1)\]The function on the right side, \( x(x+1) \), classifies the equation as non-homogeneous. Solving it involves two main steps:

• Finding the general solution to the corresponding homogeneous equation.
• Finding a particular solution to the non-homogeneous equation.

The superposition principle allows us to combine these two solutions to get the general solution to the entire non-homogeneous problem, making it a rich and practical method in many scientific fields.

Particular Solution

A Particular Solution is a specific solution to the non-homogeneous differential equation that contains no arbitrary constants. This exact solution is derived by testing a function that mirrors the form of the non-homogeneous term, \( x(x+1) \).In our problem, we assume a trial solution of the form:\[ y_p = Ax^2 + Bx + C \] where \( A, B, \) and \( C \) are constants to be determined. By substituting \( y_p \) and its derivatives into the non-homogeneous equation and equating coefficients, we determine:

• \( A = \frac{1}{2} \)
• \( B = 1 \)
• \( C = 0 \)

Thus, the particular solution is \( y_p = \frac{1}{2}x^2 + x \). This solution, when combined with the complementary (homogeneous) solution, allows us to tackle the initial value problem comprehensively.