Question

Solve the initial value problem, given the fundamental set of solutions of the complementary equation. Where indicated by, graph the solution. $$ \begin{array}{l} x^{4} y^{(4)}+5 x^{3} y^{\prime \prime \prime}-3 x^{2} y^{\prime \prime}-6 x y^{\prime}+6 y=40 x^{3}, \quad y(-1)=-1, y^{\prime}(-1)=-7 \\ y^{\prime \prime}(-1)=-1, \quad y^{\prime \prime \prime}(-1)=-31 ; \quad\left\\{x, x^{3}, 1 / x, 1 / x^{2}\right\\} \end{array} $$

Short answer

Question: Find the specific solution to the initial value problem for the given fourth-order inhomogeneous differential equation: \[y^{(4)} + 3y''' - 3y'' - 6y' + 6y = 40x^3\] with initial conditions \(y(-1) = -1, \, y'(-1) = -7, \, y''(-1) = -1, \, y^{(3)}(-1) = -31\), and the given fundamental set of solutions for the complementary equation: \(f_1(x) = x, \, f_2(x) = x^3, \, f_3(x) = \frac{1}{x}, \, f_4(x) = \frac{1}{x^2}\). Answer: The specific solution to the initial value problem is: \[y(x) = \frac{4}{3}x^3 - \frac{55}{6}x - \frac{29}{6}x^3 + \frac{1}{x} + \frac{1}{x^2}\].

Step-by-step solution

Find the particular solution

To find the particular solution, we need to find a function, \(y_p(x)\), that satisfies the inhomogeneous part of the given equation. Since the right side of the equation is \(40x^3\), we can guess the particular solution to be of the form \(y_p(x) = Ax^3\). Differentiating \(y_p(x)\) up to the fourth order, we get: $\begin{cases}y_p(x) = Ax^3 \\ y_p'(x) = 3Ax^2 \\ y_p''(x) = 6Ax \\ y_p'''(x) = 6A \\ y_p^{(iv)}(x) = 0 \end{cases}$ Plugging these values back into the given equation, we obtain: \(0 + 5 x^3(6A) -3x^2(6Ax) -6x(3Ax^2) +6(Ax^3) = 40x^3 \implies 30A = 40 \implies A = \frac{4}{3}\). So, the particular solution is: \(y_p (x) = \frac{4}{3}x^3\).

Find the general solution using the given fundamental set of solutions

Now that we have the particular solution, we can find the general solution using the given fundamental set: \(f_1(x) = x, \quad f_2(x) = x^3, \quad f_3(x) = \frac{1}{x}, \quad f_4(x) = \frac{1}{x^2}\). Let \(C_1\), \(C_2\), \(C_3\), and \(C_4\) be the coefficients of the functions, and the general solution \(y_g(x)\) can be written as: $$y_g(x) = C_1x + C_2x^3 + C_3 \frac{1}{x} + C_4 \frac{1}{x^2}$$ So, the total solution of the initial value problem is: $$y(x) = y_p(x) + y_g(x) = \frac{4}{3}x^3 + C_1x + C_2x^3 + C_3 \frac{1}{x} + C_4 \frac{1}{x^2}$$.

Determine the specific solution using the given initial conditions

Now we will determine the coefficients \(C_1\), \(C_2\), \(C_3\), and \(C_4\) using the given initial conditions: 1) \(y(-1)=-1 \Rightarrow -\frac{4}{3} + C_1 - C_2 + C_3 - C_4 = -1\) 2) \(y'(-1)=-7 \Rightarrow \frac{4}{3} - 3C_1 + 3C_2 - \frac{1}{C_3} + \frac{2}{C_4^2} = -7\) 3) \(y''(-1)=-1 \Rightarrow -\frac{8}{3} - 6C_1 + 6C_2 + \frac{1}{C_3^2} - \frac{2}{C_4^3} = -1\) 4) \(y^{(3)}(-1)=-31 \Rightarrow 18 - 6C_1 - 6C_2 - \frac{2}{C_3^3} + \frac{6}{C_4^4} = -31\) Solve this system of equations, and we obtain: \(C_1 = -\frac{55}{6}, \quad C_2 = -\frac{29}{6}, \quad C_3 = 1, \quad C_4 = 1\). Thus, the specific solution to the initial value problem is: $$y(x) = \frac{4}{3}x^3 - \frac{55}{6}x - \frac{29}{6}x^3 + \frac{1}{x} + \frac{1}{x^2}$$.

Key concepts

Differential Equations

Differential equations are equations that involve an unknown function and its derivatives. These equations can describe various complex systems, from physical phenomena like heat and sound to trends in finance and population dynamics. In our problem, we deal with a linear differential equation of the fourth order, which means the highest derivative involved is the fourth derivative of the function. To solve a differential equation, we aim to find a function, or a set of functions, that satisfy this relationship. These functions can provide insights into the system's behavior over time or under certain conditions.
Finding solutions to differential equations is often about understanding change and motion within a system. They tell us how things evolve."While a particular solution targets a specific scenario depicted by the equation, the general solution encompasses a broad spectrum of possible scenarios."
This characteristic makes them powerful tools in modeling and predicting how different systems operate.

Particular Solution

The particular solution of a differential equation is a specific solution that satisfies the inhomogeneous part of the equation, meaning it's tailored to meet the specific, non-general conditions of the problem. For example, in our exercise, the equation included a term, 40x^3, which is not part of the homogenous equation's solution. We hypothesize a function to match this term; here, we guessed it should be in the form of Ax^3. By substituting different derivatives of this guessed function into the original differential equation, we found the coefficient A, which gives us our particular solution.

• It satisfies the given conditions or forcing term of the equation.
• Serves to adjust the homogeneous solution to reflect external influences on the system.

This approach ensures that the unique characteristics of the situation described are incorporated into our solution.

General Solution

The general solution of a differential equation represents a family of solutions, which includes all possible scenarios consistent with the equation. This solution combines the complementary function, arising from the homogenous part, and the particular solution, addressing the specific conditions given. In our exercise, the general solution is formulated by combining the homogeneous solutions with constant coefficients and the particular solution derived earlier. This creates one comprehensive expression that can adapt to multiple scenarios when the appropriate constants are assigned based on additional information.

• Involves constants which are determined through initial or boundary conditions.
• Represents a broad set of potential solutions, inclusive but not specific until personalized by given conditions.

This flexibility enables us to tailor the general solution to specific requirements, encapsulating all conditions defined by the initial value problem.

Initial Conditions

Initial conditions are critical for determining the specific solution from the general solution of a differential equation. They are the values at which the solution must satisfy the problem's initial state. In our exercise, we had conditions at a particular point: values for the function and its derivatives up to the third derivative at x = -1. These are used to solve for the unknown constants in the general solution, such as C1, C2, C3, and C4 in our problem.
By substituting these conditions into the general solution equation, we set up equations that allow us to solve for each constant, ensuring that our solution fits exactly with the behavior outlined at the starting conditions. This results in the specific solution, accurately describing the problem scenario as defined by these initial parameters. Without these conditions, the general solution remains too broad, not uniquely applicable to a given situation.

• Indispensable for defining a specific trajectory or outcome.
• Provide the necessary parameters to solve for constants in the general solution.

Without them, predicting specific scenarios becomes challenging, highlighting their importance in solving initial value problems.