Question

In Exercises \(22-33\) solve the initial value problem, given the fundamental set of solutions of the complementary equation. Where indicated by, graph the solution. $$ x^{3} y^{\prime \prime \prime}-2 x^{2} y^{\prime \prime}+3 x y^{\prime}-3 y=4 x, \quad y(1)=4, \quad y^{\prime}(1)=4, \quad y^{\prime \prime}(1)=2 ; \quad\left\\{x, x^{3}, x \ln x\right\\} $$

Short answer

Question: Find the general solution of the initial value problem given by the third-order linear non-homogeneous differential equation \(x^{3}y^{\prime\prime\prime} - 2x^{2}y^{\prime\prime} + 3xy^{\prime} - 3y = 4x\), with \(y(1) = 4\), \(y'(1) = 4\), and \(y''(1) = 2\). Answer: The general solution to the given initial value problem is \(y(x) = \frac{2}{3}x^2 - \frac{10}{3}x + x^3 + x\ln x\).

Step-by-step solution

Find the particular solution using the Method of Undetermined Coefficients

Given the non-homogeneous equation: \(x^{3}y^{\prime\prime\prime} - 2x^{2}y^{\prime\prime} + 3xy^{\prime} - 3y =4x\). Since the right-hand side is a first-order polynomial, we will make an educated guess for the particular solution (\(Y_p\)) as a first-order polynomial: \(Y_p = Ax + B\). Here, \(A\) and \(B\) are coefficients to be determined. Now, we need to find the first, second, and third derivatives of \(Y_p\). \(Y_p' = A\) \(Y_p'' = 0\) \(Y_p''' = 0\) Next, we will substitute \(Y_p\), \(Y_p'\), \(Y_p''\), and \(Y_p'''\) into the given non-homogeneous equation: \(x^{3}(0) - 2x^{2}(0) + 3x(A) - 3(Ax + B) = 4x\) Now, simplifying the equation, we get: \((3A - 3A)x - 3B = 4x\) Comparing coefficients, we have: \(3A - 3A = 4 \Rightarrow 0 = 4\) This creates a contradiction, so our initial guess was incorrect. Therefore, we should try a higher-order polynomial guess for the particular solution. Let us guess \(Y_p\) as a second order polynomial: \(Y_p = Ax^2 + Bx + C\). Now we find the first, second, and third derivatives of \(Y_p\): \(Y_p' = 2Ax + B\) \(Y_p'' = 2A\) \(Y_p''' = 0\) Substituting \(Y_p\), \(Y_p'\), \(Y_p''\), and \(Y_p'''\) into the given non-homogeneous equation: \(x^{3}(0) - 2x^{2}(2A) + 3x(2Ax + B) - 3(Ax^2 + Bx + C) = 4x\) Now, simplifying the equation, we get: \(- 4Ax^2 + (6Ax + 3B - 3B)x - 3C = 4x\) Comparing coefficients, we have: \(- 4A = 0 \Rightarrow A = 0\) \(6Ax + 3B - 3B = 4x \Rightarrow 6A = 4 \Rightarrow A = \frac{2}{3}\) \(- 3C = 0 \Rightarrow C = 0\) Since \(A\) is not zero, our second guess is correct. Thus, the particular solution is \(Y_p = \frac{2}{3}x^2\).

Find the general solution using the particular solution and the fundamental set of solutions

The general solution of the non-homogeneous equation is the sum of the particular solution and a linear combination of the given fundamental set of solutions: \(y(x) = Y_p + C_1x + C_2x^3 + C_3x\ln x\) Where \(C_1\), \(C_2\), and \(C_3\) are constants. Calculating the first and second derivatives of \(y(x)\): \(y'(x) = 2\cdot\frac{2}{3}x + C_1 + 3C_2x^2 + (C_3 + C_3\ln x)\) \(y''(x) = 4\cdot\frac{2}{3} + 6C_2x + (2C_3 + C_3\ln x)\) Now we use the given initial values to find the constants: \(y(1) = \frac{2}{3} + C_1 + C_2 + C_3 = 4\) \(y'(1) =\frac{4}{3} + C_1 + 3C_2 + 2C_3 = 4\) \(y''(1) = \frac{8}{3} + 6C_2 + 3C_3 = 2\) We can solve this system of equations to find the values of \(C_1\), \(C_2\), and \(C_3\): From \(y(1) = 4\): \(C_1 + C_2 + C_3 = \frac{10}{3}\) From \(y'(1) = 4\): \(C_1 + 3C_2 + 2C_3 = \frac{4}{3}\) From \(y''(1) = 2\): \(6C_2 + 3C_3 = -\frac{2}{3}\) Solving this system, we find the values of the constants: \(C_1 = -\frac{10}{3}\) \(C_2 = 1\) \(C_3 = 1\) Thus, the general solution is: \(y(x) = \frac{2}{3}x^2 -\frac{10}{3}x + x^3 + x\ln x\) This is the solution to the initial value problem.

Key concepts

Particular Solution

In the context of an initial value problem, finding the particular solution is a crucial step towards solving differential equations. The particular solution is a single specific solution to a non-homogeneous differential equation that satisfies not just the differential equation itself, but also the initial conditions given.

When we identify a particular solution, we narrow down the infinite number of solutions that a differential equation might have to one that exactly fits the specified conditions at a certain point. This makes our answer unique and tailored to the problem at hand. The process usually involves making educated guesses about the form of the solution based on the non-homogeneous term (the right-hand side of the equation), and then determining the unknown coefficients through substitution and comparing coefficients.

The initial guess might not always be correct, as seen in the provided exercise, so one might need to adjust the guess by either increasing the polynomial degree or considering other functions based on the differential equation's coefficients and the non-homogeneous term.

Method of Undetermined Coefficients

The Method of Undetermined Coefficients is an efficient tool for finding the particular solution of non-homogeneous linear differential equations with constant coefficients. This method involves speculating a form for the particular solution, typically inspired by the non-homogeneous part of the differential equation. The chosen form includes undetermined coefficients which are then found by substituting the guess into the equation and solving for these coefficients.

However, it is important to note that this method works well only when the non-homogeneous part is a type of function for which we can easily guess the form of the particular solution, like polynomials, exponentials, sine and cosine functions, or a combination of these. Moreover, if the guessed form too closely resembles terms from the homogeneous solution, modifications such as multiplying by an appropriate power of the independent variable may be necessary. This intricacy was evident in the exercise where we had to shift from a first-order to a second-order polynomial guess due to the initial contradiction encountered.

Fundamental Set of Solutions

The fundamental set of solutions is associated with the complementary or homogeneous part of a differential equation. It consists of solutions that form a basis for the space of all solutions to the homogeneous differential equation, meaning that any solution of the homogeneous equation can be expressed as a linear combination of these fundamental solutions.

In the given exercise, the set \(\{x, x^3, x \ln x\}\) represents the fundamental solutions to the complementary equation. When combined with the particular solution, these fundamental solutions constitute a major part of the general solution to the non-homogeneous differential equation. One then uses the initial conditions to determine the specific coefficients in this linear combination, yielding a solution that satisfies not only the differential equation but also the initial conditions—rounding out the overall solution to the initial value problem.

Differential Equations

At the heart of many mathematical, scientific, and engineering problems lie differential equations—equations that relate some function with its derivatives. When we talk about differential equations, we generally classify them into two main types: homogeneous and non-homogeneous (or inhomogeneous).

Homogeneous differential equations are those where the function and its derivatives are by themselves set to zero. On the other hand, in non-homogeneous ones, there's an additional function that doesn't depend on the solution function. Solutions to differential equations are not just single functions but sets of functions, encompassing a vast spectrum of possible behaviors which reflect various physical and abstract phenomena.

To solve these equations, we utilize different strategies based on their type and complexity. Some methods are straightforward, while others, like the Method of Undetermined Coefficients described here, require more intricate approaches.