Question

Find a formula involving integrals for a particular solution of the differential equation $$ x^{3} y^{\prime \prime}-3 x^{2} y^{\prime \prime}+6 x y^{\prime}-6 y=g(x), \quad x>0 $$ Hint: Verify that \(x, x^{2},\) and \(x^{3}\) are solutions of the homogenenention.

Short answer

Question: Find the general solution of the given second-order inhomogeneous differential equation involving integrals, given that \(x, x^2,\) and \(x^3\) are solutions of the homogeneous equation. Answer: The general solution of the given second-order inhomogeneous differential equation is: $$ y(x) = C_1x + C_2x^2 + C_3x^3 + k_2x^2 \ln{x} + k_5x^2 + k_3x^2 + k_6x^3 $$ where \(C_1, C_2, C_3, k_2, k_3, k_5,\) and \(k_6\) are constant coefficients.

Step-by-step solution

Find the complementary solution

Given that \(x, x^2,\) and \(x^3\) are solutions of the homogeneous equation, the complementary function (solution of the homogeneous equation) is formed by taking a linear combination of these solutions, which can be written as: $$ y_c(x) = C_1x + C_2x^2 + C_3x^3 $$ Where \(C_1, C_2, C_3\) are constants.

Apply the method of variation of parameters

To find the particular solution of the inhomogeneous equation, we'll use the method of variation of parameters. We assume a particular solution in the form: $$ y_p(x) = u_1(x) x + u_2(x) x^2 + u_3(x) x^3 $$ where \(u_1(x), u_2(x),\) and \(u_3(x)\) are unknown functions to be determined. Now, we differentiate \(y_p(x)\) with respect to x: $$ y_p'(x) = u_1'(x) x + u_1(x) + 2u_2'(x) x^2 + 2u_2(x) x + 3u_3'(x) x^3 + 3u_3(x) x^2 $$ And differentiate again: $$ y_p''(x) = u_1''(x) x + 2u_1'(x) + 2u_2''(x) x^2 + 4u_2'(x) x + 6u_2(x) + 3u_3''(x) x^3 + 6u_3'(x) x^2 + 6u_3(x) x $$

Substitute the expressions for \(y_p, y_p', y_p''\) into the given differential equation

We substitute the expressions for \(y_p, y_p', y_p''\) into the given differential equation: $$ x^3y_p''(x) - 3x^2y_p''(x) + 6xy_p'(x) - 6y_p(x) = g(x) $$ At this stage, we impose the following conditions: $$ u_1''(x) = 0, \quad u_2''(x) x + 2u_2'(x) = 0, \quad u_3''(x) x^2 + 3u_3'(x) x = 0 $$

Solve the system of equations for the functions \(u_1(x), u_2(x), u_3(x)\)

From the imposed conditions, we can integrate once to find \(u_1'(x), u_2'(x), u_3'(x)\), and then integrate again to find \(u_1(x), u_2(x), u_3(x)\): \begin{align*} u_1'(x) &= k_1 \\ u_2'(x) &= \frac{k_2}{x} \\ u_3'(x) &= \frac{k_3}{x^2} \end{align*} Now, we integrate again to find: $$ u_1(x) = k_1x + k_4 \\ u_2(x) = k_2 \ln{x} + k_5 \\ u_3(x) = \frac{k_3}{x} + k_6 $$

Plug the solutions for \(u_1(x), u_2(x), u_3(x)\) back into the expression for \(y_p(x)\) and simplify

We have that \(y_p(x) = u_1(x)x + u_2(x)x^2 + u_3(x)x^3\). Plugging the solutions we found in Step 4: $$ y_p(x) = (k_1x + k_4) x + (k_2 \ln{x} + k_5) x^2 + (\frac{k_3}{x} + k_6) x^3 = k_1x^2 + k_4x + k_2x^2 \ln{x} + k_5x^2 + k_3x^2 + k_6x^3 $$ Since \(y_p(x)\) must satisfy the differential equation and not its homogeneous version, we set \(k_1 = k_4 = 0\). This simplifies the expression as: $$ y_p(x) = k_2x^2 \ln{x} + k_5x^2 + k_3x^2 + k_6x^3 $$

Combine the complementary solution and the particular solution to form the general solution

The general solution is given by the sum of the complementary and the particular solutions: $$ y(x) = y_c(x) + y_p(x) = C_1x + C_2x^2 + C_3x^3 + k_2x^2 \ln{x} + k_5x^2 + k_3x^2 + k_6x^3 $$ This is the general solution of the given second-order inhomogeneous differential equation involving integrals.

Key concepts

Method of Variation of Parameters

The method of variation of parameters is a powerful technique for finding a particular solution to inhomogeneous linear differential equations. Let's take our exercise as a case in point. Imagine being handed the equation \[ x^{3} y^{\prime \prime} - 3 x^{2} y^{\prime} + 6 x y^{\prime} - 6 y = g(x), \] and you've already identified the complementary solution. This method involves adjusting the parameters (initially constants) of the complementary solution to turn them into functions that cater to the inhomogeneity expressed by \( g(x) \).

In our example, we assumed the particular solution to be \( y_p(x) = u_1(x) x + u_2(x) x^2 + u_3(x) x^3 \), where \( u_1, u_2, \) and \( u_3 \) are the functions to be determined. Through a couple of differentiation steps and plugging back into the original equation, we systematically work out these functions.
By ensuring that certain derivatives (like \( u_1''(x) \)) vanish, we can simplify the integration process and find our unknowns. The method of variation of parameters may look daunting at first glance, but with a step-by-step approach, it becomes an invaluable tool for solving differential equations.

Complementary Solution

A complementary solution, or \( y_c(x) \), is essentially the solution to the homogeneous version of the differential equation. In simple terms, it addresses the part of the equation without the 'extra stuff' on the right side, denoted as \( g(x) \).

In our scenario, because \( x, x^2, \) and \( x^3 \) were already identified as solutions to the homogeneous part of the equation, the complementary solution is a linear combination of these, elegantly represented as \( y_c(x) = C_1x + C_2x^2 + C_3x^3 \). These \( C \) constants are what make the solution 'complete' when no inhomogeneity is involved. Think of it as the foundation upon which we will build the particular solution.

Particular Solution

The particular solution tackles the uniqueness brought about by the non-homogeneous component \( g(x) \). It's like saying, 'Okay, we've got our general baseline from the complementary solution; now let's handle the specifics introduced by \( g(x) \).'

For the exercise at hand, the particular solution was approached through the adjusted functions of the complementary solution's components (\( u_1(x), u_2(x), u_3(x) \)). By allowing these constants to become functions, we could integrate to resolve their forms and then plug them back to get our specific solution \( y_p(x) \). This method turns an otherwise guesswork task into a systematic process, ultimately yielding a solution that will satisfy the entire equation, including the inhomogeneous part.

Linear Differential Equation

Now, what's the stage where all this mathematical drama unfolds? It's the linear differential equation. These equations are characterized by terms of the unknown function and its derivatives multiplied by a function (or just a constant), but never by each other, maintaining linearity. Our sample problem is a second-order linear differential equation because the highest derivative is the second derivative (\( y'' \)).

The structure of linear differential equations allows for solutions to be superposed; in our case, the general solution \( y(x) \) is exactly that — a superposition of the complementary solution (\( y_c(x) \)) and the particular solution (\( y_p(x) \)). It’s fascinating how this linear structure paves the way for methods like the variation of parameters to be effective in finding solutions to complex, real-world problems modeled by such equations.