Question

In Exercises 7-29 use variation of parameters to find a particular solution, given the solutions \(y_{1}, y_{2}\) of the complementary equation. $$ x^{2} y^{\prime \prime}+x y^{\prime}-y=2 x^{2}+2 ; \quad y_{1}=x, \quad y_{2}=\frac{1}{x} $$

Short answer

To summarize, given the second-order non-homogeneous differential equation \(x^{2} y^{\prime \prime} + x y^{\prime} - y = 2 x^{2} + 2\), we were asked to apply the method of variation of parameters using the complementary solutions \(y_1 = x\) and \(y_2 = \frac{1}{x}\). After going through the method of variation of parameters, we found the particular solution to be \(Y(x) = 0\).

Step-by-step solution

Understand the problem

We are given a non-homogeneous second-order differential equation: $$ x^{2} y^{\prime \prime}+x y^{\prime}-y=2 x^{2}+2 $$ We are also given two solutions of its complementary equation: \(y_1 = x\) and \(y_2 = \frac{1}{x}\). We'll apply the method of variation of parameters using these solutions to find a particular solution of the given differential equation.

Set up the equation for the particular solution

The method of variation of parameters requires us to find two functions \(v_1(x)\) and \(v_2(x)\) such that the particular solution \(Y(x)\) is given by $$ Y(x) = v_1(x)y_1(x) + v_2(x)y_2(x) $$ So, we want to find \(v_1(x)\) and \(v_2(x)\) such that $$ Y(x) = v_1(x)x + v_2(x)\frac{1}{x} $$

Determine \(Y'(x)\) and \(Y''(x)\)

To apply the method of variation of parameters, we need to find the first and second derivative of the function \(Y(x)\). To find \(Y'(x)\), we differentiate \(Y(x)\) with respect to x: \begin{align*} Y'(x) &= \frac{d}{dx}(v_1(x)x) + \frac{d}{dx}(v_2(x) \frac{1}{x}) \\ &= x \frac{d v_1(x)}{dx} + v_1(x) + \frac{1}{x^2} \frac{d v_2(x)}{dx} - \frac{v_2(x)}{x^2} \end{align*} To find \(Y''(x)\), we differentiate \(Y'(x)\) with respect to x: \begin{align*} Y''(x) &= \frac{d}{dx}(x \frac{d v_1(x)}{dx} + v_1(x)) + \frac{d}{dx}(\frac{1}{x^2} \frac{d v_2(x)}{dx} - \frac{v_2(x)}{x^2}) \\ &= \frac{d^2 v_1(x)}{dx^2} x+ 2 \frac{d v_1(x)}{dx} - \frac{d v_1(x)}{dx} + \frac{1}{x^3} \frac{d^2 v_2(x)}{dx^2} + \frac{2 v_2(x)}{x^3} \end{align*}

Applying the method of variation of parameters

In the method of variation of parameters, we assume that the particular solution \(Y(x)\) has the imposed conditions $$ x \frac{dv_1(x)}{dx} + \frac{1}{x^2} \frac{dv_2(x)}{dx} = 0 $$ Now, we substitute the expressions for \(Y(x)\), \(Y'(x)\), and \(Y''(x)\) in the given differential equation: \begin{align*} x^2 Y''(x) + x Y'(x) - Y(x) &= 2 x^{2} + 2 \\ x^2 \left( \frac{d^2 v_1(x)}{dx^2} x + 2 \frac{d v_1(x)}{dx} - \frac{d v_1(x)}{dx} + \frac{1}{x^3} \frac{d^2 v_2(x)}{dx^2} + \frac{2 v_2(x)}{x^3} \right) \\ + x \left(x \frac{d v_1(x)}{dx} + v_1(x) + \frac{1}{x^2} \frac{d v_2(x)}{dx} - \frac{v_2(x)}{x^2}\right) \\ - \left(v_1(x)x + v_2(x)\frac{1}{x}\right) &= 2 x^2 + 2 \end{align*} Using the conditions on \(v_1(x)\) and \(v_2(x)\), we get $$ x^4 \frac{d^2 v_1(x)}{dx^2} - x^3 \frac{d v_1(x)}{dx} - x^2 v_1(x) + 2 x^3 v_2(x) = 2x^4 + 2 x^2 $$

Solve for \(v_1(x)\) and \(v_2(x)\)

Divide both sides of the equation by \(x^2\) to get $$ x^2 \frac{d^2 v_1(x)}{dx^2} - x \frac{d v_1(x)}{dx} - v_1(x) + 2 x v_2(x) = 2x^2 + 2 $$ Now, we can solve the system of differential equations: \begin{align*} x \frac{dv_1(x)}{dx} &= -\frac{1}{x^2} \frac{dv_2(x)}{dx} \\ x^2 \frac{d^2 v_1(x)}{dx^2} - x \frac{d v_1(x)}{dx} - v_1(x) + 2 x v_2(x) &= 2x^2 + 2 \end{align*} Integrating the first equation, $$ v_1(x) = -\frac{1}{x^2} v_2(x) + C_1 $$ Substituting this back into the second equation, we can find \(v_2(x)\): $$ 2 x \left(-\frac{1}{x^2} v_2(x)+C_1 \right) = 2x^2 + 2$$ Solve for \(v_2(x)\): $$ v_2(x) = -x^3 + C_2 x$$ Now substitute \(v_2(x)\) back into the equation for \(v_1(x)\): $$ v_1(x) = -\frac{1}{x^2} \left(-x^3 + C_2 x\right) + C_1 = x + C_1 $$

Write the particular solution

Finally, we substitute the found functions \(v_1(x)\) and \(v_2(x)\) into our equation for \(Y(x)\): $$ Y(x) = x \left( x + C_1 \right) + \left(-x^3 + C_2 x\right)\frac{1}{x}$$ As we are required to find a particular solution, we can set the constants \(C_1\) and \(C_2\) to zero: $$ Y(x) = x^2 - x^2 = 0 $$ So, the particular solution for the given differential equation is \(Y(x) = 0\).

Key concepts

Particular Solution

The concept of a particular solution in differential equations is crucial for understanding the overall behavior of a system described by the equation. A particular solution refers to one specific solution to a non-homogeneous differential equation, which includes the arbitrary constants determined by the initial conditions. It is the solution that uniquely fits additional constraints, such as boundary conditions or specific values at particular points.

In our exercise, we're applying the method of variation of parameters to find a particular solution of the given second-order differential equation. This involves using the given solutions of the complementary equation and adjusting them by finding suitable functions which, in this context, are labeled as v₁(x) and v₂(x). Together, these create a new solution that still satisfies the original equation but also accounts for the non-homogeneous part.

It is worth noting that particular solutions are often used in applied mathematics, such as physics and engineering, to model specific scenarios based on the conditions at hand. The skill of finding such solutions is thus crucial for practical problem-solving.

Second-order Differential Equation

A second-order differential equation is an equation that involves the unknown function, its first derivative, and its second derivative. The general form is often written as a(x)y'' + b(x)y' + c(x)y = f(x), where the coefficients a(x), b(x), and c(x) are functions of the independent variable, and f(x) represents a forcing function or the non-homogeneous component of the equation.

In the exercise provided, we're given a non-homogeneous second-order differential equation with x² as the coefficient of the second derivative, x as the coefficient of the first derivative, and -1 as the coefficient of the function y. The right-hand side, 2x² + 2, acts as the non-homogeneous part, which is what makes this equation particularly interesting to solve as it does not yield a solution easily expressible as a combination of simple functions.

Such equations frequently appear in mechanics and other areas of physics, usually representing dynamic processes, like vibrations or wave propagation, making their solution relevant for many scientific fields.

Complementary Equation

The term complementary equation is related to a homogeneous part of a second-order differential equation. It is derived by setting the non-homogeneous term (essentially the right-hand side of the equation) to zero. The solutions to this complementary equation are called the complementary functions and are often denoted as y₁, y₂, and so on.

In the context of our exercise, the complementary equation to the original non-homogeneous equation would be obtained by dropping the terms 2x² + 2, which leaves us with x²y'' + xy' - y = 0. The solutions to this, given in the exercise, are y₁ = x and y₂ = 1/x. These solutions form a fundamental set of solutions to the complementary equation and are used as the starting point for the variation of parameters technique.

This technique uses the solutions of the complementary equation and adjusts them to find a particular solution that satisfies the complete non-homogeneous equation. Understanding and solving the complementary equation is, therefore, a crucial step in many analytic methods for solving differential equations.