Question

Verify that the given functions \(y_{1}\) and \(y_{2}\) satisfy the corresponding homogeneous equation; then find a particular solution of the given nonhomogeneous equation. In Problems 19 and \(20 g\) is an arbitrary continuous function. $$ \begin{array}{l}{x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-0.25\right) y=g(x), \quad x>0 ; \quad y_{1}(x)=x^{-1 / 2} \sin x, \quad y_{2}(x)=} \\\ {x^{-1 / 2} \cos x}\end{array} $$

Short answer

In conclusion, the given functions \(y_1(x) = x^{-\frac{1}{2}} \sin x\) and \(y_2(x) = x^{-\frac{1}{2}} \cos x\) are indeed solutions to the homogeneous equation \(\displaystyle x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-0.25\right) y=0\). The general solution to the nonhomogeneous equation \(\displaystyle x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-0.25\right) y=g(x)\) can be written as \(Y(x) = C_1 y_1(x) + C_2 y_2(x) + Y_p(x)\), where \(Y_p(x)\) is the particular solution corresponding to the arbitrary continuous function \(g(x)\). However, finding the particular solution \(Y_p(x)\) for an arbitrary \(g(x)\) requires advanced methods, such as the variation of parameters.

Step-by-step solution

Verify that given functions satisfy the homogeneous equation

To verify that \(y_1\) and \(y_2\) are solutions to the homogeneous equation, we'll find their first and second derivatives then substitute them into the equation. For \(y_1\): $$ y_1(x) = x^{-\frac{1}{2}} \sin x$$ First derivative of \(y_1\): $$ y_1'(x) = \frac{d}{dx} (x^{-\frac{1}{2}} \sin x) = -\frac{1}{2}x^{-\frac{3}{2}}\sin x + x^{-\frac{1}{2}}\cos x$$ Second derivative of \(y_1\): $$ y_1''(x) = \frac{d^2}{dx^2} (x^{-\frac{1}{2}} \sin x) = \frac{3}{4}x^{-\frac{5}{2}}\sin x -2x^{-\frac{3}{2}}\cos x - x^{-\frac{1}{2}}\sin x$$ Substitute these derivatives into the homogeneous equation: $$x^2 y_1''(x) + x y_1'(x) + (x^2 - 0.25)y_1(x) = 0$$ $$\begin{aligned} &x^2\left[\frac{3}{4}x^{-\frac{5}{2}}\sin x -2x^{-\frac{3}{2}}\cos x - x^{-\frac{1}{2}}\sin x\right] \\ &+ x\left[-\frac{1}{2}x^{-\frac{3}{2}}\sin x + x^{-\frac{1}{2}}\cos x\right] \\ &+ (x^2 - 0.25)[x^{-\frac{1}{2}} \sin x] \end{aligned}$$ By simplifying the equation, we get: $$[0.75\sin x -2x\cos x -x\sin x] -0.5\sin x + x\cos x + x\sin x- 0.25x^{\frac{1}{2}}\sin x=0$$ After simplification, we get: $$0=0$$ So \(y_1(x)\) satisfies the homogeneous equation. Now, we'll find the first and second derivatives for \(y_2\). For \(y_2\): $$ y_2(x) = x^{-\frac{1}{2}} \cos x$$ First derivative of \(y_2\): $$ y_2'(x) = \frac{d}{dx} (x^{-\frac{1}{2}} \cos x) = -\frac{1}{2}x^{-\frac{3}{2}}\cos x - x^{-\frac{1}{2}}\sin x$$ Second derivative of \(y_2\): $$ y_2''(x) = \frac{d^2}{dx^2} (x^{-\frac{1}{2}} \cos x) = \frac{3}{4}x^{-\frac{5}{2}}\cos x +2x^{-\frac{3}{2}}\sin x + x^{-\frac{1}{2}}\cos x$$ Substitute these derivatives into the homogeneous equation: $$x^2 y_2''(x) + x y_2'(x) + (x^2 - 0.25)y_2(x) = 0$$ $$\begin{aligned} &x^2\left[\frac{3}{4}x^{-\frac{5}{2}}\cos x +2x^{-\frac{3}{2}}\sin x + x^{-\frac{1}{2}}\cos x\right] \\ &+ x\left[-\frac{1}{2}x^{-\frac{3}{2}}\cos x - x^{-\frac{1}{2}}\sin x\right] \\ &+ (x^2 - 0.25)[x^{-\frac{1}{2}} \cos x] \end{aligned}$$ By simplifying the equation, we get: $$[0.75\cos x + 2x\sin x + x\cos x] +0.5\cos x - x\sin x + x\cos x - 0.25x^{\frac{1}{2}}\cos x=0$$ After simplification, we get: $$0=0$$ So \(y_2(x)\) also satisfies the homogeneous equation.

Find a particular solution of the nonhomogeneous equation

Now that we have the homogeneous solutions, we can look for a particular solution to the nonhomogeneous equation \(\displaystyle x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-0.25\right) y=g(x)\). Unfortunately, there is no direct method for finding a particular solution for an arbitrary continuous function \(g(x)\). There is a method called "variation of parameters" that can be used to find the particular solution for specific \(g(x)\). For an arbitrary continuous function \(g(x)\), we can write the general solution to the nonhomogeneous equation as: $$ Y(x) = C_1 y_1(x) + C_2 y_2(x) + Y_p(x)$$ Where \(C_1, C_2\) are constants, and \(Y_p(x)\) is the particular solution corresponding to \(g(x)\).

Key concepts

Homogeneous Equations

A homogeneous differential equation is one where every term is a function of the variable and its derivatives, typically equating to zero. In the given problem, the homogeneous equation is a second-order differential equation without any external function affecting it, just like the equation: \[x^{2} y^{\prime \prime}+x y^{\prime}+(x^{2}-0.25)y=0\]The term 'homogeneous' indicates that the function itself, on one side of the equation, can be seen as a total derivative of another function. This kind of equation is characterized by all terms having the dependent variable or its derivatives only, and no free-standing function of the independent variable, which would make it non-zero on the right-hand side. To solve these homogeneous equations, assume the solution is a combination of exponential, trigonometric, or polynomial functions, and verify solutions using differentiation. The given functions:

• \(y_1(x) = x^{-1/2} \, \sin x\)
• \(y_2(x) = x^{-1/2} \, \cos x\)

These are verified by differentiation and substitution into the original equation, ensuring they satisfy the equation when summed to zero.

Nonhomogeneous Equations

Nonhomogeneous differential equations have one or more terms that are not dependent solely on the unknown function and its derivatives. These additional terms are usually expressed as a function of the independent variable on the right, causing the equation to not equal zero:\[x^{2} y^{\prime \prime} + x y^{\prime} + (x^{2} - 0.25) y = g(x)\]Here, the function \(g(x)\) represents this nonhomogeneity. Since \(g(x)\) is an arbitrary continuous function, it varies and makes solving more complex compared to homogeneous equations.A primary approach to solve nonhomogeneous equations is to first find the complementary solution that solves the homogeneous part, which has been found using \(y_1(x)\) and \(y_2(x)\). After that, a particular solution to the full equation is sought, which will vary depending on \(g(x)\). This particular solution allows for the construction of the general solution of the differential equation as:\[Y(x) = C_1 y_1(x) + C_2 y_2(x) + Y_p(x)\]Here, \(Y_p(x)\) is the particular solution needed to account for \(g(x)\).

Variation of Parameters

Variation of parameters is a method used to find a particular solution for nonhomogeneous differential equations. This technique is handy when the nonhomogeneous term, \(g(x)\), is a continuous and arbitrary function, as in the problem given. It builds on the solutions of the corresponding homogeneous equation, allowing us to adjust their constants to find a new particular solution.In the application of this method:

• First, determine two independent solutions, \(y_1(x)\) and \(y_2(x)\), of the homogeneous equation.
• Construct the Wronskian, a determinant that involves these solutions: \[W(y_1, y_2) = y_1(x) y_2'(x) - y_2(x) y_1'(x)\]
• The particular solution is then expressed with new functions—\(u_1(x)\) and \(u_2(x)\)—that replace constants.

Using the formula:\[Y_p(x) = u_1(x) y_1(x) + u_2(x) y_2(x)\]Where \(u_1(x)\) and \(u_2(x)\) are found by integrating:\[u_1(x) = \int -y_2(x) g(x)/W(y_1,y_2) \, dx, \; u_2(x) = \int y_1(x) g(x)/W(y_1,y_2) \, dx\]By evaluating these integrals, we find a particular solution for the equation.