Question

If \(\mathrm{f}(\mathrm{x})\) and \(\mathrm{g}(\mathrm{x})\) are two solutions of the differential equation \(\mathrm{q}\left(\mathrm{d}^{2} \mathrm{y} / \mathrm{d} \mathrm{x}^{2}\right)+\mathrm{x}^{2}(\mathrm{dy} / \mathrm{d} \mathrm{x})+\mathrm{y}=\mathrm{e}^{\mathrm{x}}\), then \(\mathrm{f}(\mathrm{x})-\mathrm{g}(\mathrm{x})\) is the solution of: (A) \(\mathrm{q}\left(\mathrm{d}^{2} \mathrm{y} / \mathrm{d} \mathrm{x}^{2}\right)+\mathrm{y}=\mathrm{e}^{\mathrm{x}}\) (B) \(q^{2}\left(d^{2} y / d x^{2}\right)+(d y / d x)+y=e^{x}\) (C) \(q^{2}\left(d^{2} y / d x^{2}\right)+y=e^{x}\) (D) \(q\left(d^{2} y / d x^{2}\right)+x^{2}(d y / d x)+y=0\)

Short answer

The short answer is: \(h(x) = f(x) - g(x)\) is the solution of the differential equation \(\mathrm{D}\): \(q\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d}\mathrm{x}^{2}}\right)+\mathrm{x}^{2}\left(\frac{\mathrm{dy}}{\mathrm{d}\mathrm{x}}\right)+\mathrm{y}=0\).

Step-by-step solution

Differentiate h(x)

Let h(x) = f(x) - g(x). Then, differentiate h(x) with respect to x to get: h'(x) = f'(x) - g'(x)

Differentiate h'(x) again

Now, differentiate h'(x) with respect to x again to get: h''(x) = f''(x) - g''(x)

Substitute into the given differential equation

We know that f'(x), f''(x), g'(x), and g''(x) satisfy the original differential equation: q(d²y/dx²) + x²(dy/dx) + y = e^x So we have: qf''(x) + x²f'(x) + f(x) = e^x and qg''(x) + x²g'(x) + g(x) = e^x

Find the differential equation satisfied by h(x)

Subtract the second equation from the first equation to isolate h(x), h'(x) and h''(x): [qf''(x) + x²f'(x) + f(x)] - [qg''(x) + x²g'(x) + g(x)] = 0 qh''(x) + x²h'(x) + h(x) = 0 This is the differential equation satisfied by h(x) = f(x) - g(x).

Compare with the given options

Compare the obtained differential equation for h(x) with the given options: (A) q(d²y/dx²) + y = e^x (B) q²(d²y/dx²) + (dy/dx) + y = e^x (C) q²(d²y/dx²) + y = e^x (D) q(d²y/dx²) + x²(dy/dx) + y = 0 The correct answer is (D) since it matches the differential equation we found for h(x).

Key concepts

Second-Order Differential Equation

A second-order differential equation involves the second derivative of a function and typically appears in the form \( y'' = f(x, y, y') \), where \( y'' \) represents the second derivative of \( y \) with respect to \( x \), \( y \) is the unknown function, and \( f \) can be any expression involving \( x \), \( y \) and the first derivative of \( y \), denoted by \( y' \). These equations often arise in physics and engineering, for example, in the description of oscillatory motion such as springs and circuits.

When solving a second-order differential equation, we typically seek to find the function \( y(x) \) that satisfies the equation. There can be many potential methods for solving these equations, including reduction of order, variation of parameters, or utilizing the characteristic equation for linear differential equations with constant coefficients. In the context of the given textbook exercise, we are examining how the difference of two solutions, represented as \( h(x) = f(x) - g(x) \) where \( f \) and \( g \) are individual solutions, also satisfies a second-order differential equation derived by applying prescribed operations.

Linear Differential Equation

A linear differential equation is characterized by the requirement that the unknown function and its derivatives appear linearly; this means they are each multiplied by a function of the independent variable (often \( x \) or \( t \) for time) and are not raised to any power nor multiplied by each other. Specifically, a second-order linear differential equation has the form \( ay'' + by' + cy = g(x) \) where \( a \) , \( b \) , and \( c \) are functions of the independent variable \( x \) and \( g(x) \) is the nonhomogeneous term.

In the example provided, the original differential equation \( q(d^{2}y/dx^{2}) + x^{2}(dy/dx) + y = e^{x} \) demonstrates linearity because \( y \) and its derivatives appear linearly and are not products or functions of each other. Notice that \( q \) could be a function of \( x \) which affects the coefficients but maintains the equation's linearity. Understanding the nature of linearity is crucial for choosing the appropriate method to solve these types of differential equations.

Homogeneous Differential Equation

Homogeneous differential equations refer to the case where every term in the equation is a function of the unknown function \( y(x) \) and its derivatives, with no additional terms; essentially, these equations can be set to zero. In more formal terms, a homogeneous differential equation of the second-order is of the form \( ay'' + by' + cy = 0 \) where the absence of the function \( g(x) \) (as seen in the linear differential equation section) results in the homogeneity. This zero on the right side means the solutions to the equation can be superimposed or added together to find more solutions -- a principle called the superposition principle.

In relation to the exercise, the resulting equation for \( h(x) = f(x) - g(x) \) simplifies to a homogeneous equation \( qh''(x) + x^2h'(x) + h(x) = 0 \) because the nonhomogeneous part (\( e^{x} \) in the original equation) cancels out when we subtract one solution from the other. This highlights a critical property of linearity and homogeneity – the difference between any two solutions of a linear nonhomogeneous differential equation is itself a solution of the corresponding homogeneous equation.