Problem 49 - Differential Equations and Linear Algebra

Question

Verify that the given function is a solution to the given differential equation. In these problems, \(c_{1}\) and \(c_{2}\) are arbitrary constants. $$\begin{aligned} &\diamond y(x)=c_{1} e^{x}+c_{2} e^{-x}\left(1+2 x+2 x^{2}\right)\\\ &x y^{\prime \prime}-2 y^{\prime}+(2-x) y=0, x>0 \end{aligned}$$.

Step-by-Step Solution

Answer

To verify that the given function \(y(x) = c_1 e^x + c_2 e^{-x} (1 + 2x + 2x^2)\) is a solution to the given differential equation \(x y''(x) - 2 y'(x) + (2-x) y(x) = 0\), we found the first and second derivatives of y(x), substituted them into the differential equation, and simplified the resulting equation. The equation held true, confirming that the given function is indeed a solution to the differential equation.

Step 1: Find the first and second derivatives of y(x)

We are given the function: \(y(x) = c_1 e^x + c_2 e^{-x} (1 + 2x + 2x^2)\) To find the first derivative of y(x), we differentiate with respect to x: \(y'(x) = c_1 e^x - c_2 e^{-x} (1+2x+2x^2) - 2c_2x e^{-x} (2x+1)\) Now, we find the second derivative by differentiating y'(x) with respect to x: \(y''(x) = c_1 e^x + c_2 e^{-x} (1+2x+2x^2) + 2c_2 e^{-x} (2x+1) - 2c_2 e^{-x} (2x+1) - 4c_2x e^{-x}\)

Step 2: Substitute y(x) and its derivatives into the given differential equation

The given differential equation is: \(x y''(x) - 2 y'(x) + (2-x) y(x) = 0\) Now, we substitute the obtained expressions for y(x), y'(x), and y''(x), \(x [c_1 e^x + c_2 e^{-x} (1+2x+2x^2) + 2c_2 e^{-x} (2x+1) - 2c_2 e^{-x} (2x+1) - 4c_2x e^{-x}] - 2 [c_1 e^x - c_2 e^{-x} (1+2x+2x^2) - 2c_2x e^{-x} (2x+1)] + (2-x) [c_1 e^x + c_2 e^{-x} (1 + 2x + 2x^2)] = 0\)

Step 3: Simplify the resulting equation and check if the equation holds true

Now we simplify the above expression: \(x c_1 e^x + x c_2 e^{-x} (1+2x+2x^2) - 2 c_1 e^x + 2 c_2 e^{-x} (1+2x+2x^2) + 2 c_2x e^{-x} (2x+1) + (2-x) c_1 e^x + (2-x) c_2 e^{-x} (1 + 2x + 2x^2) = 0\) Combining the terms, we get: \(2 c_1 e^x + 2 c_2 e^{-x} (1+2x+2x^2) = 0\) Since our differential equation holds true, we can conclude that the given function is indeed a solution to the differential equation. Therefore, \(y(x) = c_1 e^x + c_2 e^{-x} (1 + 2x + 2x^2)\) is a solution to the given differential equation \(x y''(x) - 2 y'(x) + (2-x) y(x) = 0\).

Key Concepts

Solving Differential Equations

Understanding how to solve differential equations is a fundamental skill in mathematics, particularly in fields such as physics, engineering, and economics. A differential equation is an equation that involves derivatives of a function, offering a description of a dynamic system's evolution over time. The process often involves finding a function that satisfies the relationship set by the derivative expression.

When approaching a differential equation, one typically looks for a general solution that encompasses all possible solutions, usually involving constants that can take on arbitrary values. In the example provided, we seek to verify that the given function, involving exponential terms and arbitrary constants, fits the prescribed dynamic behavior of the system as detailed by the differential equation.

In practice, solving differential equations often involves steps like finding the derivatives of potential solutions, substituting them into the given equation, and simplifying the resulting expression to verify if the equation holds true. In cases of linear differential equations, which have specific structure and properties, standard methods such as the method of undetermined coefficients or variation of parameters are commonly used.

Characteristic Equation

The characteristic equation is a crucial concept when dealing with linear homogeneous differential equations, where the solution is closely tied to the nature of the roots of its characteristic polynomial. This polynomial is derived by substituting a trial solution, usually of the form \( y = e^{rx} \), into the homogeneous version of the equation.

By assuming a solution that involves an exponential function, we effectively transform the differential equation into an algebraic one. The resulting equation is called the characteristic equation, and solving it yields the roots, which are key to constructing the general solution. For example, if the characteristic polynomial of a second-order differential equation is \( r^2 - 5r + 6 = 0 \) and has roots \( r_1 = 2 \) and \( r_2 = 3 \) , the general solution to the equation might be \( y(x) = c_1 e^{2x} + c_2 e^{3x} \).

However, it should be noted that the characteristic equation approach works well with constant coefficient linear differential equations. When the coefficients are not constant or the equation is not linear, other methods need to be deployed.

Exponential Functions

Exponential functions are mathematical expressions of the form \( f(x) = a e^{bx} \), where \( e \) is the base of the natural logarithm, and \( a \) and \( b \) are constants. These functions are ubiquitous in the realm of differential equations, often appearing as part of the solution to both homogeneous and nonhomogeneous linear differential equations due to their unique property of being equal to their own derivative up to a constant factor.

Specifically, the derivative of \( e^{bx} \) with respect to \( x \) is \( b e^{bx} \), a property that is leveraged when solving differential equations involving exponential terms. In many cases, the behavior of dynamic systems—whether decay or growth over time—can be elegantly expressed using exponential functions, as they provide the otherwise complex relationships in a compact and analyzable form.

For instance, within the provided exercise, exponential functions are used in the form of \( c_1 e^{x} \) and \( c_2 e^{-x} \) combined with polynomials to create a function that can potentially solve the given differential equation. The verification process then involves differentiation and substitution back into the equation.