Question

Use variation of parameters to find a particular solution, given the solutions \(y_{1}, y_{2}\) of the complementary equation. $$ x^{2} y^{\prime \prime}-4 x y^{\prime}+\left(x^{2}+6\right) y=x^{4} ; \quad y_{1}=x^{2} \cos x, \quad y_{2}=x^{2} \sin x $$

Short answer

Question: Using the variation of parameters method, find the particular solution to the given differential equation, where the given differential equation is \(x^{2}y''-4xy' +(x^2+6)y=x^4\), and the given complementary solutions are \(y_1=x^2\cos x\) and \(y_2=x^2\sin x\). Answer: The particular solution using variation of parameters is \(y_p(x) = -\frac{1}{5}x^6\cos x + \frac{1}{5}x^6\sin x\).

Step-by-step solution

Understand the given information

The given differential equation is: $$x^{2}y''-4xy'+\left(x^{2}+6\right)y=x^{4}.$$ The given complementary solutions are: $$y_1=x^2\cos x \quad \text{and} \quad y_2=x^2\sin x.$$

Calculate the Wronskian of \(y_1\) and \(y_2\)

To use variation of parameters, we need to determine the Wronskian of the complementary solutions (\(W\)), which is given by: $$W=\begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix}.$$ First, find the derivatives of \(y_1\) and \(y_2\): $$y_1' = \frac{d}{dx}(x^2\cos x) = 2x\cos x - x^2\sin x$$ $$y_2' = \frac{d}{dx}(x^2\sin x) = 2x\sin x + x^2\cos x$$ Now, compute the Wronskian: $$W = \begin{vmatrix} x^2\cos x & x^2\sin x \\ 2x\cos x - x^2\sin x & 2x\sin x + x^2\cos x \end{vmatrix} = (x^2\cos x)(2x\sin x + x^2\cos x) - (x^2\sin x)(2x\cos x - x^2\sin x)$$ $$W = 2x^3\cos^2 x + x^4\cos^2 x - 2x^3\sin^2 x + x^4\sin^2 x = x^3(2\cos^2 x - 2\sin^2 x) + x^4(\cos^2 x + \sin^2 x) = x^3 + x^4.$$

Find the particular solution using variation of parameters

The particular solution, \(y_p\), can be found using the formula: $$y_p = -y_1 \int \frac{y_2 f(x)}{W}dx + y_2 \int \frac{y_1 f(x)}{W}dx$$ where \(f(x) = x^4\) in our case. Substitute the given values and find the integrals: $$y_p = -x^2\cos x\left(\int \frac{x^2\sin x\cdot x^4}{x^3+x^4}dx\right) + x^2\sin x\left(\int \frac{x^2\cos x\cdot x^4}{x^3+x^4}dx\right)$$ After simplifying and integrating, we get: $$y_p = -x^2\cos x\left(\int x^3 dx\right) + x^2\sin x\left(\int x^3 dx\right)$$ $$y_p = -\frac{1}{5}x^6\cos x + \frac{1}{5}x^6\sin x$$ Thus, the particular solution using variation of parameters is: $$y_p(x) = -\frac{1}{5}x^6\cos x + \frac{1}{5}x^6\sin x$$

Key concepts

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. They are used to describe a wide range of phenomena in science and engineering. In the context of our exercise, we consider the differential equation \(x^{2}y''-4xy'+(x^{2}+6)y=x^{4}\). This is a second-order linear differential equation with variable coefficients. The goal is to find a solution, or set of solutions, that satisfies this equation for given boundary or initial conditions.

Solving differential equations involves determining the unknown function that makes the equation true. There are various methods for solving differential equations, such as separation of variables, characteristic equations, and variation of parameters, as used in our example. Each method is applicable under different conditions and can provide both particular and general solutions to the equations. Differential equations are pivotal in understanding the relationship between dynamically changing systems.

Wronskian

The Wronskian is a determinant used to assess the linear independence of a set of functions. It is specifically used in the context of solutions to differential equations. For two functions \(y_1\) and \(y_2\), their Wronskian \(W(y_1, y_2)\) is calculated as
\[W = \begin{vmatrix} y_1 & y_2 \ y_1' & y_2' \end{vmatrix}\].

In our solution, the Wronskian is calculated for the complementary solutions \(y_1 = x^2 \cos x\) and \(y_2 = x^2 \sin x\) to substitute into the variation of parameters formula. To calculate it, we find the derivatives of \(y_1\) and \(y_2\) and then compute the determinant. The result reveals whether \(y_1\) and \(y_2\) are linearly independent solutions, which is crucial for employing variation of parameters to find a particular solution. In this case, the Wronskian contributes to constructing the particular solution of the non-homogeneous differential equation.

Particular Solution

The particular solution to a differential equation is a solution that satisfies the non-homogeneous part of the equation. It's called 'particular' because it doesn't involve any arbitrary constants; instead, it's a specific solution tailored to the given problem. In our exercise, the particular solution \(y_p(x)\) is found using the method of variation of parameters. This technique is useful for non-homogeneous linear differential equations when complementary solutions are already known.

In the method, we use formulas involving the complementary solutions \(y_1\) and \(y_2\), the non-homogeneous term \(f(x) = x^4\), and the Wronskian to calculate the particular solution.

• The formula involves integrals: \(-y_1 \int \frac{y_2 f(x)}{W}\,dx + y_2 \int \frac{y_1 f(x)}{W}\,dx\).
• Each term requires integrating over the function components, providing the particular solution by evaluation.

By correctly applying this formula, we reach the particular solution \(y_p(x) = -\frac{1}{5}x^6\cos x + \frac{1}{5}x^6\sin x\), addressing the specific factors present in the non-homogeneous equation.

Complementary Solutions

Complementary solutions are the solutions to the homogeneous part of a differential equation, meaning they satisfy the equation when the right-hand side is zero. For the given exercise, these solutions are \(y_1 = x^2 \cos x\) and \(y_2 = x^2 \sin x\). They are critical because they form the basis upon which the particular solution is built through the variation of parameters.

They are often found using strategies like the characteristic equation or guessing and verification for simpler equations.

• These solutions are necessary because they ensure the overall solution to the differential equation captures both the homogeneous behavior and the specific non-homogeneous behavior dictated by the initial conditions or input functions.
• In linear differential equations, the general solution is typically a combination of complementary solutions and a particular solution.

Understanding complementary solutions involves not only knowing how to find them but also how they fit into the broader context of the full solution to the differential equation.