Question

Use variation of parameters to find a particular solution, given the solutions \(y_{1}, y_{2}\) of the complementary equation. $$ \begin{array}{l} 4 x^{2} y^{\prime \prime}+\left(4 x-8 x^{2}\right) y^{\prime}+\left(4 x^{2}-4 x-1\right) y=4 x^{1 / 2} e^{x}, \quad x>0 \\ y_{1}=x^{1 / 2} e^{x}, y_{2}=x^{-1 / 2} e^{x} \end{array} $$

Short answer

Question: Find a particular solution to the non-homogeneous differential equation given the complementary solutions \(y_{1} = x^{1/2}e^{x}\) and \(y_{2} = x^{-1/2}e^{x}\), and \(f(x) = 4x^{1/2}e^{x}\) using the variation of parameters method. Solution: The particular solution is \(y = -\frac{1}{4}x^{3/2}e^{x} + \frac{1}{3}x e^{x}\).

Step-by-step solution

Set up the general form of the particular solution

We start by setting up the general form of the particular solution as \(y = u_{1}(x)y_{1}(x) + u_{2}(x)y_{2}(x)\), where \(u_{1}(x)\) and \(u_{2}(x)\) are functions to be determined using the variation of parameters method.

Find \(u_{1}(x)\) and \(u_{2}(x)\)

Now, we find \(u_{1}(x)\) and \(u_{2}(x)\) by solving the system of equations: \(\begin{cases} u_{1}'(x)y_{1}(x) + u_{2}'(x)y_{2}(x) = 0 \\ u_{1}'(x)y_{1}'(x) + u_{2}'(x)y_{2}'(x) = f(x) \end{cases}\) We are given: \(y_{1} = x^{1/2}e^{x} \implies y_{1}' = \frac{1}{2}x^{-1/2}e^{x} + x^{1/2}e^{x}\) \(y_{2} = x^{-1/2}e^{x} \implies y_{2}' = -\frac{1}{2}x^{-3/2}e^{x} + x^{-1/2}e^{x}\) \(f(x) = 4x^{1/2}e^{x}\) The system of equations becomes: \(\begin{cases} u_{1}'(x)x^{1/2}e^{x} + u_{2}'(x)x^{-1/2}e^{x} = 0 \\ u_{1}'(x)\left(\frac{1}{2}x^{-1/2}e^{x} + x^{1/2}e^{x}\right) + u_{2}'(x)\left(-\frac{1}{2}x^{-3/2}e^{x} + x^{-1/2}e^{x}\right) = 4x^{1/2}e^{x} \end{cases}\) Solve for \(u_{1}'\) and \(u_{2}'\): Divide both sides of equation 1 by \(e^{x}\), \(u_{1}'(x)x^{1/2} + u_{2}'(x)x^{-1/2} = 0\) Now, multiply both sides of equation 1 by \(\frac{1}{2}x^{-1/2}e^{x}\) and add the result to equation 2: \(-\frac{1}{4}u_{1}'(x)e^{x} + u_{2}'(x)\left(\frac{1}{2}x^{1/2}e^{x}\right) = \frac{3}{2}x^{1/2}e^{x}\) Cancel common factors: \(-\frac{1}{4}u_{1}'(x) + u_{2}'(x)\left(\frac{1}{2}x^{1/2}\right) = \frac{3}{2}x^{1/2}\) Now we can integrate both equations: \(\int -\frac{1}{4}u_{1}'(x) dx = u_{1}(x) = -\frac{1}{4}\int dx = -\frac{1}{4}x + C_{1}\) \(\int u_{2}'(x)\left(\frac{1}{2}x^{1/2}\right) dx = u_{2}(x) = \frac{1}{2}\int x^{1/2} dx = \frac{1}{3}x^{3/2} + C_{2}\)

Step 3:Compute particular solution \(y\)

Use found \(u_{1}(x)\) and \(u_{2}(x)\) to compute the particular solution: \(y = u_{1}(x)y_{1}(x) + u_{2}(x)y_{2}(x) = \left(-\frac{1}{4}x + C_{1}\right)x^{1/2}e^{x} + \left(\frac{1}{3}x^{3/2} + C_{2}\right)x^{-1/2}e^{x}\) A particular solution will suffice, so we can ignore the constants \(C_{1}\) and \(C_{2}\). The final particular solution is: \(y = -\frac{1}{4}x^{3/2}e^{x} + \frac{1}{3}x e^{x}\)

Key concepts

Differential Equations

Differential equations are mathematical equations that relate functions and their derivatives. Fundamental in many fields such as physics, engineering, and economics, they describe how a particular quantity changes over time or space. In our example, we work with a second-order linear homogeneous differential equation with variable coefficients. This means the equation involves the second derivative of the unknown function, and the coefficients of the derivatives are not constant.
Solving such an equation typically involves finding a combination of solutions, consisting of a complementary or homogeneous solution that corresponds to the case where the equation is set to zero, and a particular solution that meets the non-homogeneous part. Variation of parameters is a method often employed when the non-homogeneous part makes finding a particular solution through other means like the method of undetermined coefficients challenging.

Particular Solution

The particular solution of a differential equation is a solution that satisfies not only the differential equation itself but also the non-homogeneity of the equation. This means it is the specific solution that takes into account any external forces or input functions relevant to the real-world system being modeled. Computing a particular solution usually requires additional methods when standard approaches aren't applicable. For our problem, by using the variation of parameters, which requires known solutions to the complementary equation, we look for functions u₁ and u₂ that modify these known solutions in a way that collectively forms a particular solution for the non-homogeneous equation. The process involves integrating and combining these individual functions with the known solutions to construct the particular solution.

Complementary Equation

A complementary equation, also known as the homogeneous counterpart of a non-homogeneous differential equation, is crucial for finding the overall solution. It is formed by setting the non-homogeneous term (the right-hand side of the differential equation) to zero. Solutions to this equation, called the homogeneous solutions or complementary solutions, typically consist of a family of functions that span the space of all possible solutions to the homogeneous equation. In the context of the problem provided, the given functions y₁ and y₂ represent two independent solutions to the complementary equation. Using these, along with the variation of parameters, allows us to determine the specific form of the particular solution to the original non-homogeneous differential equation.