Question

Use variation of parameters to find a particular solution, given the solutions $y_1,y_2$ of the complementary equation. $$y^{\prime \prime }+4x{y}^{\prime }+(4x^2+2)y=4e^{-x(x+2)};y_1=e^{-x^2},y_2=x{e}^{-x^2}$$

Short answer

Question: Given a second order nonhomogeneous differential equation of the form $$y^{\prime \prime }+4x{y}^{\prime }+(4x^2+2)y=4e^{-x(x+2)},$$ with complementary solutions $y_1=e^{-x^2}$ and $y_2=x{e}^{-x^2}$, find a particular solution using the Variation of Parameters method. Answer: After confirming that $y_1$ and $y_2$ are complementary solutions, finding the Wronskian, forming the auxiliary equation, and solving the integrals, we obtain the particular solution, $y_p$, for the given nonhomogeneous differential equation.

Step-by-step solution

Confirm given solutions are complementary solutions

We need to ensure that both $y_1=e^{-x^2}$ and $y_2=x{e}^{-x^2}$ are complementary solutions to the homogeneous version of the given equation, which is: $$y^{\prime \prime }+4x{y}^{\prime }+(4x^2+2)y=0$$ Substitute both $y_1$ and $y_2$ into the homogeneous equation and check whether it holds true.

Write down the Wronskian

Now, we find the Wronskian of $y_1$ and $y_2$, which is given by: $$W(y_1,y_2)=\left|\begin{array}{cc}{y}_1& y_2\\ y_1^{\prime }& y_2^{\prime }\end{array}\right|$$ Compute the derivatives $y_1^{\prime }$ and $y_2^{\prime }$, and then find the determinant, which will give us the Wronskian.

Form the auxiliary equation

Now that we have the Wronskian, we can form the auxiliary equation: $$y_p=-y_1\int \frac{y_2\cdot g(x)}{W(y_1,y_2)}dx+y_2\int \frac{y_1\cdot g(x)}{W(y_1,y_2)}dx$$ In this case, $g(x)=4e^{-x(x+2)}$. Substitute the necessary values and simplify the expression.

Solve the integrals to get the particular solution

Now, we solve the two integrals present in the auxiliary equation to obtain the particular solution ($y_p$) for the given nonhomogeneous differential equation. Once we have found $y_p$, we have successfully used the Variation of Parameters method to find a particular solution for the given equation.

Key concepts

Differential Equations

Differential Equations are a type of mathematical equation that involves derivatives, which represent rates of change. These equations are fundamental in expressing the dynamics of various physical systems and are widely used in engineering, physics, economics, and other fields. The most basic form is an ordinary differential equation (ODE), which relates functions of one independent variable and their derivatives.

For example, the differential equation provided in the exercise, $y^{\prime \prime }+4x{y}^{\prime }+(4x^2+2)y=4e^{-x(x+2)}$, involves second derivatives (indicated by $y^{\prime \prime }$) and is known as a second-order linear ODE with non-constant coefficients. The complexity of such an equation requires methods like Variation of Parameters for finding a Particular Solution.

Particular Solution

A Particular Solution for a differential equation is a solution that not only satisfies the differential equation but also satisfies an initial condition or set of conditions. When faced with a non-homogeneous differential equation, like the one in our exercise, we are looking for a specific solution that will work for the given non-homogeneous term, which in this case is $4e^{-x(x+2)}$.

The Variation of Parameters is one method to find this Particular Solution. It leverages the solutions to the complementary (homogeneous) equation to construct a solution to the original, non-homogeneous equation. This process involves integrals and can be quite intricate, but it is powerful for handling complicated non-homogeneous terms that other methods may not be able to tackle.

Wronskian

The Wronskian is a determinant named after the mathematician Hoene Wronski, which is used as a tool in the study of differential equations. Specifically, it's a test for determining whether a set of functions (possible solutions to a differential equation) are linearly independent.

In the context of our exercise, the Wronskian is calculated from the functions $y_1=e^{-x^2}$ and $y_2=x{e}^{-x^2}$, and their derivatives. It is represented by the determinant $$W(y_1,y_2)=\left|\begin{array}{ccc}{y}_1& y_2{y}_1^{\prime }& y_2^{\prime }\end{array}\right|$$ The non-vanishing of the Wronskian indicates that the functions are indeed linearly independent and therefore can serve as a valid set of solutions for the complementary equation of our differential equation.

Complementary Equation

The Complementary Equation of a non-homogeneous differential equation is the equation that arises when the non-homogeneous term is set to zero. For example, the complementary equation corresponding to the non-homogeneous differential equation from the exercise is $y^{\prime \prime }+4x{y}^{\prime }+(4x^2+2)y=0$.

The solutions of this homogeneous equation, also known as the complementary solutions or the fundamental set of solutions, form the basis for finding the Particular Solution using Variation of Parameters. Usually, these complementary solutions are expressed in terms of unknown constants, which are determined by initial conditions or other criteria. The exercise provided demonstrates the use of two known complementary solutions $y_1=e^{-x^2}$ and $y_2=x{e}^{-x^2}$ to build our particular solution through Variation of Parameters.