Question

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

Short answer

Question: Solve the given differential equation using the Variation of Parameters method. Differential Equation: $$ (\sin x) y^{\prime \prime}+(2 \sin x-\cos x) y^{\prime}+(\sin x-\cos x) y=e^{-x} $$ Complementary solutions: \(y_1 = e^{-x}\), \(y_2 = e^{-x}\cos x\) Answer: After using the Variation of Parameters method, we find that the particular solution is in the form of: $$ y_p(x) = u_1(x)e^{-x} + u_2(x)e^{-x}\cos x $$ However, the integrals to find \(u_1(x)\) and \(u_2(x)\) do not have elementary solutions. As a result, we must use numerical or approximate methods to find these functions. Once approximate values for \(u_1(x)\) and \(u_2(x)\) are found, plug them into the formula to obtain the particular solution.

Step-by-step solution

Write the complimentary solution and the differential equation

From the given information, we have two complementary solutions \(y_{1}=e^{-x}\) and \(y_{2}=e^{-x}\cos x\). The given differential equation is: $$ (\sin x) y^{\prime \prime}+(2 \sin x-\cos x) y^{\prime}+(\sin x-\cos x) y=e^{-x} $$

Set up the particular solution using the Variation of Parameters method

Using the Variation of Parameters method, we assume the particular solution to be in the form: $$ y_p(x) = u_1(x)y_1(x) + u_2(x)y_2(x) $$ Plug in the given complementary solutions to get: $$ y_p(x) = u_1(x)e^{-x} + u_2(x)e^{-x}\cos x $$

Differentiate the particular solution with respect to x

We now need to find \(y_p'(x)\) and \(y_p''(x)\). Start by finding \(y_p'(x)\): $$ y_p'(x) = u_1'(x)e^{-x} - u_1(x)e^{-x} + u_2'(x)e^{-x}\cos x - u_2(x)e^{-x}\cos x + u_2(x)e^{-x}\sin x $$ Next, find \(y_p''(x)\): $$ y_p''(x) = u_1''(x)e^{-x} - 2u_1'(x)e^{-x} + u_1(x)e^{-x} + u_2''(x)e^{-x}\cos x - 2u_2'(x)e^{-x}\cos x + 2u_2'(x)e^{-x}\sin x + u_2(x)e^{-x}\cos x - u_2(x)e^{-x}\sin x - u_2(x)e^{-x}\cos x $$

Plug in \(y_p(x)\), \(y_p'(x)\), and \(y_p''(x)\) into the given differential equation

Substitute the expressions for \(y_p(x)\), \(y_p'(x)\), and \(y_p''(x)\) into the given differential equation: $$ (\sin x)[u_1''(x)e^{-x} - 2u_1'(x)e^{-x} + u_1(x)e^{-x} + u_2''(x)e^{-x}\cos x - 2u_2'(x)e^{-x}\cos x + 2u_2'(x)e^{-x}\sin x + u_2(x)e^{-x}\cos x - u_2(x)e^{-x}\sin x - u_2(x)e^{-x}\cos x] $$ $$ + (2\sin x - \cos x)[u_1'(x)e^{-x} - u_1(x)e^{-x} + u_2'(x)e^{-x}\cos x - u_2(x)e^{-x}\cos x + u_2(x)e^{-x}\sin x] + (\sin x - \cos x)[u_1(x)e^{-x} + u_2(x)e^{-x}\cos x] = e^{-x} $$

Simplify the differential equation and find \(u_1', u_2'\)

Now simplify the differential equation and solve for \(u_1'(x)\) and \(u_2'(x)\): $$ u_1''(x)e^{-x}\sin x - u_2'(x)e^{-x}\cos^2 x = \sin x $$ Now divide both sides by \(\sin xe^{-x}\): $$ u_1''(x) - u_2'(x)\cos^2 x = 1 $$ Next, differentiate the equation \(y_p'(x)\): $$ e^{-x}u_1'(x) - e^{-x}u_1(x) + e^{-x}\cos x u_2'(x) - e^{-x}\cos x u_2(x) + e^{-x}\sin x u_2(x) = 0 $$ Divide both sides by \(e^{-x}\): $$ u_1'(x) - u_1(x) + \cos x u_2'(x) - \cos x u_2(x) + \sin x u_2(x) = 0 $$ Now, we have a system of two equations for \(u_1'(x)\) and \(u_2'(x)\): $$ u_1''(x) - u_2'(x)\cos^2 x = 1 $$ $$ u_1'(x) - u_1(x) + \cos x u_2'(x) - \cos x u_2(x) + \sin x u_2(x) = 0 $$

Solve the system of equations for \(u_1'(x)\) and \(u_2'(x)\)

Express \(u_1'(x)\) and \(u_2'(x)\) from the system of equations: $$ u_1'(x) = u_1(x) - \cos x u_2'(x) + \cos x u_2(x) - \sin x u_2(x) $$ $$ u_2'(x) = \frac{1}{\cos^2 x}(u_1''(x) - 1) $$ Now, we need to integrate the last two equations to get \(u_1(x)\) and \(u_2(x)\). Unfortunately, these integrals do not have elementary solutions, so we must resort to numeric or approximate methods to find these functions.

Find the particular solution \(y_p(x)\)

Finally, after finding approximate \(u_1(x)\) and \(u_2(x)\) values using numerical methods, plug them into the formula for \(y_p(x)\) to obtain the particular solution: $$ y_p(x) = u_1(x)e^{-x} + u_2(x)e^{-x}\cos x $$ Although not exact, this particular solution will provide a good approximation for the given differential equation.

Key concepts

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. In practice, these functions usually represent physical quantities, the derivatives represent their rates of change, and the differential equation defines a relationship between the two. For example, if we are observing the growth rate of bacteria, the rate of growth may depend on the current population, and this relationship can be represented by a differential equation.

Differential equations come in many varieties. The order of a differential equation is determined by the highest derivative it contains. The equation can be linear or nonlinear, homogeneous or nonhomogeneous. The equation provided in our exercise is a second-order, nonhomogeneous, linear differential equation with variable coefficients. The nonhomogeneous part, in this case, represented by the term \( e^{-x} \), is what drives the need for a particular solution, which will be discussed in its respective section.

When solving these equations, it's essential to understand their structure and how methods like variation of parameters help in finding solutions when traditional methods are not applicable. This technique works well when the complementary solutions are known, which guide us in constructing the particular solution that will satisfy the entire differential equation for any given nonhomogeneous term.

Particular Solution

In the context of differential equations, a particular solution is a specific solution to a nonhomogeneous differential equation. It incorporates the nonhomogeneous part of the equation, which, unlike the homogeneous equation, does not equal zero. While the complementary solution, related to the homogeneous part of the equation, reveals the general behavior of the system, the particular solution represents the system's behavior due to the nonhomogeneous part.

In our exercise, a particular solution \( y_{p}(x) \) is sought to satisfy the overall differential equation with the given nonhomogeneous term, \( e^{-x} \). To construct a particular solution via variation of parameters, we first use the known complementary solutions (\( y_{1} \) and \( y_{2} \) from the step-by-step solution) and then find functions \( u_{1}(x) \) and \( u_{2}(x) \) that modify these solutions to account for the nonhomogeneous term. This approach leads to a system of equations that must be solved to find the modifying functions, culminating in our desired particular solution.

The key takeaway here is that the particular solution is the missing puzzle piece that complements the complementary solution to provide a full solution to the nonhomogeneous differential equation. It reflects the influence of external forces or inputs on the system described by the differential equation.

Complementary Solutions

Complementary solutions, sometimes called the homogeneous solution, to a differential equation are the solutions to the related homogeneous equation, which is the equation without its nonhomogeneous part (usually set to zero). These solutions represent the natural behavior of the system, free from external influences.

In our example, the complementary solutions are \( y_{1} = e^{-x} \) and \( y_{2} = e^{-x} \cos x \). They form a basis for the solution space of the homogeneous part of the equation, which means any linear combination of these solutions is also a solution. However, these complementary solutions do not satisfy the original nonhomogeneous equation on their own. When we introduce variation of parameters, we essentially search for a way to adjust these complementary solutions—using the functions \( u_{1}(x) \) and \( u_{2}(x) \)—so that when combined, they do satisfy the original differential equation.

Understanding complementary solutions is crucial when applying the variation of parameters method. It is these very solutions, in their modified form, that work in concert with the particular solution to address the complete behavior of the system and satisfy the overall equation.