Problem 22 - Differential Equations and Linear Algebra

Question

Use the variation-of-parameters method to find the general solution to the given differential equation. $$y^{\prime \prime \prime}-6 y^{\prime \prime}+9 y^{\prime}=12 e^{3 x}$$ Suggest a better method for solving this problem.

Step-by-Step Solution

Answer

In summary, to solve the given differential equation using the variation of parameters method: 1. The complementary function is found as \(y_c(x)=C_1+C_2 e^{3x}+C_3 xe^{3x}\). 2. We assume a particular solution \(y_p(x) = u(x) e^{3x}\) and find u(x) by integrating thrice: \(u(x) = \frac{1}{3} e^{3x} + \frac{1}{2}ax^2 + bx + c\). 3. The general solution is obtained by adding the complementary and particular solutions: \(y(x) = (C_1+C_2 e^{3x}+C_3 xe^{3x}) + \left(\frac{1}{3} + \frac{1}{2}ax^2 + bx + c\right)e^{3x}\). A better method to solve this problem could be the method of undetermined coefficients, which is more convenient for differential equations with constant coefficients.

Step 1: Find the complementary function

The given differential equation is homogeneous with constant coefficients: \[y^{\prime \prime \prime}-6 y^{\prime \prime}+9 y^{\prime}=0\] To find the complementary function, let's solve for the auxiliary equation: \[m^{3}-6m^{2}+9m = 0\] Factoring out m, we obtain: \[m(m^{2}-6m+9)=m(m-3)^{2}=0\] The roots are \(m_1=0, m_2=3,\) and \(m_3=3\). Thus, the complementary function \(y_c(x)\) is: \[y_c(x)=C_1+C_2 e^{3x}+C_3 xe^{3x}\]

Step 2: Assume a particular solution

Let's assume a particular solution in the form of \(y_p(x)= u(x) e^{3x}\), where u(x) is a function to be determined.

Step 3: Differentiate

Differentiate \(y_p(x)\) thrice: \[y^{\prime}_p(x) = u^{\prime} e^{3x}+3u e^{3x}\] \[y^{\prime \prime}_p(x) = u^{\prime \prime} e^{3x}+6u^{\prime} e^{3x}+6u e^{3x}\] \[y^{\prime \prime \prime}_p(x) = u^{\prime \prime \prime} e^{3x}+9u^{\prime \prime} e^{3x}+18u^{\prime} e^{3x}+6u e^{3x}\] Substitute these expressions into the given differential equation: \[(u^{\prime \prime \prime} +9u^{\prime \prime} +18u^{\prime} +6u) e^{3x} - 6(u^{\prime \prime}+6u^{\prime} +6u)e^{3x} + 9(u^{\prime} +3u)e^{3x} = 12 e^{3x}\]

Step 4: Solve for u(x)

Simplify the equation and solve for u(x): \[u^{\prime \prime \prime} e^{3x} = 12 e^{3x}\] Now, we need to find u(x) by integrating thrice: \[u^{\prime \prime}(x) = \int 12 e^{3x} dx = 4 e^{3x} + a\] \[u^{\prime}(x) = \int (4 e^{3x} + a) dx = \frac{4}{3} e^{3x} + ax + b\] \[u(x) = \int \left(\frac{4}{3} e^{3x} + ax + b\right) dx = \frac{1}{3} e^{3x} + \frac{1}{2}ax^2 + bx + c\] Thus, the particular solution is: \[y_p(x) = u(x) e^{3x} = \left(\frac{1}{3} + \frac{1}{2}ax^2 + bx + c\right)e^{3x}\]

Step 5: Write the general solution

Finally, the general solution is the sum of the complementary function and the particular solution: \[y(x) = y_c(x) + y_p(x) = (C_1+C_2 e^{3x}+C_3 xe^{3x}) + \left(\frac{1}{3} + \frac{1}{2}ax^2 + bx + c\right)e^{3x}\]

Step 6: Suggestion for a better method

Since the given differential equation has constant coefficients, we may find it more convenient to solve the problem using the method of undetermined coefficients. This approach allows us to directly substitute a trial function into the differential equation and solve for the constants.

Key Concepts

Third-Order Linear Differential Equation

When dealing with a third-order linear differential equation, we're looking at a formula that involves the third derivative of a function. In this case, the function we're interested in is typically denoted as \( y(x) \), and the equation has the general form \( y^{\prime\prime\prime} + a_2 y^{\prime\prime} + a_1 y^{\prime} + a_0 y = g(x) \), where \( a_0 \), \( a_1 \), and \( a_2 \) are constants, and \( g(x) \) is a known function called the nonhomogeneous term.

Our specific problem involves one step further simplification as \( a_0 \) is not present and \( g(x) \) is an exponential function. A third-order equation can describe complex physical systems and require specific methods for finding solutions. Variation-of-parameters is one such method that allows us to tackle third-order nonhomogeneous equations by considering the complementary solution and the particular solution separately.

Complementary Function

In the context of differential equations, the complementary function, often denoted as \( y_c(x) \) corresponds to the solution of the homogeneous part of the equation - basically when the nonhomogeneous term \( g(x) \) is zero. For the third-order linear differential equation, it is the solution of the equation \( y^{\prime\prime\prime} - 6y^{\prime\prime} + 9y^{\prime} = 0 \) derived from the original problem. \\ This solution is the sum of solutions derived from the roots of the auxiliary equation and includes constants \( C_1 \), \( C_2 \) and \( C_3 \) which will be determined by boundary conditions or initial values. In our exercise, we found the complementary function to be \( y_c(x) = C_1 + C_2 e^{3x} + C_3 xe^{3x} \) where the presence of the repeated root leads to the introduction of an extra \( x \) in the solution.

Method of Undetermined Coefficients

The method of undetermined coefficients is an alternative approach to the variation of parameters for solving nonhomogeneous linear differential equations with constant coefficients. It is generally easier to apply when the nonhomogeneous term \( g(x) \) is a simple function like polynomials, exponentials, sines or cosines, or a combination thereof.

The user assumes a form for the particular solution \( y_p(x) \) by examining \( g(x) \) and then determine the coefficients by substituting \( y_p(x) \) into the differential equation. For example, if \( g(x) \) were \( e^{3x} \), as in our exercise, we would propose a solution that takes the form of \( Ae^{3x} \) and find \( A \) by direct substitution.

The efficiency of this method in our case comes from the fact that the nonhomogeneous term matches the form of the assumed particular solution, which simplifies the constant's determination.

Auxiliary Equation

Last but not least, let's talk about the auxiliary equation, which plays a central role in finding the complementary function for linear differential equations with constant coefficients. The auxiliary equation, sometimes also called the characteristic equation, is a polynomial equation obtained by substituting \( y = e^{mx} \) into the homogeneous differential equation and then equating the resulting polynomial in \( m \) to zero.

The roots of this polynomial correspond to the exponential solutions of the homogeneous differential equation. In the context of our problem, the auxiliary equation is \( m^3 - 6m^2 + 9m = 0 \) and it factors to \( m(m-3)^2 = 0 \), indicating that we have a zero root and a repeated root of multiplicity two. These roots guide us in constructing the general form of the complementary function, taking into account that repeated roots introduce additional terms involving powers of \( x \). It's fascinating how the roots of a simple polynomial can dictate the shape of the solutions to a differential equation!