Chapter 7: Problem 74
Solve the differential equation using either the method of undetermined coefficients or the variation of parameters. $$y^{\prime \prime}+6 y^{\prime}+9 y=e^{-x}$$
Short Answer
Expert verified
The solution is \( y = C_1 e^{-3x} + C_2 x e^{-3x} + \frac{1}{4}xe^{-x} \).
Step by step solution
01
Identify the Type of Differential Equation
The given differential equation is \( y'' + 6y' + 9y = e^{-x} \). It is a second-order linear non-homogeneous differential equation with constant coefficients.
02
Find the Complementary Solution
To find the complementary solution, solve the associated homogeneous equation \( y'' + 6y' + 9y = 0 \). The characteristic equation is \( r^2 + 6r + 9 = 0 \), which can be written as \((r+3)^2 = 0 \). Thus, \( r = -3 \) is a repeated root.
03
Write the Complementary Solution
For a repeated root \( r = -3 \), the complementary solution is \( y_c = C_1 e^{-3x} + C_2 x e^{-3x} \), where \( C_1 \) and \( C_2 \) are arbitrary constants.
04
Identify the Method of Undetermined Coefficients
Since the right-hand side of the equation is \( e^{-x} \), a form that is suitable for the method of undetermined coefficients, choose this method to find the particular solution.
05
Choose the Form of the Particular Solution
Propose a particular solution of the form \( y_p = Axe^{-x} \) because \( e^{-x} \) is part of the complementary solution's form. Adjust the form by multiplying by \( x \) due to the overlap.
06
Differentiate the Particular Solution
Differentiate \( y_p = Axe^{-x} \) to find \( y_p' = A(e^{-x} - xe^{-x}) \) and \( y_p'' = A(-2e^{-x} + xe^{-x}) \).
07
Substitute into the Original Equation
Substitute \( y_p, y_p', y_p'' \) into the differential equation: \( (A(-2e^{-x} + xe^{-x})) + 6(A(e^{-x} - xe^{-x})) + 9(Axe^{-x}) = e^{-x} \).
08
Simplify and Solve for A
Simplify the expression: \( -2A e^{-x} + Axe^{-x} + 6A e^{-x} - 6Axe^{-x} + 9Axe^{-x} = e^{-x} \). Collect like terms: \( (4A)e^{-x} + (4Ax)e^{-x} = e^{-x} \). Equate coefficients: \( 4A = 1 \) and \( 4Ax = 0 \). Thus, \( A = \frac{1}{4} \).
09
Write the Particular Solution
Thus, the particular solution is \( y_p = \frac{1}{4}xe^{-x} \).
10
Write the General Solution
The general solution of the differential equation is the sum of the complementary and particular solutions: \( y = C_1 e^{-3x} + C_2 x e^{-3x} + \frac{1}{4}xe^{-x} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Method of Undetermined Coefficients
The method of undetermined coefficients is a technique used to find a particular solution to a non-homogeneous linear differential equation. This method works well when the non-homogeneous term (right-hand side of the equation) is a simple function, such as a polynomial, an exponential, or a trigonometric function. The key idea is to assume a form for the particular solution that resembles the non-homogeneous term. For instance, if the non-homogeneous term is of the form \(e^{-x}\), then the guessed particular solution \(y_p\) could initially be \(Ae^{-x}\), where \(A\) is a constant to be determined.
However, if a term like \(e^{-x}\) already appears in the complementary solution, as is the case here, we adjust by incorporating a factor of \(x\) to prevent overlap, leading to the form \(y_p = Axe^{-x}\). The coefficients such as \(A\) are then determined by substituting the proposed solution into the original differential equation and solving for the unknown coefficients.
However, if a term like \(e^{-x}\) already appears in the complementary solution, as is the case here, we adjust by incorporating a factor of \(x\) to prevent overlap, leading to the form \(y_p = Axe^{-x}\). The coefficients such as \(A\) are then determined by substituting the proposed solution into the original differential equation and solving for the unknown coefficients.
Complementary Solution
The complementary solution of a differential equation refers to the solution of the associated homogeneous equation. This type of solution specifically addresses the equation without the non-homogeneous term, allowing us to understand the natural behavior of the system without external forces.
For the given homogeneous equation \(y'' + 6y' + 9y = 0\), we solve the characteristic equation derived from this differential equation. The characteristic equation is formed as \(r^2 + 6r + 9 = 0\). On solving, it simplifies to \((r+3)^2 = 0\), indicating that \(r = -3\) is a repeated root.
In cases of repeated roots, the complementary solution is expressed as \(y_c = C_1 e^{-3x} + C_2 x e^{-3x}\), where \(C_1\) and \(C_2\) are constants determined by initial conditions. The inclusion of the term \(C_2 x e^{-3x}\) compensates for the repeated nature of the root, ensuring linear independence between terms.
For the given homogeneous equation \(y'' + 6y' + 9y = 0\), we solve the characteristic equation derived from this differential equation. The characteristic equation is formed as \(r^2 + 6r + 9 = 0\). On solving, it simplifies to \((r+3)^2 = 0\), indicating that \(r = -3\) is a repeated root.
In cases of repeated roots, the complementary solution is expressed as \(y_c = C_1 e^{-3x} + C_2 x e^{-3x}\), where \(C_1\) and \(C_2\) are constants determined by initial conditions. The inclusion of the term \(C_2 x e^{-3x}\) compensates for the repeated nature of the root, ensuring linear independence between terms.
Particular Solution
Finding the particular solution to a differential equation involves addressing the non-homogeneous part of the equation. The particular solution accounts for the external inputs or forces reflected in the non-homogeneous term. In our example, the non-homogeneous component is \(e^{-x}\). By employing the method of undetermined coefficients, we proposed a particular solution of the form \(y_p = Axe^{-x}\), suitably adjusted to accommodate repetition in the complementary function.
To establish the particular solution, substitute \(y_p\) along with its derivatives back into the original differential equation to determine the unknown coefficients. Simplifying and equating coefficients yields the constant \(A = \frac{1}{4}\). This gives us the particular solution \(y_p = \frac{1}{4}xe^{-x}\).
Finally, the general solution of the differential equation integrates both the complementary and particular solutions, written as \(y = C_1 e^{-3x} + C_2 x e^{-3x} + \frac{1}{4}xe^{-x}\). This comprehensive solution encompasses both natural and forced responses of the system.
To establish the particular solution, substitute \(y_p\) along with its derivatives back into the original differential equation to determine the unknown coefficients. Simplifying and equating coefficients yields the constant \(A = \frac{1}{4}\). This gives us the particular solution \(y_p = \frac{1}{4}xe^{-x}\).
Finally, the general solution of the differential equation integrates both the complementary and particular solutions, written as \(y = C_1 e^{-3x} + C_2 x e^{-3x} + \frac{1}{4}xe^{-x}\). This comprehensive solution encompasses both natural and forced responses of the system.