Question

Use the method of variation of parameters to find a particular solution of the given differential equation. Then check your answer by using the method of undetermined coefficients. $$ y^{\prime \prime}+2 y^{\prime}+y=3 e^{-t} $$

Short answer

Question: Find a particular solution of the second-order differential equation \(y^{\prime\prime} + 2y^{\prime} + y = 3e^{-t}\) using both the method of variation of parameters and undetermined coefficients. Verify that both methods give the same result. Answer: Using both the method of variation of parameters and undetermined coefficients, the particular solution is found to be \(y_p(t) = 3e^{-t}\). Both methods yield the same result, confirming the correctness of the solution.

Step-by-step solution

Identify the homogeneous equation and find its complementary solution

Firstly, we need to identify the homogeneous equation, which can be written as: $$ y^{\prime \prime} + 2y^{\prime} + y = 0 $$ To find the complementary solution, we need to solve the characteristic equation: $$ r^2 + 2r + 1 = 0 $$ This equation can be factored into \((r + 1)^2 = 0\), which gives us a single root, \(r = -1\). Therefore, the complementary solution is: $$ y_c(t) = C_1 e^{-t} + C_2 te^{-t} $$

Apply the method of variation of parameters

Now we will use the method of variation of parameters to find a particular solution of the non-homogeneous equation. We assume a particular solution of the form: $$ y_p(t) = u_1(t) e^{-t} + u_2(t) te^{-t} $$ Take the first and second derivatives of \(y_p(t)\): $$ y_p^{\prime}(t) = u_1'(t) e^{-t} - u_1(t) e^{-t} + u_2'(t) te^{-t} + e^{-t}u_2(t) - te^{-t}u_2(t) $$ $$ y_p^{\prime\prime}(t) = u_1^{\prime\prime}(t)e^{-t} + 2 u_1^{\prime}(t)e^{-t} - 2u_1^{\prime}(t)e^{-t} - u_1(t)e^{-t} + u_2^{\prime\prime}(t)te^{-t} + 2u_2^{\prime}(t)e^{-t} - 2u_2(t)e^{-t} - 2u_2^{\prime}(t)e^{-t} + u_2(t)e^{-t} $$ Substitute \(y_p^{\prime\prime}(t)\), \(y_p^{\prime}(t)\), and \(y_p(t)\) into the original non-homogeneous equation, and we get: $$ u_1^{\prime\prime}(t)e^{-t} + u_2^{\prime\prime}(t)te^{-t} = 3 e^{-t} $$

Solve for \(u_1^{\prime}(t)\) and \(u_2^{\prime}(t)\)

To solve for \(u_1^{\prime}(t)\) and \(u_2^{\prime}(t)\), we will use the following conditions: $$ u_1^{\prime\prime}(t)e^{-t} = 0 $$ $$ u_2^{\prime\prime}(t)te^{-t} = 3e^{-t} $$ Integrate both equations to get \(u_1^{\prime}(t)\) and \(u_2^{\prime}(t)\): $$ u_1^{\prime}(t) = C_1 $$ $$ u_2^{\prime}(t) = 3 $$

Integrate again to solve for \(u_1(t)\) and \(u_2(t)\)

Integrate the above equations again to find \(u_1(t)\) and \(u_2(t)\): $$ u_1(t) = C_1t + C_2 $$ $$ u_2(t) = 3t + C_3 $$ Substitute back into the assumed particular solution: $$ y_p(t) = (C_1t + C_2) e^{-t} + (3t + C_3) te^{-t} $$

Apply the method of undetermined coefficients

To verify the solution, we will now use the method of undetermined coefficients. Assume a particular solution of the form: $$ y_p(t) = Ae^{-t} $$ Then the first and second derivatives are: $$ y_p^{\prime}(t) = -Ae^{-t} $$ $$ y_p^{\prime\prime}(t) = Ae^{-t} $$ Substitute these back into the original non-homogeneous equation: $$ Ae^{-t} - 2Ae^{-t} + Ae^{-t} = 3e^{-t} $$ Solve for \(A\), we get \(A = 3\). Therefore, the particular solution is: $$ y_p(t) = 3e^{-t} $$

Compare the solutions

Comparing the results from both methods, we see that the particular solutions are the same, \(3e^{-t}\). This confirms that our solution is correct and both methods yield the same particular solution.

Key concepts

Differential Equations

Differential equations are mathematical equations that involve an unknown function and its derivatives. They are used to describe various phenomena in physics, engineering, biology, and economics. When solving a differential equation, the goal is to find the function that satisfies the given relationship between the function and its derivatives. In the context of this exercise, we are working with a second-order linear differential equation.
In general, solutions to differential equations are found in two parts: the complementary (or homogeneous) solution, which solves the associated homogeneous equation (where the right-hand side is zero), and the particular (or non-homogeneous) solution, which is a specific solution to the non-homogeneous equation (with the non-zero term on the right-hand side). The overall solution to the differential equation is the sum of these two parts. To improve clarity and understanding, focusing on the distinction between homogeneous and non-homogeneous parts is crucial.

Method of Undetermined Coefficients

The 'method of undetermined coefficients' is a straightforward technique often utilized to find the particular solution of linear non-homogeneous differential equations with constant coefficients. The method is best applied when the non-homogeneous part (the right-hand side of the equation) is a simple function, such as a polynomial, an exponential, or a sine or cosine function.
To use this method, we make a guess about the form of the particular solution based on the non-homogeneous part of the differential equation, and then solve for the coefficients that make it work. It's like fitting puzzle pieces into pre-defined slots. In the sample problem, our guess for the particular solution is based on the exponential form of the source term, and by substituting this guess into the equation, we determine the undetermined coefficient. When writing about this method, explaining the logic behind choosing the guess and linking it to common source functions can greatly enhance comprehension.

Characteristic Equation

The 'characteristic equation' is integral to solving homogeneous linear differential equations with constant coefficients. It's a bridge to transform the differential equation into an algebraic one, which we can solve more comfortably.
By substituting an exponential trial solution into the homogeneous part of the differential equation, we arrive at the characteristic equation. The roots of this equation provide insight into the form of the complementary solution. For example, if the characteristic equation yields real, distinct roots, the complementary solution will consist of exponential terms corresponding to these roots. When the roots are complex or repeated, as in the case of the provided exercise, we have to adapt our approach and include sines and cosines for complex roots or polynomial corrections for repeated roots. Illustrating how the roots of the characteristic equation determine the structure of the complementary solution is an effective strategy to demystify this concept.