Question

For the given differential equation, $$ y^{\prime \prime}+2 y^{\prime}+2 y=\cos t+e^{-t} $$

Short answer

Answer: The general solution to the given differential equation is $\displaystyle y(t) = (C_1 e^{-t} \cos t + C_2 e^{-t} \sin t) -\frac{1}{5} \cos t - \frac{2}{5} \sin t + \frac{1}{2} e^{-t}$, where $C_1$ and $C_2$ are constants.

Step-by-step solution

Solve the complementary equation

We will first find the complementary function to the given inhomogeneous differential equation. The complementary equation is given by: $$ y^{\prime \prime}+2 y^{\prime}+2 y=0 $$ To solve this homogeneous equation, consider the characteristic equation given by: $$ r^2 + 2r + 2 = 0 $$ This equation has two complex roots, \(r_1 = -1 + i\) and \(r_2 = -1 - i.\) Hence, the complementary function is given by: $$ y_c(t) = C_1 e^{-t} \cos t + C_2 e^{-t} \sin t $$ where \(C_1\) and \(C_2\) are constants to be determined from the initial conditions.

Finding a particular solution

Given the inhomogeneous right-hand side (\(\cos t + e^{-t}\)), we will use the method of undetermined coefficients to find a particular solution (guess an appropriate form with undetermined coefficients and solve for those coefficients). We see that our guess for a particular solution could be in the form: $$ y_p(t) = A\cos t + B\sin t + Ce^{-t} $$ Taking the first derivative, we get: $$ y_p^{\prime}(t) = -A\sin t + B\cos t - Ce^{-t} $$ Taking the second derivative, we get: $$ y_p^{\prime\prime}(t) = -A\cos t - B\sin t + Ce^{-t} $$ Now substitute \(y_p(t),\) \(y_p^{\prime}(t),\) and \(y_p^{\prime\prime}(t)\) into the given inhomogeneous differential equation: $$ (-A\cos t - B\sin t + Ce^{-t}) + 2(-A\sin t + B\cos t - Ce^{-t}) + 2(A\cos t + B\sin t + Ce^{-t}) = \cos t + e^{-t} $$ This simplifies to: $$ (-A + 2B) \cos t + (-B - 2A) \sin t + (2C)e^{-t} = \cos t + e^{-t} $$ Comparing coefficients, we have: $ \begin{cases} -A + 2B=1\\ -2A - B=0\\ 2C=1 \end{cases} $ By solving these equations, we find \(A = -\frac{1}{5}\), \(B = -\frac{2}{5}\), and \(C=\frac{1}{2}\). Thus, the particular solution is given by: $$ y_p(t) = -\frac{1}{5} \cos t - \frac{2}{5} \sin t + \frac{1}{2} e^{-t} $$

Finding the general solution

Now, we can combine our complementary function and our particular solution to find the general solution to the given differential equation. The general solution is given by: $$ y(t) = y_c(t) + y_p(t) = (C_1 e^{-t} \cos t + C_2 e^{-t} \sin t) + (-\frac{1}{5} \cos t - \frac{2}{5} \sin t + \frac{1}{2} e^{-t}) $$ This is the general solution to the given differential equation.

Key concepts

Complementary Function

In differential equations, the complementary function represents the solution to the associated homogeneous equation. Given the inhomogeneous differential equation, we remove the non-homogeneous part and solve for what is known as the complementary function or homogeneous solution.
The complementary function, denoted as \(y_c(t)\), solves the equation without the external force or input, meaning we focus on solving:

• \(y^{\prime \prime} + 2y^{\prime} + 2y = 0\)

To find this solution, convert it into a characteristic equation using trial solutions like \(y(t) = e^{rt}\), yielding:

• \(r^2 + 2r + 2 = 0\)

The roots of this equation are complex: \(r_1 = -1 + i\) and \(r_2 = -1 - i\). This gives rise to a complementary function in the form:

• \(y_c(t) = C_1 e^{-t} \cos t + C_2 e^{-t} \sin t\)

The functions \(e^{-t} \cos t\) and \(e^{-t} \sin t\) capture the natural response of the system, while \(C_1\) and \(C_2\) are constants to be defined given specific initial conditions.

Particular Solution

The particular solution of an inhomogeneous differential equation corresponds to the specific answer to the non-homogeneous part of the equation. Unlike the complementary function, which solves the homogeneous part, the particular solution addresses the unique external inputs or forces.
For the equation \(y^{\prime \prime} + 2y^{\prime} + 2y = \cos t + e^{-t}\), the particular solution \(y_p(t)\) is assumed to have a form resembling the inhomogeneous terms:

• \(y_p(t) = A\cos t + B\sin t + Ce^{-t}\)

By substituting \(y_p(t)\), along with its derivatives, back into the differential equation, we align the coefficients of like terms. This leads to a system of equations:

• \(-A + 2B = 1\)
• \(-2A - B = 0\)
• \(2C = 1\)

Solving these equations provides the values for \(A\), \(B\), and \(C\): \(-\frac{1}{5}\), \(-\frac{2}{5}\), and \(\frac{1}{2}\), respectively. Hence, the particular solution is:

• \(y_p(t) = -\frac{1}{5} \cos t - \frac{2}{5} \sin t + \frac{1}{2} e^{-t}\)

Inhomogeneous Differential Equation

An inhomogeneous differential equation is characterized by having a non-zero right-hand side, typically representing external forces or inputs. This type of equation can usually be expressed in the form:

• \(Ly = f(t)\)

Here, \(L\) is a differential operator, and \(f(t)\) is a non-zero function representing sources of inhomogeneity. The given equation \(y^{\prime \prime} + 2y^{\prime} + 2y = \cos t + e^{-t}\) is typical of this form.
Inhomogeneous equations require both the particular solution and the complementary function for their complete solution. The complementary function addresses the homogeneous part, while the particular solution tackles the specific characteristics of the non-homogeneous inputs \(\cos t\) and \(e^{-t}\).
This allows us to model systems where, alongside their natural behavior captured by the homogeneous part, there is an external driving force or input, making the inhomogeneous differential equation a more realistic representation of many physical systems.

Method of Undetermined Coefficients

The method of undetermined coefficients is a systematic technique used to find the particular solution of inhomogeneous linear differential equations. The approach involves "guessing" a specific form for the particular solution based on the nature of the non-homogeneous term and then determining the coefficients by substitution.
For our equation, \(y^{\prime \prime} + 2y^{\prime} + 2y = \cos t + e^{-t}\), we recognize the right-hand side terms and decide a trial solution:

• \(y_p(t) = A\cos t + B\sin t + Ce^{-t}\)

The key steps are:

• Assuming the form of \(y_p(t)\) that includes terms similar to \(\cos t\) and \(e^{-t}\).
• Taking derivatives of this assumed form and substituting them into the original equation.
• Matching coefficients of like terms from both sides of the equation to solve for the undetermined coefficients \(A, B,\) and \(C\).

The method excels at handling equations where the non-homogeneous part has simple polynomial, exponential, or trigonometric forms. This approach simplifies finding a particular solution without resorting to complex integrals.