Question

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

Short answer

Question: Find the general solution of the given differential equation: $$ y^{\prime \prime} + y^{\prime} = \cos t. $$ Answer: The general solution of the given differential equation is $$ y(t) = A + Be^{-t} - \dfrac{1}{2}\cos t + \dfrac{1}{2}\sin t, $$ where \(A\) and \(B\) are constants.

Step-by-step solution

Find the complementary function

To find the complementary function, we need to solve the homogeneous equation: $$ y^{\prime \prime} + y^{\prime} = 0. $$ This is a second-order linear homogeneous equation with constant coefficients. We start by assuming a solution of the form \(y=e^{rt}\), where \(r\) is a constant. Substituting this into the equation, we get: $$ r^2 e^{rt} + re^{rt} = 0. $$ Factor out \(e^{rt}\): $$ (r^2 + r)e^{rt} = 0. $$ As \(e^{rt}\) cannot be zero, we are left with the quadratic equation: $$ r^2 + r = 0. $$ Factoring, we get: $$ r(r+1) = 0. $$ This has the roots \(r_1 = 0\) and \(r_2 = -1\). Therefore, the complementary function is given by the linear combination of these solutions: $$ y_c(t) = Ae^{0t} + Be^{-t} = A + Be^{-t}, $$ where \(A\) and \(B\) are constants.

Find a particular integral

Now, we need to find a particular integral of the inhomogeneous equation: $$ y^{\prime \prime} + y^{\prime} = \cos t. $$ As the right side of the equation is a cosine function, we will assume a solution of the form \(y_p(t) = C\cos t + D\sin t\), where \(C\) and \(D\) are constants. Differentiating \(y_p\) with respect to \(t\), we have: $$ y_p^{\prime}(t) = -C\sin t + D\cos t $$ And differentiating once again with respect to \(t\), we obtain: $$ y_p^{\prime \prime}(t) = -C\cos t - D\sin t $$ Now, substitute \(y_p(t)\), \(y_p^{\prime}(t)\) and \(y_p^{\prime \prime}(t)\) back into the differential equation: $$ (-C\cos t - D\sin t) + (-C\sin t + D\cos t) = \cos t $$ We can rewrite this equation as: $$ (C+D)\sin t + (-C-D)\cos t = \cos t $$ For this equation to hold for all values of \(t\), we can equate the coefficients: $$ \begin{cases} C+D = 0 \\ -C-D = 1 \end{cases} $$ Solving these equations, we find \(C = -\dfrac{1}{2}\) and \(D = \dfrac{1}{2}\). Hence, the particular integral of the given differential equation is: $$ y_p(t) = -\dfrac{1}{2}\cos t + \dfrac{1}{2}\sin t $$

Combine the complementary function and particular integral

Finally, the general solution of the given differential equation is the sum of the complementary function and the particular integral: $$ y(t) = y_c(t) + y_p(t) = A + Be^{-t} - \dfrac{1}{2}\cos t + \dfrac{1}{2}\sin t $$ This is the general solution to the given differential equation.

Key concepts

Second-Order Linear Homogeneous Differential Equation

A second-order linear homogeneous differential equation is a type of equation that can be written in the form \[y'' + p(t)y' + q(t)y = 0\] where \(y\) is the unknown function of \(t\), \(y'\) is the first derivative of \(y\) with respect to \(t\), and \(y''\) is the second derivative of \(y\). Homogeneous means that the equation is set to zero. One key aspect to understand about these equations is that their solutions can be combined to form more solutions, a property known as superposition.

When solving these types of equations, we often first assume a trial solution of the form \(y = e^{rt}\). Upon substituting this into the differential equation and solving for \(r\), we find characteristic roots. These roots could be real and distinct, real and repeated, or complex conjugates. Each type of root provides a different form for the complementary function, which represents the general solution to the homogeneous equation.

Complementary Function

The complementary function, often denoted as \(y_c(t)\), is an essential concept in solving second-order linear homogeneous differential equations. It captures all possible solutions to the homogeneous part of the differential equation, which is the equation without the external forcing term (in other words, the right-hand side set to zero).

The complementary function is derived from the solution to the characteristic equation, which comes from assuming an exponential solution. For instance, if the characteristic equation's roots are \(r_1\) and \(r_2\), the complementary function is formed by a linear combination of exponential terms based on these roots. For example, if the roots are real and distinct, like in the original exercise where \(r_1 = 0\) and \(r_2 = -1\), the complementary function is \(y_c(t) = A + Be^{-t}\), with \(A\) and \(B\) being arbitrary constants that will be determined by initial conditions or boundary values.

Particular Integral

The particular integral, \(y_p(t)\), is the part of the solution to a non-homogeneous differential equation that corresponds directly to the form of the non-homogeneity—the right-hand side of the equation. Unlike the complementary function, it is not a general solution but one specific solution that makes the entire differential equation hold true.

To find the particular integral, we use the method of undetermined coefficients, involving guessing a solution based on the form of the forcing function. For example, if the forcing function is a sine or cosine, as in our original exercise, we might guess \(y_p(t) = C\cos t + D\sin t\). We then differentiate this guess, substitute it back into the differential equation, and solve for the coefficients \(C\) and \(D\) by matching the coefficients of like terms on both sides of the equation. The resulting expressions for \(C\) and \(D\) yield the particular integral.

Method of Undetermined Coefficients

The method of undetermined coefficients is a technique used for finding the particular integral of linear differential equations with constant coefficients. The key to this method is making an educated guess for the form of the particular solution based on the non-homogeneous term (forcing function) in the equation.

To apply this method correctly, one needs to identify the form of the non-homogeneous term and then propose a function with undetermined coefficients that might form a solution. This guess is informed by the type of function—for instance, if the forcing function is a polynomial, exponential, sine, or cosine. After the guess has been made, it is necessary to differentiate it as many times as the order of the differential equation and substitute these derivatives into the equation. The coefficients are found by equating the coefficients of like terms on both sides, which gives a system of equations that can be solved to determine the exact values of the coefficients. This process ensures that the particular integral satisfies the non-homogeneous equation completely.