Question

Find the general solution of the given differential equation. $$ y^{\prime \prime}-y^{\prime}-2 y=\cosh 2 t \quad \text { Hint } \cosh t=\left(e^{\prime}+e^{-t}\right) / 2 $$

Short answer

Question: Find the general solution of the given differential equation: \(y^{\prime\prime} - y^{\prime} - 2y = \cosh(2t)\) Answer: The general solution of the given differential equation is: \(y(t) = (C_1 + \frac{1}{2}) e^{2t} + C_2 e^{-t} - \frac{1}{2} e^{-2t}\), where \(C_1\) and \(C_2\) are constants.

Step-by-step solution

Solve the homogeneous differential equation

To solve the homogeneous equation, we consider the differential equation without the right-hand side: $$ y^{\prime\prime} - y^{\prime} - 2y = 0 $$ We should find the complementary function by solving this equation using the characteristic equation: $$ r^2 - r - 2 = 0 $$ This quadratic equation can be factored as (r - 2)(r + 1) = 0, which gives us two roots r1 = 2 and r2 = -1. So, the complementary function is: $$ y_c(t) = C_1 e^{2t} + C_2 e^{-t} $$ where C_1 and C_2 are constants.

Find a particular solution

Now, we need to find a particular solution for our non-homogeneous differential equation: $$ y^{\prime\prime} - y^{\prime} - 2y = \cosh(2t) $$ Using the hint provided, we know that \(\cosh(2t) = \frac{e^{2t} + e^{-2t}}{2}\). Let's assume a particular solution in the form: $$ y_p(t) = A e^{2t} + B e^{-2t} $$ where A and B are constants to be determined. Now, we need to find the first and second derivative of the particular solution and plug them back into the non-homogeneous differential equation. The first and second derivatives of the particular solution are: $$ y_p^{\prime}(t) = 2A e^{2t} - 2B e^{-2t} $$ $$ y_p^{\prime\prime}(t) = 4A e^{2t} + 4B e^{-2t} $$ Now we plug these back into our main non-homogeneous differential equation: $$ 4A e^{2t} + 4B e^{-2t} - (2A e^{2t} - 2B e^{-2t}) - 2(A e^{2t} + B e^{-2t}) = \frac{e^{2t} + e^{-2t}}{2} $$ Simplifying the equation, we get: $$ A e^{2t} - B e^{-2t} = \frac{e^{2t} + e^{-2t}}{2} $$ Comparing the exponentials, we find: $$ A = \frac{1}{2}, \quad B = - \frac{1}{2} $$ So our particular solution is: $$ y_p(t) = \frac{1}{2} e^{2t} - \frac{1}{2} e^{-2t} $$

Combine the complementary function and the particular solution

Now, we combine the complementary function and the particular solution to get the general solution of the given differential equation: $$ y(t) = y_c(t) + y_p(t) $$ Plugging in the values we got: $$ y(t) = C_1 e^{2t} + C_2 e^{-t} + \frac{1}{2} e^{2t} - \frac{1}{2} e^{-2t} $$ Therefore, the general solution of the given differential equation is: $$ y(t) = (C_1 + \frac{1}{2}) e^{2t} + C_2 e^{-t} - \frac{1}{2} e^{-2t} $$

Key concepts

Homogeneous Differential Equations

Homogeneous differential equations are a fundamental class of differential equations where the equation is set equal to zero. These equations often describe systems in physics and engineering where the input or the 'forcing function' is absent. To solve a homogeneous differential equation, one typically looks for solutions that consist of terms multiplied by constants, which are often represented by exponentials.
For instance, the homogeneous version of our original equation is written as:\[\begin{equation}y^{\text{\prime \prime}} - y^{\text{\prime}} - 2y = 0\text{\end{equation}\]}To find the solution of this equation, we apply the characteristic equation method. This involves turning the differential equation into a polynomial equation by substituting a trial solution of the form \(e^{rt}\), where \(r\) is a constant that we are trying to find. By doing so, we end up with a quadratic equation whose solutions will dictate the form of our complementary function, often denoted by \(y_c(t)\). This complementary function represents the general solution to the homogeneous equation, which, when combined with a particular solution of the non-homogeneous equation, gives the general solution to the original problem.
Understanding how to solve the homogeneous differential equation is crucial because it provides the foundation upon which we construct the complete solution when faced with a non-homogeneous equation.

Characteristic Equation

The characteristic equation is the cornerstone in solving linear homogeneous differential equations with constant coefficients. It is derived from substituting a trial solution, typically of the form \(e^{rt}\), into the homogeneous differential equation, thereby transforming the differential equation into an algebraic equation.
In our case, the characteristic equation is:\[\begin{equation}r^2 - r - 2 = 0\text{\end{equation}\]}Upon solving this quadratic equation, we get roots that are crucial to forming the complementary solution. In the textbook example, our roots r1 and r2 are 2 and -1, respectively. The significance of these roots cannot be overstated; they provide the exponents for the exponential terms in our complementary function:\[\begin{equation}y_c(t) = C_1 e^{2t} + C_2 e^{-t}\text{\end{equation}\]}where \(C_1\) and \(C_2\) are arbitrary constants. The ability to solve the characteristic equation accurately is essential as it sets the stage for the rest of the problem-solving process and ensures that the complementary function accurately represents any solution to the homogeneous problem without external influences.

Particular Solution

While the complementary function encapsulates the solution to the homogeneous equation, the particular solution, denoted by \(y_p(t)\), is specific to the non-homogeneous equation. It is sought after to satisfy the entire equation, including the non-zero right-hand side, which represents an external input or forcing function.
In our exercise, we assume a particular solution of a similar form to the forcing function, which, in this case, is the hyperbolic cosine function. The specific assumption of the solution form depends on the nature of the non-homogeneous component. Here, the form was:\[\begin{equation}y_p(t) = A e^{2t} + B e^{-2t}\text{\end{equation}\]}After determining the particular solution, we use its first and second derivatives and substitute them back into the original non-homogeneous differential equation. From there, we equate terms involving similar exponential functions on both sides of the equation to solve for the coefficients \(A\) and \(B\). The outcome is our particular solution that perfectly captures the impact of the forcing function within the dynamic being studied.
It's the combination of the particular solution and the complementary function that gives us the overall general solution. This step is vital as it ensures that our final answer addresses both the inherent dynamics of the system (incorporated in the complementary function) and the external influences represented by the non-homogeneous part of the equation.