Question

For the given differential equation, $$ y^{\prime \prime}+4 y=2 \sinh t \cosh t+\cosh ^{2} t $$

Short answer

Answer: The general solution is given by \(y(t) = A\cos{2t} + B\sin{2t} + 2\sinh t - \frac{1}{2}\cosh t\), where A and B are arbitrary constants.

Step-by-step solution

Find the homogeneous solution

First, we'll find the solution to the homogeneous differential equation by setting the RHS to zero: $$ y'' + 4y = 0. $$ This is a second-order linear homogeneous differential equation. To solve it, we guess a solution of the form \(y(t) = e^{rt}\), where r is a constant to be determined. So, we have: $$ y'(t) = re^{rt}, \quad y''(t) = r^2e^{rt}. $$ Now, we substitute this into our homogeneous differential equation: $$ r^2e^{rt} + 4e^{rt} = 0. $$ Factoring out \(e^{rt}\), we obtain $$ (r^2 + 4)e^{rt} = 0. $$ Since \(e^{rt}\) can never be zero, we only need to solve the equation \(r^2 + 4 = 0\). We find that \(r = \pm 2i\). Therefore, the homogeneous solution is given by: $$ y_h(t) = A\cos{2t} + B\sin{2t}, $$ where A and B are arbitrary constants.

Find the particular solution

To find a particular solution \(y_p(t)\), we will use the method of undetermined coefficients. Seeing that the RHS contains hyperbolic sine and cosine functions, we will assume a solution of the form $$ y_p(t) = C\sinh t + D\cosh t. $$ Now we find the first and second derivatives of our guess: $$ y_p'(t) = C\cosh t + D\sinh t, \quad y_p''(t) = C\sinh t + D\cosh t. $$ Plugging \(y_p(t)\), \(y_p'(t)\), and \(y_p''(t)\) into our original differential equation, we get $$ (C\sinh t + D\cosh t) + 4(C\sinh t + D\cosh t) = 2\sinh t \cosh t + \cosh^2 t. $$ Simplifying and equating coefficients, we obtain the following system of equations: 1. \(C + 4D = 0\) 2. \(C + 4D = 2\) Solving this system of equations, we find that \(C = 2\) and \(D = -\frac{1}{2}\). Thus, our particular solution is: $$ y_p(t) = 2\sinh t - \frac{1}{2}\cosh t. $$

Determine the general solution

Now that we have both the homogeneous and particular solutions, we can add them together to obtain the general solution. This is given by: $$ y(t) = y_h(t) + y_p(t) = A\cos{2t} + B\sin{2t} + 2\sinh t - \frac{1}{2}\cosh t. $$ This is the general solution to the given differential equation.

Key concepts

Homogeneous Differential Equation

At the heart of solving second-order linear differential equations is understanding homogeneous differential equations. These equations take on a general form like \[ ay'' + by' + cy = 0 \]where the terms on the right-hand side equal zero. The key feature of a homogeneous equation is that any linear combination of its solutions is also a solution—a property known as the superposition principle. To solve these equations, we often look for solutions that contain exponential functions, sine, or cosine, depending on the nature of the characteristic equation's roots. If the roots are complex, as in the exercise, sine and cosine functions come to the fore, due to Euler's formula relating exponential functions to trigonometric functions.

Specifically, the exercise provided has a second-order linear homogeneous differential equation of the form \[ y'' + 4y = 0 \]where the solution comprises a combination of trigonometric functions because the characteristic equation's roots are imaginary numbers.

Method of Undetermined Coefficients

The method of undetermined coefficients is a powerful technique for finding a particular solution to non-homogeneous differential equations. It works best when the non-homogeneous part (the right-hand side of the equation) is a type of function for which the method is suited, such as polynomials, exponentials, sines, cosines, and their products. When applying this method, you make an educated guess of the solution's form, referred to as the ansatz, by looking at the type of functions present in the non-homogeneous part.

In our exercise, the presence of hyperbolic sine and cosine functions guides us to guess a form similar to them. Then, we determine unknown coefficients by substituting the ansatz into the differential equation and matching the coefficients on both sides. It's a bit like solving a puzzle where matching the correct pieces reveals the shape of a particular solution.

Particular Solution

A particular solution to a differential equation is specific to the non-homogeneous part—it doesn't include the arbitrary constants found in the homogeneous solution. In essence, it 'fits' the non-homogeneous equation once, without aiming for the generality of the homogeneous solution. Determining the particular solution is essential because it ensures that the overall general solution works for any non-homogeneous term present.

When added to the homogeneous solution, the particular solution makes up the general solution to the entire differential equation. The exercise involved using the method of undetermined coefficients to arrive at a particular solution of the form \[ y_p(t) = 2\text{sinh}(t) - \frac{1}{2}\text{cosh}(t) \]which, when combined with the homogeneous solution, gives us the complete picture and enables us to solve for any given initial conditions.

Hyperbolic Functions

Hyperbolic functions, including hyperbolic sine (\text{sinh}) and hyperbolic cosine (\text{cosh}), show up in differential equations and many areas of mathematics. They resemble trigonometric functions, but they relate to hyperbolas, not circles. Hyperbolic functions are defined using exponential functions:

• \[ \text{sinh}(x) = \frac{e^x - e^{-x}}{2} \]
• \[ \text{cosh}(x) = \frac{e^x + e^{-x}}{2} \]

These functions have useful properties like their trigonometric counterparts—for example, the hyperbolic identity \[ \text{cosh}^2(x) - \text{sinh}^2(x) = 1 \]reflects the Pythagorean identity in trigonometry. In our exercise, the non-homogeneous part involves hyperbolic functions, and so our particular solution incorporates them, mirroring the way trigonometric functions would be included if the non-homogeneous part were periodic.