Question

In each exercise, (a) Find the general solution of the differential equation. (b) If initial conditions are specified, solve the initial value problem. $$ y^{(4)}-y=0 $$

Short answer

Answer: The general solution of the given differential equation is: $$ y(x) = c_1e^{x} + c_2e^{-x} + c_5\cos(x) + c_6\sin(x) $$ where \(c_1, c_2, c_5\), and \(c_6\) are constants.

Step-by-step solution

Find the Characteristic Equation

To find the characteristic equation, we will replace the derivatives in the given equation, \(y^{(4)} - y = 0\), with their respective powers of \(r\). That is: $$ r^4 - 1 = 0 $$ This is the characteristic equation related to the given differential equation.

Find the Roots of the Characteristic Equation

Now we need to find the roots of the characteristic equation, which looks like a quartic equation. It can be rewritten as: $$ (r^4 - 1) = (r^2 + 1)(r^2 - 1) = (r^2 + 1)(r - 1)(r + 1) $$ The roots of this equation are \(r_1 = 1, r_2 = -1, r_3 = i, r_4 = -i\).

Write down the General Solution

With the roots of the characteristic equation found, we can now write the general solution for the given differential equation as a linear combination of the form (principles of superposition): $$ y(x) = c_1e^{r_1x} + c_2e^{r_2x} + c_3e^{r_3x} + c_4e^{r_4x} $$ Substitute the roots to get: $$ y(x) = c_1e^{x} + c_2e^{-x} + c_3e^{ix} + c_4e^{-ix} $$ Since the differential equation is real, the general solution must also be real. To achieve this, we rewrite the solution given the complex exponential terms, using Euler's formula: $$ y(x) = c_1e^{x} + c_2e^{-x} + c_3(\cos(x) + i\sin(x)) + c_4(\cos(-x) + i\sin(-x)) $$ Next, we simplify the cosine and sine terms: $$ y(x) = c_1e^{x} + c_2e^{-x} + (c_3 + c_4)\cos(x) + (c_3 - c_4)i\sin(x) $$ From here, we introduce two new constants \(c_5 = c_3 + c_4\) and \(c_6 = c_3 - c_4\) to get our final general solution: $$ y(x) = c_1e^{x} + c_2e^{-x} + c_5\cos(x) + c_6\sin(x) $$

Solve the Initial Value Problem (if applicable)

In case initial conditions are provided, we can utilize them to find the specific values of the constants \(c_1, c_2, c_5, c_6\) in order to solve the initial value problem. For this exercise, initial conditions are not given, so the general solution is our final answer.

Key concepts

Initial Value Problem

When we come across a differential equation, often we're provided with initial conditions, which serve as a starting point to find a unique solution. An initial value problem pairs a differential equation with a set of initial conditions. Imagine a car's journey being described by its velocity; to know its exact position, we also need its starting point. Similarly, to pin down a specific solution to our differential equation, we need to know where to start - that is, the initial value. When initial conditions are not specified, we can only talk about the general solution, which contains arbitrary constants representative of the family of all potential solutions.

Characteristic Equation

Diving into the heart of solving linear homogeneous differential equations, we encounter the characteristic equation. Think of it as a detective's clue to catch the potential culprits, the solutions to our problem. By substituting derivatives with powers of some variable (usually r), we transform our differential equation into an algebraic one. This change of form simplifies the complexity of the differential equation, revealing a clearer path towards the general solution.

Roots of the Characteristic Equation

The solution to our detective story unfolds by finding the roots of the characteristic equation. These roots are the suspects we were after - the key values that will shape the form of our general solution. Just as in a line-up, each root contributes to the final outcome, with each one corresponding to a particular part of the solution. Finding these roots might involve factoring or using more sophisticated algebraic techniques, depending on the complexity of the equation.

General Solution of Differential Equation

After catching our culprits, the roots, it's time to meet the general solution of the differential equation. It's the mathematical equivalent of a sketch artist's drawing, merging all possibilities into a single portrayal. The general solution integrates all the components associated with the roots we found, creating a formula that encompasses all possible specific solutions. Expressing the general solution often involves a sum of functions, each multiplied by an arbitrary constant that reflects the diverse range of solutions this equation may have.

Euler's Formula

Sometimes the roots of the characteristic equation are complex, and this is where Euler's formula steps in, a stroke of mathematical genius that connects the exponential function to trigonometric functions. Euler's formula, expressed as \( e^{ix} = \text{cos}(x) + i\text{sin}(x) \), allows us to rewrite complex exponentials in terms of cosine and sine, providing a real representation for the solutions to our differential equation. This beautiful equation reminds us that even in complexity, there's harmony and elegance to be found.