Question

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

Short answer

Question: Find the general solution of the given differential equation $$y^{\prime \prime \prime} - 4y^{\prime} = 0 $$. Answer: The general solution of the given differential equation is $$y(x) = C_1 + C_2 e^{2x} + C_3 e^{-2x} $$, where $$C_1$$, $$C_2$$, and $$C_3$$ are constants.

Step-by-step solution

Find the characteristic equation

For the given differential equation, we have: $$ y^{\prime \prime \prime}-4 y^{\prime} = 0 $$ To find the characteristic equation, we will replace the highest order derivative with $$r^3$$, second-highest order derivative with $$r^2$$, and so on. In this case, we obtain: $$ r^3 - 4r = 0 $$

Solve the characteristic equation

Now, we will solve the equation $$ r^3 - 4r = 0 $$ for r: $$ r(r^2 - 4) = 0 $$ This equation can be further factored as: $$ r(r-2)(r+2) = 0 $$ The roots of the equation are: $$ r_1 = 0, r_2 = 2, r_3 = -2 $$

Write the general solution

Based on the roots obtained, the general solution of the differential equation can be written as: $$ y(x) = C_1 e^{0x} + C_2 e^{2x} + C_3 e^{-2x} $$ Simplifying, we get: $$ y(x) = C_1 + C_2 e^{2x} + C_3 e^{-2x} $$ If no initial conditions are provided, this is the general solution to the differential equation.

(Optional) Step 4: Solve the initial value problem (if applicable)

If initial conditions are provided, say $$ y(0) = y_0, y'(0) = y_0', y''(0) = y_0'' $$, we can use them to find the values of $$ C_1, C_2 $$ and $$ C_3 $$. Substitute the initial conditions into the general solution and its derivatives, and solve the resulting system of equations for $$ C_1, C_2 $$, and $$ C_3 $$. This will give the specific solution to the given initial value problem.

Key concepts

Characteristic Equation

Understanding the characteristic equation is essential for solving higher-order linear homogeneous differential equations like the one in our exercise. But what is a characteristic equation? It is a tool we use to find the possible values of 'r' that make a certain equation, derived from our differential equation, equal to zero.

Let's break down the steps you'd follow to solve for the characteristic equation using our example, which is given by the differential equation: \(y^{\textprime \textprime \textprime}-4y^{\textprime}=0\). To form the characteristic equation, we systematically replace each derivative of y with powers of r. The third derivative, \(y^{\textprime \textprime \textprime}\), is replaced by \(r^3\), and the first derivative, \(y^{\textprime}\), by r. This substitution leads us to the algebraic equation \(r^3 - 4r = 0\), which is our characteristic equation.

Solving for r in the characteristic equation gives us the roots of the equation—critical values that help us to construct the general solution. Making these algebraic manipulations more approachable can be a turning point for many students wrapping their heads around differential equations.

Initial Value Problem

An initial value problem (IVP) is a type of differential equation along with specific values, called initial conditions, provided for the unknown function at a given point. These initial conditions allow us to find the one particular solution that not only satisfies the differential equation but also passes through the given points in the function's domain.

In our exercise, once we've found the general solution to the differential equation, we would move on to use the initial conditions to solve the IVP—if such conditions were provided. For instance, conditions like \(y(0) = y_0\), \(y^{\textprime}(0) = y_0^{\textprime}\), \(y^{\textprime\textprime}(0) = y_0^{\textprime\textprime}\) would give us specific values to plug into our general solution. By doing so, we can solve for the constants in the general solution, thus tailoring it to fit the initial conditions. This step is vital for real-world applications where specific behavior of the solution is crucial.

General Solution of Differential Equation

The general solution of a differential equation is a formula that includes all possible solutions of the equation. It incorporates arbitrary constants, which represent the infinite number of solutions that a differential equation may have. Take our sample exercise, where finding the general solution is our objective. Once we've solved the characteristic equation and found the roots, we use them to construct the general solution.

In our case, the roots \(r_1 = 0\), \(r_2 = 2\), and \(r_3 = -2\) lead us to a general solution of the form \(y(x) = C_1 e^{0x} + C_2 e^{2x} + C_3 e^{-2x}\). Here, \(C_1\), \(C_2\), and \(C_3\) are the arbitrary constants. Simplifying the terms, particularly \(e^{0x}\) which is equal to 1, we get \(y(x) = C_1 + C_2 e^{2x} + C_3 e^{-2x}\). This elegant representation of the general solution neatly holds the key to all particular solutions dictated by initial conditions.

Being able to seamlessly integrate the characteristic roots to form such a solution is a skill that's foundational in the study of differential equations. The strength of the general solution lies in its versatility - it's the master key that has the potential to unlock every door in the domain of its associated differential equation.