Question

In each exercise, (a) Find the general solution of the differential equation. (b) If initial conditions are specified, solve the initial value problem. $$ 2 y^{\prime \prime \prime}-y^{\prime \prime}=0 $$

Short answer

Question: Determine the general solution to the third-order linear homogeneous differential equation: \(2 y^{\prime \prime \prime}-y^{\prime \prime}=0\) Answer: The general solution to the given differential equation is \(y(x) = C_1 + C_2 e^{\frac{1}{2}x} + C_3 x e^{\frac{1}{2}x}\), where \(C_1\), \(C_2\), and \(C_3\) are arbitrary constants.

Step-by-step solution

Assume a general form of the solution

Let's assume a general solution of the given differential equation: $$y(x) = e^{rx}$$ where r is a constant.

Calculate its derivatives

Now, we need to find the first, second, and third derivatives of this assumed solution. First derivative: $$y'(x) = (e^{rx})' = r e^{rx}$$ Second derivative: $$y''(x) = (re^{rx})' = r^2 e^{rx}$$ Third derivative: $$y'''(x) = (r^2e^{rx})' = r^3 e^{rx}$$

Substitute the derivatives into the given equation

Now, we need to substitute these derivatives back into the given differential equation: $$2 y^{\prime \prime \prime}-y^{\prime \prime}=0$$ Substituting the derivatives, we get: $$2(r^3 e^{rx}) - (r^2 e^{rx}) = 0$$

Solve the resulting equation

After substituting the derivatives, we can try to simplify and solve the resulting equation. First, factor out the common term, \(e^{rx}\). This leaves us with: $$e^{rx}(2r^3 - r^2) = 0$$ Since \(e^{rx}\) is never equal to zero, we only need to focus on solving the equation: $$2r^3 - r^2 = 0$$ To solve this equation, first factor out \(r^2\): $$r^2(2r - 1) = 0$$ This equation has two roots: \(r_1 = 0\) and \(r_2 = \frac{1}{2}\).

Construct the general solution

Now that we have the roots, we can construct the general solution. Since we have a third-order differential equation, we need three linearly independent solutions. We already have two solutions based on our roots: \(e^{0x} = 1\) and \(e^{\frac{1}{2}x}\). The third solution will be in the form of \(x e^{\frac{1}{2}x}\). Therefore, the general solution is given by: $$y(x) = C_1 + C_2 e^{\frac{1}{2}x} + C_3 x e^{\frac{1}{2}x}$$ where \(C_1\), \(C_2\), and \(C_3\) are arbitrary constants. Since no initial conditions are given in the problem, our final answer is the general solution: $$y(x) = C_1 + C_2 e^{\frac{1}{2}x} + C_3 x e^{\frac{1}{2}x}$$

Key concepts

General Solution

When dealing with differential equations, the General Solution represents a family of potential solutions that satisfy the equation. Such a solution contains arbitrary constants, reflecting the different possible solutions based on initial or boundary conditions not specified yet. In our example, the differential equation is \(2y'''-y''=0\).

We suppose a solution of the form \(y(x) = e^{rx}\), standard for linear differential equations with constant coefficients. By substituting derivatives back into the equation, we solve for \(r\). Once we determine the values of \(r\), or the roots of the characteristic equation, we form the general solution as a combination of these solutions.

For a third-order equation like ours, we expect three linearly independent solutions. If roots are real and distinct, we simply use these exponential functions in the general solution. Nevertheless, if roots are repeated, we might introduce polynomial terms multiplied by exponentials to maintain independence.

Initial Value Problem

The term Initial Value Problem (IVP) refers to a differential equation paired with specific values, usually given for the function and its derivatives at a particular point. These conditions allow us to pinpoint a single, unique solution from the many contained in the general solution. Unlike our example, where no initial conditions were given, these initial values help us determine the arbitrary constants \(C_1, C_2, C_3\) in the general solution.

When working on an IVP, follow these steps:

• Find the general solution as usual.
• Substitute the initial conditions to formulate equations for each.
• Solve these equations to find specific values for the constants.

This process transitions from the general solution to a definitive solution that meets the specified initial conditions.

Characteristic Equation

The solution of many linear differential equations starts with deriving the Characteristic Equation. This equation emerges when substituting \(y(x) = e^{rx}\) into the original differential equation, resulting in a polynomial equation typically in terms of \(r\).

In our problem, substituting gives us \(2r^3 - r^2 = 0\). Solving this polynomial helps uncover the roots (\(r_1 = 0\) and \(r_2 = \frac{1}{2}\)), which guide the formulation of solutions:

• If the roots are real and distinct, each provides a simple exponential solution.
• Repeated roots lead to solutions involving multiplication by polynomial terms (e.g., \(x\)).
• Complex roots give rise to oscillatory solutions involving sine and cosine functions.

The characteristic equation reveals the core structure of the solution space, providing a systematic method for constructing the general solution.

Exponential Function

Exponential Functions are pivotal in solving linear differential equations, particularly with constant coefficients, because they simplify the calculations and solutions. The assumed form \(y(x) = e^{rx}\) is simple yet powerful, integral in linking derivatives smoothly thanks to the nature of exponential functions. Each derivative introduces just a multiplying factor of \(r\).

For instance, in our exercise:

• First derivative: \(y'(x) = r e^{rx}\)
• Second derivative: \(y''(x) = r^2 e^{rx}\)
• Third derivative: \(y'''(x) = r^3 e^{rx}\)

This simplicity helps in clearly formulating the characteristic equation, from which we derive the roots critical to the general solution. Exponential solutions efficiently model many phenomena, from population dynamics to financial growth, due to their consistent growth rate.