Problem 41 - Calculus 3

Question

Solve the boundary-value problem, if possible. $$ y^{\prime \prime}+10 y^{\prime}+34 y=0, \quad y(0)=6, \quad y(\pi)=2 $$

Step-by-Step Solution

Answer

No solution exists due to inconsistent boundary conditions.

Step 1: Characteristic Equation

To solve the differential equation, we first write the characteristic equation for the homogeneous equation \( y'' + 10y' + 34y = 0 \). The characteristic equation is given by \( r^2 + 10r + 34 = 0 \).

Step 2: Solve the Characteristic Equation

Next, solve the quadratic equation \( r^2 + 10r + 34 = 0 \) using the quadratic formula: \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 10, \) and \( c = 34 \). This gives \( r = \frac{-10 \pm \sqrt{10^2 - 4 \cdot 1 \cdot 34}}{2 \cdot 1} = \frac{-10 \pm \sqrt{100 - 136}}{2} = \frac{-10 \pm \sqrt{-36}}{2} \).

Step 3: Simplify the Roots

The roots are \( r = \frac{-10 \pm 6i}{2} \), which simplify to \( r = -5 \pm 3i \). These are complex roots indicating the general solution of the form: \( y(x) = e^{-5x}(C_1 \cos(3x) + C_2 \sin(3x)) \).

Step 4: Apply Initial Condition \( y(0) = 6 \)

The boundary condition \( y(0) = 6 \) requires substitution into the general solution: \( y(0) = e^{0}(C_1 \cos(0) + C_2 \sin(0)) = C_1 = 6 \). Thus, \( C_1 = 6 \).

Step 5: Apply Boundary Condition \( y(\pi) = 2 \)

Substitute \( x = \pi \) into the general solution: \( y(\pi) = e^{-5\pi}(6 \cos(3\pi) + C_2\sin(3\pi)) = 2 \). Since \( \cos(3\pi) = -1 \) and \( \sin(3\pi) = 0 \), we have \( e^{-5\pi} \times 6 \times (-1) = 2 \). This simplifies to \(-6e^{-5\pi} = 2 \), leading to an inconsistency because \( e^{-5\pi} \) is a positive number, hence the positive side of the equation cannot equal \(2\).

Step 6: Conclude the Inconsistency

Because of the inconsistency derived from the boundary conditions, there is no solution to this boundary value problem. The requested conditions cannot be satisfied simultaneously due to the structure of the differential equation and prescribed boundary values.

Key Concepts

Differential equations

Differential equations are fundamental tools in mathematics that involve functions and their derivatives. They provide a way to mathematically model a wide variety of phenomena in engineering, physics, and other sciences. For the given exercise, we are dealing with a second-order linear homogeneous differential equation of the form:

• \( y'' + 10y' + 34y = 0 \)

A second-order equation like this can describe systems with acceleration or curvature. Since it is homogeneous, there are no terms without the variable \( y \), meaning the output will inherently depend on the function \( y \) itself, and its derivatives.
Solving differential equations often involves finding the characteristic equation, analyzing the roots, and applying initial or boundary conditions to find the constants of integration that customize the solution to the problem's needs.

Characteristic equation

The characteristic equation transforms a differential equation into an algebraic form that is often easier to solve. For second-order linear differential equations like the one at hand, the characteristic equation is derived by assuming a solution of the form \( y = e^{rx} \). Substituting into the differential equation gives:

• \( r^2 + 10r + 34 = 0 \)

This quadratic equation is pivotal because it reveals the nature of the solutions through its roots. The roots of a characteristic equation can be real or complex, and this will directly influence the form of the general solution.
In our problem, the roots are determined using the quadratic formula, highlighting why such transforms are indispensable in solving differential equations efficiently. Understanding the characteristic equation allows us to make deductions about the system’s behavior.

Complex roots

Complex roots emerge when the discriminant \( b^2 - 4ac \) of the characteristic equation is negative, indicating that the square root of a negative number is necessary. In the exercise, the calculation led to:

• \( r = -5 \pm 3i \)

These roots reveal that the solution involves oscillatory components, which are characteristic of sine and cosine functions. Therefore, the general solution is expressed as:

• \( y(x) = e^{-5x}(C_1 \cos(3x) + C_2 \sin(3x)) \)

This form accounts for both exponential decay (from \( e^{-5x} \)) and periodicity (from \( \cos(3x) \) and \( \sin(3x) \)).
Such solutions are common when describing physical systems involving harmonic oscillators or circuits involving reactive components like inductors and capacitors.

Initial condition

Boundary or initial conditions are necessary to determine specific solutions from the general solution of a differential equation. An initial condition specifies the exact value of the solution at a particular point, here given as:

• \( y(0) = 6 \)

This allows us to substitute into the general solution to solve for one of the constants, making the solution unique. However, a boundary value like:

• \( y(\pi) = 2 \)

also needs to fit into the framework set by the differential equation’s general solution.
Unfortunately, in this case, it's discovered that the boundary conditions lead to an inconsistency. The boundary condition \( y(\pi) = 2 \) becomes infeasible due to the nature of the exponential term \( e^{-5\pi} \). This illustrates an important aspect of boundary-value problems: they can sometimes impose requirements that are impossible to satisfy, revealing the need for careful selection of conditions in models.