3.10. Problems for Practice

Solve:

d²y/dx² + 3 dy/dx + 2y = 0, y(0) = 1, y′(0) = 0

In this problem, we are asked to solve a second-order differential equation with constant coefficients. This equation is in standard form, where the second derivative is given along with first derivative and the function itself. To solve it, we first need to find the characteristic equation by substituting a trial solution (often in the form of e^(rx)). The roots of the characteristic equation will help us understand the behavior of the solution based on the nature of those roots. Once we find the roots, we apply the initial conditions provided to determine the constants in our general solution.

Imagine you're trying to design a spring that bounces back after being compressed. The differential equation represents how the position of the spring changes over time, influenced by its stiffness and mass. By solving this equation, we can predict how the spring will behave whencompressed.

Solve and classify the roots:

y′′ + y′ + y = 0

This problem involves solving a homogeneous linear differential equation and classifying the nature of its roots. First, we write the characteristic equation derived from the differential equation. Then, we need to determine the discriminant (D) from the characteristic equation to classify the roots as real and distinct, repeated, or complex. This classification informs us how the solution behaves over time, indicating whether it oscillates, grows, or decays.

Think of a pendulum. Depending on how much energy you give it and the friction involved, the way it swings back and forth can change dramatically. By classifying the roots from our equation, we determine if our pendulum will swing steadily, slowly stop, or spin wildly.

A cantilever beam’s deflection satisfies:

d⁴y/dx⁴ = 0 • Integrate twice and show that the solution involves a second-order homogeneous ODE. Then solve assuming zero shear and moment at the free end.

In this problem, we analyze the deflection of a cantilever beam, which leads us to a fourth-order differential equation. We are instructed to integrate this equation twice to reduce it to a second-order ODE. The integration steps will yield a general solution, which can then be solved under the assumption of specific boundary conditions (zero shear and moment at the free end). This practical application of mathematical models helps us understand how engineering structures respond to loads.

Consider a diving board. When someone stands on the end, the board bends. By understanding how to calculate its deflection, engineers can ensure it won't break under pressure, much like solving this problem helps engineers predict the behavior of a beam under load.

Solve:

y′′ − 6y′ + 13y = 0, y(0) = 0, y′(0) = 2

This problem presents another second-order linear differential equation. To solve it, we again find the characteristic equation. Following this, we determine the nature of the roots, which will impact the shape of the general solution. The initial conditions given will allow us to calculate specific values for the constants in the general solution. This reinforces the concept of how specific constraints on a problem lead to unique solutions in mathematics and engineering.

Imagine sound waves generated by a tuning fork. The way they emit and oscillate depends on how the fork is struck and how much energy it has, which is similar to how we predict system behavior in our differential equations based on their roots and initial conditions.

--