Problem 12 - Calculus: Early Transcendentals

Question

Solve the initial value problem \(y^{\prime \prime}+100 y=0, y(0)=5, y^{\prime}(0)=50 .\)

Step-by-Step Solution

Answer

The solution is \(y(t) = 5 \cos(10t) + 5 \sin(10t)\).

Step 1: Identify the type of differential equation

The given differential equation is \(y'' + 100y = 0\). This is a second-order linear homogeneous differential equation with constant coefficients.

Step 2: Write down the characteristic equation

For a differential equation of the form \(y'' + eta y = 0\), the characteristic equation is \(r^2 + eta = 0\). Thus, for \(y'' + 100y = 0\), the characteristic equation is \(r^2 + 100 = 0\).

Step 3: Solve the characteristic equation

The characteristic equation is \(r^2 + 100 = 0\). Solving for \(r\), we find \(r^2 = -100\), leading to \(r = \pm 10i\).

Step 4: Write the general solution

Since the roots of the characteristic equation are complex, \(r = 0 \pm 10i\), the general solution is \(y(t) = C_1 \cos(10t) + C_2 \sin(10t)\).

Step 5: Apply the initial conditions

The initial conditions given are \(y(0) = 5\) and \(y'(0) = 50\).- For \(y(0) = 5\), substitute into the general solution: \(5 = C_1 \cos(0) + C_2 \sin(0)\) which simplifies to \(5 = C_1\).- Next, we find \(y'(t) = -10C_1 \sin(10t) + 10C_2 \cos(10t)\). Applying \(y'(0) = 50\): \(50 = 10C_2 \cos(0) - 10C_1 \sin(0)\), which simplifies to \(50 = 10C_2\). Thus, \(C_2 = 5\).

Step 6: Write the particular solution

Substituting the constants \(C_1 = 5\) and \(C_2 = 5\) back into the general solution, we get the particular solution: \(y(t) = 5 \cos(10t) + 5 \sin(10t)\).

Key Concepts

Characteristic Equation

When examining second-order linear differential equations like the provided example, the "characteristic equation" is a fundamental concept. It helps us to find the solutions of the differential equation by reducing it to solving a simpler algebraic equation. For a differential equation in the form of \(y'' + \beta y = 0\), the characteristic equation is \(r^2 + \beta = 0\). In this exercise, \(y'' + 100y = 0\), the characteristic equation becomes \(r^2 + 100 = 0\).

• The form of the differential equation dictates the structure of the characteristic equation.
• Here, \(\beta\) is replaced by the constant present in the original equation, which is 100 in this instance.

Understanding the characteristic equation is crucial because its roots determine the form of the solution to the differential equation. Whether these roots are real, complex, or repeated significantly affects the nature of the solution.

Complex Roots

Roots of the characteristic equation reveal much about the behavior of the solutions. If these roots are complex, the solutions will involve trigonometric functions. In our problem, the characteristic equation \(r^2 + 100 = 0\) has complex roots. By solving it, we find \(r = \pm 10i\).

• Complex roots arise when the discriminant (the part under the square root in the quadratic formula) is negative.
• These roots come in conjugate pairs, here being \(r = 0 \pm 10i\).

The presence of complex roots means the general solution will involve sine and cosine functions: \(y(t) = C_1 \cos(10t) + C_2 \sin(10t)\). This mix of trigonometric functions corresponds to oscillatory behavior, which is characteristic in solutions to differential equations with imaginary roots.

Initial Value Problems

Initial value problems (IVPs) are essential in determining the specific solution to a differential equation tailored to given starting conditions. With the problem \(y'' + 100y = 0\), the initial values provided are \(y(0) = 5\) and \(y'(0) = 50\). These conditions are vital to pinning down the constants \(C_1\) and \(C_2\) in the general solution.

• Initial values are necessary to solve for unknown constants in the general solution.
• This ensures the particular solution fits the given scenario at specific points in time.

Plugging the initial conditions into the general solution \(y(t) = C_1 \cos(10t) + C_2 \sin(10t)\) determines the constants \(C_1 = 5\) and \(C_2 = 5\). As a result, the particular solution becomes \(y(t) = 5 \cos(10t) + 5 \sin(10t)\). Understanding IVPs is crucial as they allow solutions to be uniquely defined, adapting to real-world situations and making them immensely useful in modeling physical phenomena.