Question

The function \(y(t)\) is a solution of the initial value problem \(y^{\prime \prime}+a y^{\prime}+b y=0, y\left(t_{0}\right)=y_{0}\), \(y^{\prime}\left(t_{0}\right)=y_{0}^{\prime}\), where the point \(t_{0}\) is specified. Determine the constants \(a, b, y_{0}\), and \(y_{0}^{\prime} .\) $$ y(t)=3 e^{-2 t} \cos (t-\pi / 2), \quad t_{0}=0 $$

Short answer

The values of the constants are \(a = -4\), \(b = -8\), \(y_0 = -3\), and \(y'_0 = 6\).

Step-by-step solution

First Derivative

Differentiate \(y(t)\) with respect to \(t\) to find the first derivative: $$ y^{\prime}(t) = \frac{d}{dt} \left(3 e^{-2t} \cos(t - \frac{\pi}{2})\right) $$ Using the product rule and chain rule, this becomes: $$ y^{\prime}(t) = -6 e^{-2t} \cos(t - \frac{\pi}{2}) - 3e^{-2t} \sin(t - \frac{\pi}{2}) $$

Second Derivative

Differentiate \(y^{\prime}(t)\) with respect to \(t\) to find the second derivative: $$ y^{\prime\prime}(t) = \frac{d^{2}}{dt^{2}} \left(-6 e^{-2t} \cos(t - \frac{\pi}{2}) - 3e^{-2t} \sin(t - \frac{\pi}{2})\right) $$ Using the product rule and chain rule again, we find: $$ y^{\prime\prime}(t) = 12 e^{-2t} \cos(t - \frac{\pi}{2}) + 6e^{-2t} \sin(t - \frac{\pi}{2}) + 6e^{-2t} \sin(t - \frac{\pi}{2}) - 6e^{-2t} \cos(t - \frac{\pi}{2}) $$ Simplifying the second derivative, we have: $$ y^{\prime\prime}(t) = 12 e^{-2t} \sin(t - \frac{\pi}{2}) $$ #Step 2: Determine the constants \(a\) and \(b\)#

Plug in the derivatives and original function

Substitute \(y(t)\), \(y^{\prime}(t)\), and \(y^{\prime\prime}(t)\) into the given differential equation: $$ 12 e^{-2t} \sin(t - \frac{\pi}{2}) + a(-6 e^{-2t} \cos(t - \frac{\pi}{2}) - 3e^{-2t} \sin(t - \frac{\pi}{2})) + b(3 e^{-2t} \cos(t - \frac{\pi}{2})) = 0 $$

Compare coefficients

Now, we compare the coefficients of the trigonometric functions to find the values of \(a\) and \(b\): $$ 12 + (-3a) = 0 \Rightarrow a = -4 \\ -6a + 3b = 0 \Rightarrow -6(-4) + 3b = 0 \\ 24 + 3b = 0 \Rightarrow b = -8 $$ #Step 3: Determine \(y_0\) and \(y'_0\)#

Find \(y_0\) by plugging in \(t_0\)

Plug in the initial condition \(t_0 = 0\) into \(y(t)\): $$ y(0) = 3 e^{-2(0)} \cos(0 - \frac{\pi}{2}) = 3(-1) = -3 $$ So, \(y_0 = -3\).

Find \(y'_0\) by plugging in \(t_0\)

Plug in the initial condition \(t_0 = 0\) into \(y^{\prime}(t)\): $$ y^{\prime}(0) = -6 e^{-2(0)} \cos(0 - \frac{\pi}{2}) - 3e^{-2(0)} \sin(0 - \frac{\pi}{2}) = -6(-1) - 3(0) = 6 $$ So, \(y'_0 = 6\). We have found all four constants: \(a = -4\), \(b = -8\), \(y_0 = -3\), and \(y_0' = 6\).

Key concepts

Initial Value Problem

An initial value problem in mathematics refers to finding a particular solution to a differential equation given specific initial conditions. In the exercise provided, we have an initial value problem of a second-order differential equation. This means that the function and its derivatives must satisfy certain predefined values, known as initial conditions at a particular starting point, which is represented by \(t_0\).

• The given differential equation is written as \(y'' + ay' + by = 0\).
• The initial conditions are \(y(t_0) = y_0\) and \(y'(t_0) = y_0'\).

These initial conditions help us in determining a unique solution curve from the family of solutions possible for the differential equation.
Understanding initial value problems is crucial, especially in real-world applications where the start condition is pivotal, such as predicting the motion of a pendulum or modeling population growth.

Second Order Differential Equation

A second order differential equation involves the second derivative of a function. These equations can be more complex than first-order ones due to the higher derivative involved.
In our exercise, the equation \(y''+ ay' + by = 0\) characterizes a linear homogeneous second order differential equation with constant coefficients, where \(a\) and \(b\) are constants determined through calculations. Solving such equations often requires finding the characteristic equation and solving for its roots, which finally aids in constructing the general solution.
The beauty of such equations lies in their ability to describe numerous physical phenomena, including vibrations, electronics circuits, and mechanical systems. Efficiently solving these requires a blend of analytical skills and knowledge of initial conditions.

Trigonometric Functions

Trigonometric functions are essential mathematical functions that relate the angles of a triangle to the lengths of its sides. In this exercise, trigonometric functions like \(\sin\) and \(\cos\) appear in the solution to the differential equation.

• They help in expressing sinusoidal forms which are often solutions to differential equations with oscillatory components.
• In the given function \(y(t) = 3e^{-2t} \cos(t - \frac{\pi}{2})\), the cosine function is shifted by \(\frac{\pi}{2}\).

Understanding trigonometric relationships and transformations is crucial for interpreting and solving differential equations effectively. Trigonometric identities and amplitude-shift properties often simplify complex calculations in differential equations.

Product Rule

The product rule is a fundamental rule in calculus used for finding the derivative of a product of two functions. In our task, the product rule plays a vital role in differentiating the function \(y(t) = 3e^{-2t} \cos(t - \frac{\pi}{2})\).
The product rule is expressed as:\[(u \, v)' = u' \, v + u \, v'\]
Here, \(u = 3e^{-2t}\) and \(v = \cos(t - \frac{\pi}{2})\), so the derivatives are added using the product rule during calculus operations.

• This ensures that both parts of the product are differentiated accurately.
• It accounts for changes in both functions being multiplied, providing the total rate of change.

The product rule is indispensable when solving problems involving multiple function multiplication, enhancing our ability to tackle first and second derivative calculations efficiently.