Question

The function $y(t)$ is a solution of the initial value problem $y^{\prime \prime }+a{y}^{\prime }+by=0,y\left(t_0\right)=y_0$, $y^{\prime }\left(t_0\right)=y_0^{\prime }$, where the point $t_0$ is specified. Determine the constants $a,b,y_0$, and $y_0^{\prime }.$ $$y(t)=e^{1-1/6}\cos 2t-e^{t-1/6}\sin 2t,t_0=\pi /6$$

Short answer

Question: Determine the constants a, b, y_0, and y_0' for the initial value problem: Solution: Based on the step-by-step solution provided, the constants for the initial value problem are: a = -2/(e^(π/6 - 1/6)) b = -8/(e^(1 - 1/6)) y_0 = e^(1 - 1/6)cos(π/3) - e^(π/6 - 1/6)sin(π/3) y_0' = -2e^(1 - 1/6)sin(π/3) - (e^(π/6 - 1/6)(2sin(π/3)) + e^(π/6 - 1/6)(-2cos(π/3)))

Step-by-step solution

Find the first derivative of y(t)

To find the first derivative of the given function, we will use the chain rule and the product rule of derivatives. These rules, together with the derivatives of exponential, sine, and cosine functions, will provide us with the first derivative: $$y(t)=e^{1-1/6}\cos (2t)-e^{t-1/6}\sin (2t)$$ $$y^{\prime }(t)=e^{1-1/6}(-2\sin (2t))-(e^{t-1/6}(2\sin (2t))+e^{t-1/6}(-2\cos (2t)))$$

Find the second derivative of y(t)

Now, we will find the second derivative by taking the derivative of y'(t) with respect to t: $$y^{″}(t)=-4e^{1-1/6}\cos (2t)-2e^{t-1/6}(2\cos (2t)-4\sin (2t)-4\sin (2t)+8\cos (2t))$$

Substitute the given value of t_0 into y(t), y'(t), and y''(t)

Substitute the value of $t_0=\pi /6$ into the functions: $$y(\pi /6)=e^{1-1/6}\cos (\pi /3)-e^{\pi /6-1/6}\sin (\pi /3)$$ $$y^{\prime }(\pi /6)=-2e^{1-1/6}\sin (\pi /3)-(e^{\pi /6-1/6}(2\sin (\pi /3))+e^{\pi /6-1/6}(-2\cos (\pi /3)))$$ $$y^{″}(\pi /6)=-4e^{1-1/6}\cos (\pi /3)-2e^{\pi /6-1/6}(2\cos (\pi /3)-4\sin (\pi /3)-4\sin (\pi /3)+8\cos (\pi /3))$$

Use the initial value problem equation to find the constants a and b

Substitute y''(t), y'(t), y(t) and t_0 into the given equation: $$y^{\prime \prime }+a{y}^{\prime }+by=0$$ $$-4e^{1-1/6}\cos (\pi /3)-2e^{\pi /6-1/6}(2\cos (\pi /3)-4\sin (\pi /3)-4\sin (\pi /3)+8\cos (\pi /3))+a(-2e^{1-1/6}\sin (\pi /3)-(e^{\pi /6-1/6}(2\sin (\pi /3))+e^{\pi /6-1/6}(-2\cos (\pi /3))))+b(e^{1-1/6}\cos (\pi /3)-e^{\pi /6-1/6}\sin (\pi /3))=0$$ Solve for a and b: $$a=\frac{-2}{e^{\pi /6-1/6}},b=\frac{-8}{e^{1-1/6}}$$

Determine the constants y_0 and y_0'

From the initial condition, we can determine the initial values y_0 and y_0': $$y_0=y(t_0)=e^{1-1/6}\cos (\pi /3)-e^{\pi /6-1/6}\sin (\pi /3)$$ $$y_0^{\prime }=y^{\prime }(t_0)=-2e^{1-1/6}\sin (\pi /3)-(e^{\pi /6-1/6}(2\sin (\pi /3))+e^{\pi /6-1/6}(-2\cos (\pi /3)))$$ So, the constants are: $$a=\frac{-2}{e^{\pi /6-1/6}},b=\frac{-8}{e^{1-1/6}},y_0=e^{1-1/6}\cos (\pi /3)-e^{\pi /6-1/6}\sin (\pi /3),y_0^{\prime }=-2e^{1-1/6}\sin (\pi /3)-(e^{\pi /6-1/6}(2\sin (\pi /3))+e^{\pi /6-1/6}(-2\cos (\pi /3)))$$

Key concepts

Differential Equations

Differential equations play a crucial role in modeling real-world phenomena where the change in one quantity relates to the change in another. In particular, an initial value problem involves not only a differential equation but also an initial condition that specifies values of the function and its derivatives at a certain point. These problems are pivotal in fields such as physics, engineering, and economics because they provide the information required to uniquely determine the solution to a differential equation.

In the given exercise, the second-order linear homogeneous differential equation involves finding a function that satisfies a relationship between itself and its derivatives. This problem specifically asks to find a unique solution for given initial conditions. Identifying constants such as 'a' and 'b' in the equation helps to further define the specific solution needed to model a particular situation.

Derivatives Calculus

The fundamental tool in calculus, derivatives quantify how a function's value changes as its input changes. They are foundational in solving differential equations, as they represent the rates of change. Calculus provides rules such as the product rule, quotient rule, and chain rule to handle the differentiation of a variety of functions including polynomials, exponentials, and trigonometric functions.

In our problem, derivatives are calculated using the chain rule and product rule applied to exponential and trigonometric functions. This step-by-step computation of the first and second derivatives of the given y(t) is essential in preparing to apply the initial conditions. Understanding how to apply these rules correctly ensures accuracy in deriving further solutions in differential equations.

Exponential Functions

Exponential functions such as the natural exponential function, denoted as 'e' raised to a power, appear frequently in many contexts including compound interest, population growth, and radioactive decay. These functions have the unique property that the rate of change is directly proportional to the function's current value, which perfectly aligns with the concepts of differential equations.

In the exercise, we see an exponential function combined with trigonometric functions. This particular combination, especially when associated with derivatives, often indicates a solution to a differential equation describing oscillating systems like springs or electrical circuits. The behavior of exponential functions, including growth and decay, is pivotal to understanding the general solutions of our differential equation.