/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 Show that \(y=\sin 2 t\) satisfi... [FREE SOLUTION] | 91Ó°ÊÓ

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Show that \(y=\sin 2 t\) satisfies. $$ \frac{d^{2} y}{d t^{2}}+4 y=0 $$

Short Answer

Expert verified
The function satisfies the equation because substituting gives \(0 = 0\).

Step by step solution

01

Understanding the Problem

To show that the given function \(y = \sin 2t\) satisfies the differential equation \(\frac{d^2 y}{dt^2} + 4y = 0\), we need to calculate the second derivative of \(y\) with respect to \(t\), and then substitute \(y\) and its second derivative into the equation to check if it equals zero.
02

First Derivative

Calculate the first derivative of \(y = \sin 2t\). Use the chain rule: the derivative of \(\sin 2t\) is \(2\cos 2t\). Thus, \(\frac{dy}{dt} = 2\cos 2t\).
03

Second Derivative

Now, find the second derivative of \(y\). Differentiate \(2\cos 2t\) using the chain rule: the derivative of \(\cos 2t\) is \(-2\sin 2t\). Multiplying by the constant 2 from the first derivative gives \(\frac{d^2y}{dt^2} = -4\sin 2t\).
04

Substituting into the Equation

Substitute \(\frac{d^2 y}{dt^2} = -4\sin 2t\) and \(y = \sin 2t\) into the differential equation \(\frac{d^2 y}{dt^2} + 4y = 0\). This yields \(-4\sin 2t + 4\sin 2t = 0\).
05

Conclusion

Since substituting into the original equation results in \(0 = 0\), the function \(y = \sin 2t\) satisfies the differential equation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Second Derivative
In calculus, derivatives are fundamental, allowing us to understand how functions change. The first derivative of a function gives the rate of change or slope of the function. The second derivative, however, provides information on the curvature or concavity of the original function.
When we compute the second derivative, we once again differentiate the first derivative. This tells us whether the original function is curving upwards or downwards. For example, if the second derivative is positive, the function is concave up like a cup, and if it is negative, it's concave down like a frown.
In our particular example with the equation for the second derivative, \[ \frac{d^{2} y}{d t^{2}}\]we determined this by differentiating \( 2\cos 2t \) which resulted in \( -4\sin 2t \). Understanding how to find the second derivative is essential for verifying solutions to differential equations, like determining if \( y=\sin 2t \) satisfies \( \frac{d^{2} y}{d t^{2}}+4 y=0 \).
Sine Function
The sine function is a key player in trigonometry and calculus due to its periodic nature. It oscillates between -1 and 1, which makes it a perfect fit for modeling various wave-like phenomena, such as sound waves, light waves, and even tides.
In the context of differential equations, the sine function's properties are very handy. Specifically, when we use \( y = \sin 2t \), we're employing a function whose derivatives are other trigonometric functions. These derivatives are predictable, making them easier to manage.
  • The first derivative of \( \sin u \) with respect to a variable \( u \) is \( \cos u \).
  • The second derivative then becomes \( -\sin u \), showing how the slope changes.
By understanding these basics, solving equations involving sine becomes less daunting, as we recognize patterns in the behavior of derivatives and their impacts.
Chain Rule
The chain rule is an essential tool in calculus, particularly when dealing with composite functions. It allows us to differentiate a function that is "nested" within another. In simple terms, if you have a function inside another function, the chain rule helps you take its derivative.
Using the chain rule effectively means you'll handle functions like \( \sin 2t \) with ease. Here's how it works:
  • Certainly, \( \sin 2t = \sin(u)\) where \( u = 2t \).
  • According to the chain rule, the derivative of \( \sin u \) with respect to \( t \) is \( \cos u \cdot \frac{du}{dt} \).
  • In this case, \( \frac{du}{dt} = 2 \), leading to \( 2\cos 2t \).
The chain rule simplifies the process of differentiating complex functions by breaking them into manageable pieces. By mastering this rule, dealing with differential equations becomes more straightforward, as exemplified in finding the derivatives of \( \sin 2t \).

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Most popular questions from this chapter

Solve the initial value problem. $$y^{\prime \prime}+6 y^{\prime}+10 y=0, \quad y(0)=0, \quad y^{\prime}(0)=2$$

Analyze the phase plane of the differential equations for \(x, y \geq 0 .\) Show the nullclines and equilibrium points, and sketch the direction of the trajectories in each region. $$\begin{aligned}&\frac{d x}{d t}=x\left(1-x-\frac{y}{3}\right)\\\&\frac{d y}{d t}=y\left(1-y-\frac{x}{2}\right)\end{aligned}$$

Are the statements true or false? Give an explanation for your answer. The system of differential equations \(d x / d t=-x+x y^{2}\) and \(d y / d t=y-x^{2} y\) requires initial conditions for both \(x(0)\) and \(y(0)\) to determine a unique solution.

Give an explanation for your answer. A system of two differential equations has at most two nullclines.

Two companies share the market for a new technology. They have no competition except each other. Let \(A(t)\) be the net worth of one company and \(B(t)\) the net worth of the other at time \(t .\) Suppose that net worth cannot be negative and that \(A\) and \(B\) satisfy the differential equations $$\begin{aligned}&A^{\prime}=2 A-A B\\\&B^{\prime}=B-A B\end{aligned}$$ (a) What do these equations predict about the net worth of each company if the other were not present? What effect do the companies have on each other? (b) Are there any equilibrium points? If so, what are they? (c) Sketch a slope field for these equations (using a computer or calculator), and hence describe the different possible long-run behaviors.

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