Problem 4
Question
Show that \(y=\sin x-\cos x\) is a solution of the initial-value problem \(\cos x \frac{d y}{d x}+y \sin x=1, y(0)=-1\) on the interval \((-\infty, \infty) .\)
Step-by-Step Solution
Verified Answer
The function \(y = \sin{x} - \cos{x}\) has a derivative \(\frac{dy}{dx} = \cos{x} + \sin{x}\). Substituting these into the given differential equation and simplifying, we obtain the trigonometric identity \(\cos^2{x} + \sin^2{x} = 1\), thus confirming that the function is a solution to the differential equation. Furthermore, we check the initial condition \(y(0)=-1\) and find that it is satisfied. Therefore, \(y = \sin{x} - \cos{x}\) is a solution to the initial-value problem on the interval \((-\infty, \infty)\).
1Step 1: Differentiate y with respect to x
Let's first find the derivative of the function \(y = \sin{x} - \cos{x}\) with respect to \(x\). The derivative can be calculated as follows:
\[\frac{dy}{dx} = \frac{d}{dx} (\sin{x} - \cos{x}) = \cos{x} + \sin{x}.\]
2Step 2: Substitute y and dy/dx in the differential equation
Now, let's substitute the function \(y = \sin{x} - \cos{x}\) and its derivative \(\frac{dy}{dx} = \cos{x} + \sin{x}\) into the given differential equation:
\[\cos{x} \left(\cos{x} + \sin{x}\right) + (\sin{x} - \cos{x}) \sin{x} = 1.\]
3Step 3: Simplify the equation
Now, let's simplify the left-hand side of the equation:
\[\cos^2{x} + \sin{x} \cos{x} + \sin^2{x} - \sin{x} \cos{x} = 1.\]
Notice that the terms \(\sin{x} \cos{x}\) cancel each other out. Thus, the equation simplifies to:
\[\cos^2{x} + \sin^2{x} = 1.\]
4Step 4: Verify the trigonometric identity
The left-hand side of the simplified equation is a very well-known trigonometric identity \(\cos^2{x} + \sin^2{x} = 1\). Therefore, the given function \(y = \sin{x} - \cos{x}\) is a solution to the differential equation \(\cos{x} \frac{dy}{dx} + y \sin{x} = 1\).
5Step 5: Check the initial condition
Finally, let's check if the given initial condition \(y(0) = -1\) is satisfied by the function \(y = \sin{x} - \cos{x}\):
\[y(0) = \sin{0} - \cos{0} = 0 - 1 = -1.\]
The initial condition \(y(0) = -1\) is satisfied.
6Step 6: Conclusion
Since the function \(y = \sin{x} - \cos{x}\) is a solution to the given differential equation \(\cos{x} \frac{dy}{dx} + y \sin{x} = 1\), and it satisfies the initial condition \(y(0)=-1\), we can conclude that \(y = \sin{x} - \cos{x}\) is a solution to the initial-value problem on the interval \((-\infty, \infty)\).
Key Concepts
Differential EquationsTrigonometric IdentitiesDerivative Calculations
Differential Equations
Differential equations form a critical part of mathematical modeling because they can represent how physical quantities change over time or space. In our exercise, the focus is on solving an initial-value problem, which is a specific type of differential equation that not only requires finding a general solution but also a particular one that satisfies a given initial condition.
An initial-value problem includes a differential equation and one or more conditions specifying the value of the unknown function and its derivatives at a particular point. These problems are fundamental in engineering, physics, and other sciences, as they can describe complex systems such as motion, heat transfer, and fluid dynamics. The initial condition in our example, where the initial value of the function is provided as \( y(0) = -1 \), is crucial because it anchors the general solution to a specific real-world scenario.
An initial-value problem includes a differential equation and one or more conditions specifying the value of the unknown function and its derivatives at a particular point. These problems are fundamental in engineering, physics, and other sciences, as they can describe complex systems such as motion, heat transfer, and fluid dynamics. The initial condition in our example, where the initial value of the function is provided as \( y(0) = -1 \), is crucial because it anchors the general solution to a specific real-world scenario.
Trigonometric Identities
Understanding trigonometric identities is essential for solving many calculus problems, including differential equations. Trigonometric identities are equations involving trigonometric functions that are true for all values of the variables where both sides of the identity are defined.
One of the most foundational and frequently used identities is \( \[ \cos^2{x} + \sin^2{x} = 1\] \), which reflects the inherent relationship between the sine and cosine functions. This identity arises from the Pythagorean theorem applied to the unit circle. In the exercise, this identity helps us to simplify the equation and verify that the proposed function \( y = \sin{x} - \cos{x} \) satisfies the differential equation. Without such identities, solving differential equations involving trigonometric functions could become exceedingly difficult.
One of the most foundational and frequently used identities is \( \[ \cos^2{x} + \sin^2{x} = 1\] \), which reflects the inherent relationship between the sine and cosine functions. This identity arises from the Pythagorean theorem applied to the unit circle. In the exercise, this identity helps us to simplify the equation and verify that the proposed function \( y = \sin{x} - \cos{x} \) satisfies the differential equation. Without such identities, solving differential equations involving trigonometric functions could become exceedingly difficult.
Derivative Calculations
Derivative calculations are a cornerstone of calculus, representing the rate at which a quantity changes. In the context of this problem, we are computing the derivative of \( y = \sin{x} - \cos{x} \) to find \( \frac{dy}{dx} \). The differentiation step transforms the original function into its rate of change, which is then used to solve the differential equation.
For functions involving basic trigonometric functions like sine and cosine, the derivatives are relatively straightforward: the derivative of \( \sin{x} \) is \( \cos{x} \), and the derivative of \( \cos{x} \) is \( -\sin{x} \). Remembering these basic derivatives, along with the rules for differentiation, such as the sum rule and the constant multiple rule, allows you to tackle a wide array of problems. In our exercise, correctly calculating the derivative is crucial to show how the original function operates within the dynamics specified by the differential equation.
For functions involving basic trigonometric functions like sine and cosine, the derivatives are relatively straightforward: the derivative of \( \sin{x} \) is \( \cos{x} \), and the derivative of \( \cos{x} \) is \( -\sin{x} \). Remembering these basic derivatives, along with the rules for differentiation, such as the sum rule and the constant multiple rule, allows you to tackle a wide array of problems. In our exercise, correctly calculating the derivative is crucial to show how the original function operates within the dynamics specified by the differential equation.
Other exercises in this chapter
Problem 4
You are given the phase curve associated with a system of predator-prey equations, where \(x(t)\) denotes the prey (caribou) population, in hundreds, and \(y(t)
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Determine whether the differential equation is linear. $$ y^{2} \frac{d x}{d y}+3 x=\tan y $$
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In Exercises \(5-16\), solve the differential equation. $$ \frac{d y}{d x}+2 y=e^{2 x} $$
View solution Problem 6
Solve the differential equation. $$ x \frac{d y}{d x}+3 y=2 $$
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