Question

The differential equation \((2 y+t \cos y) y^{\prime}+\sin y=0\) is exact, and thus there exists a function \(H(t, y)\) such that \(\partial H / \partial t=\sin y\) and \(\partial H / \partial y=2 y+t \cos y\). Antidifferentiating the second equation with respect to \(y\), we ultimately arrived at \(H(t, y)=y^{2}+t \sin y+\) \(C_{1}\); see equation (10). Show that the same result would be obtained if we began the solution process by antidifferentiating the first equation, \(\partial H / \partial t=\sin y\), with respect to \(t\).

Short answer

In this problem, we were given an exact differential equation with two partial derivatives and were asked to verify that the same function H(t, y) would be obtained if we began integrating the first equation with respect to t instead of y. We followed the steps to integrate the first equation, differentiate with respect to y, compare the result, and find the unknown function. The final solution for H(t, y) is the same as the one obtained by starting with the second equation, proving that the same result is obtained regardless of the starting process, which is H(t, y) = t sin y + y^2 + C_1.

Step-by-step solution

Rewrite the given equations

Since it is given that it's an exact differential equation, we have the following two equations: 1. \(\frac{\partial H}{\partial t} = \sin y\) 2. \(\frac{\partial H}{\partial y} = 2y + t\cos y\) Now, we need to integrate the first equation with respect to t, instead of the second equation with respect to y which was given.

Integrate the first equation

We will integrate the first equation with respect to t: \(\int \frac{\partial H}{\partial t} \, dt = \int \sin y \, dt\) Since y is considered as a constant with respect to t, we integrate as follows: \(H(t, y) = t\sin y + C_2(y)\) Here, \(C_2(y)\) is an unknown function of y which will be found in the next step.

Differentiate H(t,y) with respect to y

Now we need to differentiate H(t, y) with respect to y to match it with the second equation: \(\frac{\partial H}{\partial y} = \frac{\partial}{\partial y}(t\sin y + C_2(y))\) Differentiate term by term: \(\frac{\partial H}{\partial y} = t\cos y + C^{\prime}_2(y)\) Here, \(C^{\prime}_2(y)\) denotes the first derivative of \(C_2(y)\) with respect to y.

Compare the obtained result with the second equation

The second equation: \(\frac{\partial H}{\partial y} = 2y + t\cos y\) By comparing this equation with the result from step 3, we observe that: \(t\cos y + C^{\prime}_2(y) = 2y + t\cos y\) it's clear that: \(C^{\prime}_2(y) = 2y\)

Integrate to find \(C_2(y)\)

Now, integrate \(C^{\prime}_2(y)\) with respect to y to find the function \(C_2(y)\): \(C_2(y) = \int 2y \, dy = y^2 + C_1\) Here, \(C_1\) is a constant of integration.

Find the final solution for H(t, y)

Replace \(C_2(y)\) in the expression for H(t, y) from step 2: \(H(t, y) = t\sin y + y^2 + C_1\) This is the same result as \(H(t, y) = y^2 + t\sin y + C_1\), which was obtained by integrating the second equation with respect to y.

Key concepts

Partial Derivatives

Partial derivatives are a fundamental part of understanding exact differential equations. They involve taking the derivative of a function with respect to one variable while keeping the other variables constant. For a function \(H(t, y)\), \(\frac{\partial H}{\partial t}\) denotes the partial derivative with respect to \(t\), treating \(y\) as a constant.

In this exercise, we have two important partial derivatives:

• \(\frac{\partial H}{\partial t} = \sin y\)
• \(\frac{\partial H}{\partial y} = 2y + t\cos y\)

These equations suggest how \(H(t, y)\) changes with respect to \(t\) or \(y\) independently. By setting these partial derivatives, we aim to find the original function \(H(t, y)\) which satisfies both conditions. This process relies heavily on the concept of holding one variable constant and differentiating with respect to the other. Understanding partial derivatives is key to navigating through exact differential equations effectively, as they form the basis for further integrations required to solve such problems.

Antidifferentiation

Antidifferentiation, also known as integration, is the process of finding a function whose derivative is the given function. It is essentially the reverse of differentiation. In exact differential equations, we often have to integrate one of the partial derivatives to construct the original function.

In this exercise, we integrated the equation \(\frac{\partial H}{\partial t} = \sin y\) with respect to \(t\). This was a strategic choice, as we initially assessed its counterpart, \(\frac{\partial H}{\partial y}\), integrating with respect to \(y\). The keyword here is choice, which shows there might be more than one path to the solution. Despite which partial equation we begin with, the correct process yields the same \(H(t, y)\).

The result we got was:

• \(H(t, y) = t\sin y + C_2(y)\)

This indefinite integration with respect to \(t\) acknowledges an unknown function \(C_2(y)\) due to the constant-like behavior of \(y\), which further needs determining. Understanding antidifferentiation is crucial for reversing differentiation and constructing the full function from its derivatives.

Integration Techniques

Integration techniques are vital for successfully finding solutions to exact differential equations. When integrating a partial derivative, especially from a function contoured by multiple variables, each variable holds a unique stance—constant or variable depending on the context.

In this problem, while integrating \(\sin y\) with respect to \(t\), the essential technique used was:

• Treat \(y\) as a constant, because integration is with respect to \(t\).
• Apply the basic integration rule: \(\int \sin y \, dt = t\sin y + C\)

This result mirrored the earlier step of integrating \(\frac{\partial H}{\partial y}\) with respect to \(y\), highlighting flexibility and interchangeability in the order of operations.

Later, for finding \(C_2(y)\), we deployed another integration technique, integrating the derivative \(C'_2(y) = 2y\) with respect to \(y\). This added depth to solving the equation:

• \(C_2(y) = \int 2y \, dy = y^2 + C_1\)

Mastering these integration techniques ensures a smoother path through solving sophisticated differential equations; hence offering different angles to approach and solve the same problem efficiently.