Question

Determine the general form of the function \(M(t, y)\) or \(N(t, y)\) that will make the given differential equation exact. $$ N(t, y) y^{\prime}+t^{2}+y^{2} \sin t=0 $$

Short answer

In summary, the given differential equation would be exact if the function N(t, y) has the general form: $$ N(t, y) = -2y \cos t + f(y) $$ where \(f(y)\) is an arbitrary function of \(y\) only.

Step-by-step solution

Identifying the given differential equation

In this problem, the given differential equation is: $$ N(t, y)y'+t^2+y^2 \sin t = 0 $$ Already, we know that \(M(t, y) = t^2 + y^2 \sin t\) and we are given that \(N(t, y) y' = -M(t, y)\), which implies that finding \(M(t, y)\) is not necessary. Therefore, our task is to find the form of \(N(t, y)\) that makes the given equation exact.

Finding partial derivatives

To find the general form for \(N(t, y)\), we will compute the partial derivatives of \(M\) and \(N\) with respect to \(y\) and \(t\), respectively: $$ \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(t^2 + y^2 \sin t) $$ and $$ \frac{\partial N}{\partial t} = \frac{\partial}{\partial t}(N(t, y)). $$ Now, let's compute the partial derivative of \(M\) with respect to \(y\): $$ \frac{\partial M}{\partial y} = 2y \sin t $$

Finding the general form of \(N(t, y)\)

Recall that for the given equation to be exact, we must have \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial t}\). Using the partial derivative of \(M(t, y)\) that we found, we can write the condition as: $$ 2y \sin t = \frac{\partial N}{\partial t} $$ Integrating both sides with respect to \(t\), we get: $$ N(t, y) = \int 2y \sin t \, dt $$ Now, you can integrate the right side: $$ N(t, y) = 2y(-\cos t) + f(y) $$ Here, we have the general form of \(N(t, y)\) that makes the given differential equation exact: $$ N(t, y) = -2y \cos t + f(y) $$ where \(f(y)\) is an arbitrary function of \(y\) only.

Key concepts

Partial Derivatives

Partial derivatives are a fundamental concept in multivariable calculus. When dealing with functions of several variables, a partial derivative with respect to one variable measures how the function changes as that specific variable is varied, while keeping all other variables constant.
This concept is crucial when working with exact differential equations because:

• We use partial derivatives to verify if a differential equation is exact.
• For an equation of the form \(M(t, y)dt + N(t, y)dy = 0\), the partial derivatives \(\frac{\partial M}{\partial y}\) and \(\frac{\partial N}{\partial t}\) must be equal for the equation to be exact.

In our solution, calculating the partial derivative \(\frac{\partial M}{\partial y} = 2y \sin t\) helped establish the condition needed for exactness.

Integration

Integration is the process of finding the antiderivative or the integral of a function. It's a key step when solving differential equations, especially in determining unknown functions that make equations exact.
In the context of exact differential equations:

• Once we find the necessary partial derivatives, we use integration to determine the missing functions.
• For example, from the condition \(2y \sin t = \frac{\partial N}{\partial t}\), we integrate \(2y \sin t\) with respect to \(t\) to find \(N(t, y)\).

Performing integration here gives \(N(t, y) = -2y \cos t + f(y)\), which is essential in making the differential equation exact.

Function of Arbitrary Constants

Functions of arbitrary constants, often recognized as functions of arbitrary variables in this context, represent the undetermined part of an integral in a solution to a differential equation. This concept is most visible post-integration.
When integrating partial derivatives:

• We include arbitrary functions in the integral, as these account for other possible solutions that lead to exactness.
• In our solution, \(f(y)\) is such an arbitrary function relying solely on \(y\), indicating the generality of potential solutions.

This flexibility is crucial, as it implies a family of solutions that could make the original equation exact, reflecting the solution's comprehensiveness.