Problem 19 - Advanced Engineering Mathematics

Question

Let \(M=4 t^{3} y-15 t^{2}-y\) and \(N=t^{4}+3 y^{2}-t\) so that \(M_{y}=4 t^{3}-1=N_{t} .\) From \(f_{t}=4 t^{3} y-15 t^{2}-y\) we obtain \(f=t^{4} y-5 t^{3}-t y+h(y), h^{\prime}(y)=3 y^{2},\) and \(h(y)=y^{3} .\) A solution is \(t^{4} y-5 t^{3}-t y+y^{3}=c\)

Step-by-Step Solution

Answer

The solution to the problem is the function \(f(t, y) = t^4y - 5t^3 - ty + y^3 = c\).

Step 1: Understand the Problem

We are given expressions for \(M\) and \(N\), where \(M = 4t^3y - 15t^2 - y\) and \(N = t^4 + 3y^2 - t\). We also have the condition \(M_y = 4t^3 - 1 = N_t\). We need to derive the solution for the function \(f(t, y)\).

Step 2: Calculate the Partial Derivative of M with respect to y

The partial derivative of \(M\) with respect to \(y\) is calculated by differentiating \(M\) only with respect to \(y\). Since \(M = 4t^3y - 15t^2 - y\), the derivative is: \(M_y = 4t^3 - 1\). This matches the given condition that \(M_y = N_t\).

Step 3: Derive the Function f from f_t

We are given \(f_t = 4t^3y - 15t^2 - y\). To integrate with respect to \(t\), treat \(y\) as a constant: \[ f = \int (4t^3y - 15t^2 - y) \, dt = t^4y - 5t^3 - ty + h(y) \]where \(h(y)\) is an arbitrary function of \(y\).

Step 4: Determine h(y) using f_y and the Given Information

Since \(h'(y) = 3y^2\), integrate with respect to \(y\) to find \(h(y)\): \[ h(y) = \int 3y^2 \, dy = y^3 + C \]where \(C\) is a constant that, without a specific boundary condition, we set \(C = 0\).

Step 5: Construct the Solution for the Function f

Substitute \(h(y) = y^3\) back into the expression for \(f\):\[ f(t, y) = t^4y - 5t^3 - ty + y^3 \]This matches the form of \(f\) given by the solution \(t^4 y - 5t^3 - t y + y^3 = c\), where \(c\) is a constant integration constant representing different level surfaces of the solution.

Key Concepts

Partial Derivative

A partial derivative is a derivative where we hold some variables constant and differentiate with respect to one variable. It is particularly useful in functions of multiple variables, such as \(M(t, y) = 4t^3y - 15t^2 - y\).
In the exercise, we need to find the partial derivative of \(M\) with respect to \(y\). This means differentiating \(M\) while treating \(t\) as a constant. Thus, \(M_y = 4t^3 - 1\).
This calculation checks if this partial derivative equals \(N_t\), as given by the condition \(4t^3 - 1 = N_t\). When both are equal, the condition is satisfied, assisting us in progressing towards the solution of the function \(f(t, y)\). Partial derivatives help in understanding how a function changes as one particular variable varies.

Integration

Integration is the process used to find a function when its derivative is known. In this context, it involves finding the function \(f(t, y)\) from its partial derivative \(f_t\). The integration involves reversing the process of differentiation.
Given \(f_t = 4t^3y - 15t^2 - y\), we can integrate with respect to \(t\), while treating \(y\) as constant. This integration results in:

• \( f = \int (4t^3y - 15t^2 - y) \ dt = t^4y - 5t^3 - ty + h(y) \)

The expression \(h(y)\) is an arbitrary function of \(y\), showing that an extra component of the function depends solely on \(y\). The concept of integration helps us reconstruct the function \(f(t, y)\) from its derivatives.

Arbitrary Functions

An arbitrary function is a function that can take various forms without a specific structure dictated purely by the given data. In this exercise, \(h(y)\) is an arbitrary function added during integration \(f = t^4 y - 5 t^3 - t y + h(y)\).
This term captures any additional complexity that might only be a function of \(y\) since \(h(y)\) doesn’t depend on \(t\). Determining \(h(y)\) involves further differentiating and integrating as per the given conditions: \(h'(y) = 3y^2\). By integrating \(h'(y)\), we find \(h(y) = y^3 + C\). This explains why such functions can impact solutions in different mathematical contexts like solving partial differential equations.

Boundary Conditions

Boundary conditions are additional constraints or criteria used to determine the constants that arise during integrations, such as \(C\) in the function \(h(y) = y^3 + C\).
In this exercise, while working on determining \(h(y)\), we do not have specified boundary conditions, and hence we chose the constant \(C = 0\) for simplicity. Different boundary conditions would lead to different particular solutions by altering the value of \(C\).
In practical applications, boundary conditions are vital as they allow us to find specific solutions out of many possible options, ensuring that the solution fits the context or real-world problem we are examining. For this exercise, the complete function \(f(t, y) = t^4y - 5t^3 - ty + y^3\) arises from such considerations, embodying a general form that could cater to diverse boundary conditions.