Problem 24 - Advanced Engineering Mathematics

Question

Let \(M=t / 2 y^{4}\) and \(N=\left(3 y^{2}-t^{2}\right) / y^{5}\) so that \(M_{y}=-2 t / y^{5}=N_{t} .\) From \(f_{t}=t / 2 y^{4}\) we obtain \(f=\frac{t^{2}}{4 y^{4}}+h(y)\) \(h^{\prime}(y)=\frac{3}{y^{3}},\) and \(h(y)=-\frac{3}{2 y^{2}} .\) The solution is \(\frac{t^{2}}{4 y^{4}}-\frac{3}{2 y^{2}}=c .\) If \(y(1)=1\) then \(c=-5 / 4\) and a solution of the initial-value problem is \(\frac{t^{2}}{4 y^{4}}-\frac{3}{2 y^{2}}=-\frac{5}{4}\)

Step-by-Step Solution

Answer

The solution of the initial-value problem is \( \frac{t^2}{4y^4} - \frac{3}{2y^2} = -\frac{5}{4} \).

Step 1: Given Functions

We are given two functions, \( M = \frac{t}{2y^4} \) and \( N = \frac{3y^2 - t^2}{y^5} \), that relate to the problem. We will consider their partial derivatives regarding specific variables.

Step 2: Set Partial Derivatives Equal

The problem states that the partial derivative of \( M \) with respect to \( y \), \( M_y \), is equal to the partial derivative of \( N \) with respect to \( t \), \( N_t \). We set \( M_y = -\frac{2t}{y^5} = N_t \).

Step 3: Solve for \( f(t, y) \)

From \( f_t = \frac{t}{2y^4} \), integrate with respect to \( t \) to find \( f(t, y) \). The integration gives \( f(t, y) = \frac{t^2}{4y^4} + h(y) \), where \( h(y) \) is a function of \( y \) only.

Step 4: Determine \( h(y) \)

Using the derivative \( h'(y) = \frac{3}{y^3} \), integrate with respect to \( y \) to find \( h(y) = -\frac{3}{2y^2} + C \), where \( C \) is a constant.

Step 5: Form the Solution

Substitute \( h(y) \) into \( f(t, y) \) to complete the expression: \( f(t, y) = \frac{t^2}{4y^4} - \frac{3}{2y^2} = c \), where \( c \) is a constant.

Step 6: Initial Condition

Apply the initial condition \( y(1) = 1 \) to find the constant \( c \). Setup: \( \frac{1^2}{4 \cdot 1^4} - \frac{3}{2 \cdot 1^2} = c \) simplifies to \( \frac{1}{4} - \frac{3}{2} = c \).

Step 7: Solve for Constant \( c \)

Calculate the value: \( \frac{1}{4} - \frac{3}{2} = -\frac{5}{4} \), which implies that \( c = -\frac{5}{4} \).

Step 8: Solution of Initial-Value Problem

Substitute \( c = -\frac{5}{4} \) into the solution for the initial-value problem: \( \frac{t^2}{4y^4} - \frac{3}{2y^2} = -\frac{5}{4} \).

Key Concepts

Partial Derivatives

Partial derivatives are an essential tool in multivariable calculus. They represent the rate of change of a function with respect to one variable while keeping others constant.
This is particularly important when dealing with functions of multiple variables, like the ones in our exercise. Consider the function \( M = \frac{t}{2y^4} \). The partial derivative of \( M \) with respect to \( y \), denoted as \( M_y \), focuses on how \( M \) changes as \( y \) changes, with \( t \) held constant. In our case, \( M_y = -\frac{2t}{y^5} \).
Similarly, for the function \( N = \frac{3y^2 - t^2}{y^5} \), the partial derivative \( N_t \) reflects how \( N \) changes concerning \( t \), with \( y \) being constant. Here, \( N_t = -\frac{2t}{y^5} \) as well.
When these partial derivatives are equal, it suggests a special relationship and symmetry between \( M \) and \( N \), helpful for solving differential equations.

Integration

Integration is the reverse process of differentiation. It's used here to find a function from its derivative.
In this context, we are given that \( f_t = \frac{t}{2y^4} \), meaning the partial derivative of \( f \) with respect to \( t \) is given, and we're tasked to find \( f(t, y) \).By integrating \( \frac{t}{2y^4} \) with respect to \( t \), we determine that:

• The integral of \( \frac{t}{2y^4} \) with respect to \( t \) gives the expression \( \frac{t^2}{4y^4} \).
• Since \( f(t, y) \) might include parts not dependent on \( t \), we introduce \( h(y) \), a function solely of \( y \), to encompass these constants of integration.

This process allows us to properly reconstruct the function \( f \) from its partial derivative.

Initial Value Problem

An initial value problem involves finding a solution to a differential equation that meets certain specified conditions at a particular point.
These conditions help uniquely determine the constants involved in the solution.In the current exercise, we are given the condition \( y(1) = 1 \). This data point assists in solving for the constant \( c \) in our equation \( \frac{t^2}{4y^4} - \frac{3}{2y^2} = c \). Applying the initial value,

• Substitute \( t = 1 \) and \( y = 1 \) into the equation:
• Set up the relation: \( \frac{1}{4} - \frac{3}{2} = c \).
• Simplify to find that \( c = -\frac{5}{4} \).

The calculated constant \( c = -\frac{5}{4} \) ensures that the solution meets the initial condition provided.

Constant Determination

When solving differential equations, determining the correct constants is crucial for finding the specific solution to a given problem.
This involves the use of initial conditions, which give precise information to solve for these constants.In this problem, determining the constant \( c \) involves substituting known values into a derived equation to ensure it satisfies the given conditions.

• From the equation \( \frac{t^2}{4y^4} - \frac{3}{2y^2} = c \), substitute the initial condition values: \( t = 1 \), \( y = 1 \).
• By solving the resulting equation, \( \frac{1}{4} - \frac{3}{2} = c \), we calculate \( c = -\frac{5}{4} \).
• This constant ensures the solution is particular to the initial condition \( y(1) = 1 \).

Accurately determining these constants helps us to solve the exact problem posed.