Problem 23 - Advanced Engineering Mathematics

Question

Solve the given initial-value problem. $$ (4 y+2 t-5) d t+(6 y+4 t-1) d y=0, \quad y(-1)=2 $$

Step-by-Step Solution

Answer

The solution is \(2yt + t^2 - 5t + 3y^2 - y = 16\).

Step 1: Check if the differential equation is exact

We have the differential equation \((4y + 2t - 5) dt + (6y + 4t - 1) dy = 0\). First, identify \(M = 4y + 2t - 5\) and \(N = 6y + 4t - 1\). Check for exactness by comparing the partial derivatives: \(\frac{\partial M}{\partial y} = 4\) and \(\frac{\partial N}{\partial t} = 4\). Since these are equal, the equation is exact.

Step 2: Find the potential function \( \\psi(t,y) \)

To find the potential function \( \psi(t, y) \), integrate \(M\) with respect to \(t\), giving \(\psi(t, y) = \int (4y + 2t - 5) \, dt = 2yt + t^2 - 5t + g(y)\). Next, differentiate \( \psi(t, y)\) with respect to \(y\) to find \(g'(y)\). Then solve \( \frac{\partial \psi}{\partial y} = 6y + 4t - 1\) to find \(g(y)\). By integrating \( 6y - 1\), we get \(3y^2 - y + C\). So, \(\psi(t, y) = 2yt + t^2 - 5t + 3y^2 - y + C\).

Step 3: Use initial condition to find C

With the initial condition \(y(-1) = 2\), substitute \(t = -1\) and \(y = 2\) into \(\psi(t, y) = C\): \(2(-1)(2) + (-1)^2 - 5(-1) + 3(2)^2 - 2 = C\). This solves to \(C = 16\). Hence, \(\psi(t, y) = C\) becomes \(2yt + t^2 - 5t + 3y^2 - y = 16\).

Step 4: General solution and verifying

The general solution of the differential equation is \(2yt + t^2 - 5t + 3y^2 - y = 16\). Verify the solution by substituting the initial condition \(y(-1) = 2\) into this equation. The left side evaluates to 16, agreeing with the right side, confirming the solution.

Key Concepts

Initial-Value Problem

An initial-value problem is a type of differential equation that comes with a specific condition known as an initial condition. This condition lets us find a unique solution to the differential equation, tailored to a particular moment in time. In our exercise, the initial condition given is \( y(-1) = 2 \).
This information tells us that when the variable \( t \) is equal to \(-1\), the value of \( y \) should be \( 2 \).
Without this condition, we could end up with infinitely many solutions.
By using the initial condition, we're able to pin down the constant \( C \) in our solution, ensuring it fits perfectly through the specified point.

• Initial condition: Provides a specific coordinate (in this case, \((-1, 2)\)) through which the solution curve must pass.
• Uniqueness: The initial condition guarantees a unique solution instead of a general form that may include multiple solutions.

Understanding initial-value problems like this one is crucial in physics and engineering, where conditions at a particular time and point significantly impact how the solution behaves later.

Potential Function

The concept of a potential function in the context of exact differential equations is a tool that streamlines finding a solution. It's a function from which both parts of the given differential equation can be derived.
For our problem, the task was to find such a potential function \( \psi(t, y) \) such that:

• The partial derivative of \( \psi \) with respect to \( t \) corresponds to the function \(M(t, y)\) (which was found as \(4y + 2t - 5\)).
• The partial derivative with respect to \( y \) is equivalent to \(N(t, y)\) (which was \(6y + 4t - 1\)).

An essential step was integrating \(M(t, y)\) with respect to \(t\), which involved treating \(y\) as a constant, and adding an arbitrary function \(g(y)\) to account for any part of the potential function that only depends on \(y\). This ultimately led to discovering \( \psi(t, y) = 2yt + t^2 - 5t + 3y^2 - y + C \).

The potential function is key because it encompasses all solutions of the differential equation by setting \( \psi(t, y) = C \), where \(C\) is a constant determined by our initial condition.

Partial Derivatives

Partial derivatives are a fundamental part of multivariable calculus. When dealing with functions of several variables, partial derivatives provide the rate of change concerning one variable while keeping the others constant.
In our exact differential equation:

• \( \frac{\partial M}{\partial y} \) represents the partial derivative of \(M(t, y)\) with respect to \(y\).
• \( \frac{\partial N}{\partial t} \) is the derivative of \(N(t, y)\) concerning \( t \).

These two derivatives were used to assess whether the given differential equation was exact. A differential equation is termed exact if \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial t} \), suggesting that the functions originate from the same potential function.

In our solution, both partial derivatives were discovered to be equal, confirming the differential equation's exactness. This allowed further steps towards finding the potential function, showing partial derivatives' critical role in identifying exactness and solving these equations.

Integration Techniques

Integration techniques are vital in solving differential equations and finding potential functions. It involves determining the antiderivative of functions to construct solutions. In our problem, we used integration to find the potential function by handling each part of the differential equation.
First, integrating \(M(t, y) = 4y + 2t - 5\) with respect to \( t \) resulted in part of the potential function:

• \( \int (4y + 2t - 5) \, dt = 2yt + t^2 - 5t + g(y) \)

The integration treats \( y \) as a constant while \( t \) is the variable of integration. Next, to find \(g(y)\), the function solely in terms of \(y\), we differentiate the intermediate potential function with respect to \( y \) and compare it to \( N(t, y) = 6y + 4t - 1 \). This comparison led us to integrate again to solve for \(g(y) = 3y^2 - y + C\).
By this careful integration process and constant adjustments using initial conditions, we finalised the potential function, completing the solution. Understanding and applying these techniques is crucial for working through differential equations and creating accurate solutions.