Problem 26 - Advanced Engineering Mathematics

Question

Let \(M=y^{2}+y \sin x\) and \(N=2 x y-\cos x-1 /\left(1+y^{2}\right)\) so that \(M_{y}=2 y+\sin x=N_{x} .\) From \(f_{x}=y^{2}+y \sin x\) we obtain \(f=x y^{2}-y \cos x+h(y), h^{\prime}(y)=\frac{-1}{1+y^{2}},\) and \(h(y)=-\tan ^{-1} y .\) The solution is \(x y^{2}-y \cos x-\tan ^{-1} y=c\) If \(y(0)=1\) then \(c=-1-\pi / 4\) and a solution of the initial-value problem is \(x y^{2}-y \cos x-\tan ^{-1} y=-1-\frac{\pi}{4}\)

Step-by-Step Solution

Answer

Solution: \( xy^2 - y \cos x - \tan^{-1} y = -1 - \frac{\pi}{4} \).

Step 1: Identify the Given Functions

The problem provides us with two main functions: \( M = y^2 + y \sin x \) and \( N = 2xy - \cos x - \frac{1}{1+y^2} \). We also have a condition \( M_y = 2y + \sin x = N_x \).

Step 2: Verify the Given Condition

Differentiate \( M \) with respect to \( y \) to get \( M_y = 2y + \sin x \). Differentiate \( N \) with respect to \( x \) to get \( N_x = 2y + \sin x \). This confirms the condition \( M_y = N_x \).

Step 3: Derive the Potential Function

To find the potential function \( f \), integrate \( M \) with respect to \( x \) assuming constant terms in \( y \). This gives: \[ f(x, y) = xy^2 - y \cos x + h(y) \]. Then differentiate \( f \) with respect to \( y \) to equate it with \( N \) to find \( h'(y) \).

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

From the condition \( h'(y) = \frac{-1}{1+y^2} \), integrate to find \( h(y) = -\tan^{-1} y \), adding a constant that will later be incorporated into the general solution.

Step 5: Write the General Solution

Substitute \( h(y) = -\tan^{-1} y \) into the expression for \( f(x, y) \, : \, f(x, y) = xy^2 - y \cos x - \tan^{-1} y = c \), where \( c \) is an arbitrary constant.

Step 6: Use the Initial Condition to Find \( c \)

Given the initial condition \( y(0) = 1 \), substitute \( x = 0 \) and \( y = 1 \) into the equation and solve for \( c \) :\[ 0 \cdot 1^2 - 1 \cdot \cos(0) - \tan^{-1}(1) = c \rightarrow -1 - \frac{\pi}{4} = c \].

Step 7: State the Particular Solution

The equation with this specific constant \( c \) is \[ xy^2 - y \cos x - \tan^{-1} y = -1 - \frac{\pi}{4} \], which is the particular solution to the initial-value problem given \( y(0) = 1 \).

Key Concepts

Initial Value Problem

An initial value problem is a type of differential equation where the solution is determined by an initial condition. This means we are given a specific value for the function at a particular point, often denoted as the condition at time zero or at an initial point.
To solve an initial value problem, we integrate the differential equation to find a general solution, which includes a constant of integration.
Then, we use the initial condition to solve for this constant, ensuring our solution matches the specified value at the initial point. For example, in our exercise, after deriving the general solution of the differential equation with an arbitrary constant, the initial condition provided was used to find the specific constant value.

Potential Function

The potential function is central in solving differential equations, especially when dealing with conservative vector fields. It is a scalar function whose gradient yields the fields represented by the system of equations.
In our problem, to find the potential function, we integrated one of the differential equations while assuming the other variables as constants. This integration step combined with partial differentiation ensures the function consistently applies across the system.
Once we integrated to obtain the potential function up to an unknown function of one variable, further calculations, like comparing with the second equation, allowed us to fully define the potential function.

Integration

Integration is a key mathematical process used to find functions when given their derivatives, and it applies extensively in solving differential equations. In the context of the exercise, integration helped us find potential functions.
The process involves integrating one equation partially, while holding other variables constant. This effectively adds an unknown function of one variable.

• For the given function, we integrated with respect to x, suspecting that some terms may only depend on y.
• After integrating, we determined the potential function using an additional step of differentiation or by leveraging known results for the inverse derivatives of trigonometric functions.

The integration ensured the derived function encompassed all necessary components for a solution.

Partial Differentiation

Partial differentiation involves taking the derivative of a multi-variable function with respect to one variable while holding others constant. It is an essential technique when dealing with functions of several variables, as encountered frequently in differential equations.
In our problem, partial differentiation was used to verify the compatibility of our derived functions and to ensure the condition \( M_y = N_x \) was accurately represented. This check is crucial whenever constructing or verifying potential functions for consistency.
By equivalently differentiating the potential function and matching it with known expressions, partial differentiation verifies the interrelations of components within a complex equation system.