Problem 13 - Advanced Engineering Mathematics

Question

Letting \(y=u x\) we have $$\begin{aligned} \left(x+u x e^{u}\right) d x-x e^{u}(u d x+x d u) &=0 \\ d x-x e^{u} d u &=0 \\ \frac{d x}{x}-e^{u} d u &=0 \\ \ln |x|-e^{u}=c. \end{aligned}$$ Using \(y(1)=0\) we find \(c=-1 .\) The solution of the initial-value problem is \(\ln |x|=e^{y / x}-1\).

Step-by-Step Solution

Answer

The solution is \( \ln |x| = e^{y/x} - 1 \).

Step 1: Substitute the given substitution

We start by using the substitution \( y = ux \), which implies \( dy = u dx + x du \). Substitute these into the original differential equation.

Step 2: Substitute into the differential equation

Substitute \( y = ux \) and \( dy = udx + xdu \) into the equation \((x + ux e^u) dx - x e^u (udy + xdu) = 0\). This simplifies to \( (x + ux e^u) dx - x e^u u dx - x e^u x du = 0 \).

Step 3: Simplify and factor the equation

Simplify the equation by combining like terms: \( dx - x e^u du = 0 \). Notice that the term \( (x + ux e^u) - x e^u u \) results in just \( dx \) - \( x e^u du \).

Step 4: Separate variables and integrate

Rearrange the simplified equation to \( \frac{dx}{x} = e^u du \). Integrate both sides: \( \int \frac{dx}{x} = \int e^u du \), yielding \( \ln |x| = e^u + c \).

Step 5: Apply the initial condition

Use the initial condition \( y(1) = 0 \). This implies \( u = 0 \) when \( x = 1 \), so substitute into the equation \( \ln |1| = e^0 + c \) which simplifies to \( 0 = 1 + c \). Solving for \( c \) gives \( c = -1 \).

Step 6: Substitute back and solve for x

Return to the equation with the determined constant: \( \ln |x| = e^u + c \), substitute \( c = -1 \): \( \ln |x| = e^{y/x} - 1 \). This completes the solution to the initial-value problem.

Key Concepts

Separation of Variables

The technique of separation of variables is a powerful tool for solving differential equations. It allows for the simplification of a problem by dividing variables into two separate groups. This process helps in finding a solution more easily.
In this exercise, we started with an equation that linked different variables and their derivatives. By carefully rearranging terms, we separated the variables into distinct sides of the equation.
This made it possible to integrate each side individually. Integration could then be conducted without interference from the opposing variable. The goal was to isolate variables, such as all terms containing "\(x\)" on one side and all terms containing "\(u\)" on the other. This allowed for straightforward integration.

• This problem was simplified to \( \frac{dx}{x} = e^u \, du \), offering a clearer path to solving the problem.
• Ultimately, separating variables allows us to solve complex differential equations step by step, making them more manageable.

Initial Value Problems

Initial value problems require not only solving the differential equation but also applying specific conditions to find a unique solution. These conditions are given in the form of initial values. The role of initial values is crucial as they tailor the general solution to meet certain criteria.
In our problem, the condition given was \( y(1) = 0 \). This particular condition means that when \( x = 1 \), \( y \) should equal 0. Utilizing this information helps in calculating the constant of integration.

• Without initial conditions, solutions to differential equations can be indefinitely broad.
• By applying these conditions, we can determine values like "\( c = -1 \)," making our final solution specific and applicable to the problem.

Initial value problems help blend calculus techniques with specific problem scenarios, providing precise solutions.

Substitution Method

The substitution method is a clever tactic to simplify complex differential equations. This method involves introducing a new variable to express an existing one, creating a simpler form of the original equation.
In this exercise, we replaced \( y \) with \( u x \). This allowed a transformation that made the equation easier to manage.
The effect of substitution is significant. It often changes the dynamic of the equation, offering a fresh perspective on solving it. By expressing one variable in terms of others, the problem becomes less tangled.

• The substitution \( y = u x \) led to derivative relationships like \( dy = u dx + x du \).
• This setup enabled further manipulations and simplifications necessary to tackle the problem.

Using substitution is like simplifying a puzzle by swapping pieces, providing a path to a solution that might not be initially obvious.

Integration

Integration is a fundamental technique in calculus used to solve differential equations. It essentially involves finding the antiderivative or integral of a function.
In our exercise, after separating variables, we integrated each side to solve for "\( x \)" and "\( e^u \)." This process reversed differentiation, revealing relationships between the variables.
The goal of integration in this context is to arrive at an equation that includes an integration constant. This constant is determined using initial conditions to provide the specific solution.

• In our solution, after integrating, we obtained \( \ln |x| = e^u + c \).
• Applying the initial condition, we determined the specific value of \( c \), completing our equation as \( \ln |x| = e^{y/x} - 1 \).

Integration transforms equations and connects calculus with real-world applications, making it a powerful analytic tool.