Problem 9 - Advanced Engineering Mathematics

Question

In Problems 1-22, solve the given differential equation by separation of variables. $$ y \ln x \frac{d x}{d y}=\left(\frac{y+1}{x}\right)^{2} $$

Step-by-Step Solution

Answer

\(x \ln x - x = \frac{y^2}{2} + 2y + \ln |y| + C\) is the implicit solution.

Step 1: Rearrange the Equation

The given equation is \(y \ln x \frac{dx}{dy} = \left(\frac{y+1}{x}\right)^{2}\). Start by separating the variables \(x\) and \(y\). Multiply both sides by \(\frac{dy}{y}\) to isolate the differential \(dx\). This results in \(\ln x \cdot dx = \frac{(y+1)^2}{y} \cdot\frac{dy}{x^2}\).

Step 2: Separate Variables

Separate the variables by integrating each side with respect to its own variable. Rewrite the equation as \(x^2 \ln x \, dx = \frac{(y+1)^2}{y} \, dy\). Move all expressions in terms of \(x\) to one side and those in terms of \(y\) to the other side resulting in \(\ln x \, dx = \frac{(y+1)^2}{y} \, dy\).

Step 3: Integrate Both Sides

Integrate both sides separately. Integrate \(\ln x \, dx\) and \(\frac{(y+1)^2}{y} \, dy\):\[ \int \ln x \, dx = \int \frac{(y+1)^2}{y} \, dy \]The integration results in separate expressions for each side.

Step 4: Solve the Integrals

Perform integration:- The integral \(\int \ln x \, dx\) involves integration by parts and equals to \(x \ln x - x + C_1\).- For \(\int \frac{(y+1)^2}{y} \, dy\), simplify the expression to \((y + 2 + \frac{1}{y})\) which results in \(\frac{y^2}{2} + 2y + \ln |y| + C_2\).

Step 5: Combine the Solutions

The left side integration yields \(x \ln x - x + C_1\) and the right side gives \(\frac{y^2}{2} + 2y + \ln |y| + C_2\). Combine these to form the equation:\[ x \ln x - x = \frac{y^2}{2} + 2y + \ln |y| + C \] where \(C = C_2 - C_1\) is the constant of integration.

Step 6: Present Final Solution

The final form of the solution after separating variables and integrating is:\[ x \ln x - x = \frac{y^2}{2} + 2y + \ln |y| + C \]This implicit solution relates \(x\) and \(y\).

Key Concepts

Separation of Variables

Separation of variables is a powerful technique used to solve differential equations by dividing and isolating terms involving different variables on opposite sides of the equation. This method simplifies complex equations by allowing you to integrate each variable separately. In practice, you aim to rearrange the equation so that all terms involving one variable, say \(x\), are multiplied by the \(dx\) differential, and terms involving the other variable, say \(y\), are multiplied by the \(dy\) differential.

• First, identify the variables and differentials within the equation. In our example, we have the differential equation: \(y \ln x \, \frac{dx}{dy} = \left(\frac{y+1}{x}\right)^2\).
• Next, manipulate the equation to isolate these variables on either side. Multiplying both sides by \(\frac{dy}{y}\), we get: \(\ln x \cdot dx = \frac{(y+1)^2}{y} \, dy\).
• This produces two integrals, \(\int \ln x \, dx\) and \(\int \frac{(y+1)^2}{y} \, dy\), allowing us to solve them independently.

This technique lays the groundwork for finding an implicit solution, which connects the variables in a relationship involving a constant of integration.

Integration by Parts

Integration by parts is an essential tool in calculus for solving integrals where the standard method of integration is not easily applicable. It's particularly useful when dealing with the product of functions, such as \(\int u \, dv\), by using the formula:\[\int u \, dv = uv - \int v \, du\]This formula helps reduce complex integrals into simpler ones. Here’s how it works in our problem:

• Choose the function \(u\) (which in our case is \(\ln x\)) and \(dv\) (which is \(dx\)). Differentiate \(u\) to get \(du\) and integrate \(dv\) to get \(v\).
• Our example integral \(\int \ln x \, dx\) is solved by selecting \(u = \ln x\) and \(dv = dx\). Then, \(du = \frac{1}{x} \, dx\) and \(v = x\).
• Substituting these into the integration by parts formula yields \(x \ln x - \int x \cdot \frac{1}{x} \, dx = x \ln x - x + C_1\).

Using integration by parts transforms a challenging integral into a more manageable one, allowing us to progress further in solving the differential equation.

Implicit Solutions

Implicit solutions are forms in which the solution to a differential equation is given as a relationship between the variables rather than explicitly solving for one variable in terms of the other. This approach often emerges in complicated equations where finding an explicit solution is either difficult or impossible.In our context, after integrating both sides, we arrive at an implicit solution:\[ x \ln x - x = \frac{y^2}{2} + 2y + \ln |y| + C \]Here’s why implicit solutions are valuable:

• They offer flexibility in solving complex relationships, especially when the equation involves nonlinear terms.
• Since both sides of the equation have been integrated separately, you need only ensure continuity across the solution by applying initial conditions or extra constraints when available, to determine the constant \(C\).
• Implicit solutions help describe the relationship between \(x\) and \(y\) comprehensively, even without giving a direct formula \(x = f(y)\) or \(y = g(x)\).

In problem-solving contexts, implicit solutions are both revealing and comprehensive, giving a complete picture of the relationship between the variables from the differential equation.