Question

Finding a General Solution Using Separation of Variables In Exercises \(1-14,\) find the general solution of the differential equation. $$ y \ln x-x y^{\prime}=0 $$

Short answer

The general solution to the differential equation \(y \ln x-x y'=0\) is \(y = x^x e^{-x} e^C\)

Step-by-step solution

Rewrite the equation

First, rewrite the differential equation in a form that isolates the derivative on one side of the equation. You can do this by dividing each side of the equation by \(y \ln(x)\): \[y' = \frac{y \ln(x)}{x}\]

Separate the variables

Now, dividing each side by \(y\) and multiplying each side by \(dx\), the result is: \[\frac{dy}{y} = \ln(x) dx\]

Integrate both sides

The next step is to integrate both sides of the equation: \[\int \frac{1}{y} dy = \int \ln(x) dx\]

Evaluate the integrals

On the left side of the equation, the integral of \(1/y\) with respect to \(y\) is \(\ln|y|\). The right side integral requires integration by parts to solve. Let \(u=\ln x\) and \(dv=dx\). The integral becomes: \[x\ln x - \int dx = x\ln x - x + C\]

Combine the integrals results

Putting the results of both sides together gives us: \[\ln|y| = x\ln x - x + C\]

Solve for y

Finally, to solve for \(y\), exponentiate both sides of the equation. Then, take the absolute value off by assuming \(y>0\). This gives you the general solution: \[y = e^{x\ln x - x + C} = x^x e^{-x} e^C\]

Key concepts

Differential Equations

A differential equation is an equation that involves an unknown function and its derivatives. In our exercise, the function represents a relationship between variables, specifically, how one quantity changes concerning another. Such equations are vital in modeling real-world phenomena, from physics to engineering. To solve differential equations, we often utilize methods like separation of variables, which is especially suitable for equations where the variables can be moved to opposite sides of the equation.

In this exercise, we encounter the differential equation: \( y \ln x - x y' = 0 \).

• The term \(y \ln x\) represents the function of \(y\) in terms of \(x\).
• The term \(x y'\) is another way of expressing the rate of change (the derivative).

Here, we aim to isolate the derivative, an essential step towards finding a solution. Remember, solving a differential equation means finding a function that satisfies the relationship given by the equation.

Integration by Parts

Integration by parts is a technique based on the product rule for differentiation. It helps us integrate products of functions, particularly when one function is easily differentiated, and the other is easily integrated.

The integration by parts formula is:\[ \int u \, dv = uv - \int v \, du \]In our solution, when integrating \( \ln(x) \), we use integration by parts as follows:

• Choose \( u = \ln(x) \) because its derivative \( du = \frac{1}{x} dx \) is simpler than the integral of \( dv = dx \).
• The integral \( \int dv = x \) becomes simpler, giving us a straightforward path to the solution.

Performing integration by parts transforms the original equation into something more manageable, ensuring that each part of the function is appropriately accounted for.

General Solution

A general solution of a differential equation is a solution that contains all possible solutions. It typically includes a constant, denoted by \(C\), that represents an entire family of solutions.

In our exercise, after manipulating and integrating the given differential equation, we arrive at:\[ \ln|y| = x\ln x - x + C \]

• This form showcases how integration introduces an arbitrary constant into the solution.
• Exponentiating both sides transforms the logarithmic term back into \(y\), giving us:
  \( y = e^{x\ln x - x + C} = x^x e^{-x} e^C \).

The constant \(e^C\) can be redefined as another constant because \(e^C\) itself is just another constant. As a result, our solution effectively describes all potential solutions for the differential equation under given conditions.

Exponential Functions

Exponential functions are powerful mathematical expressions where the variable is the exponent. They're characterized by growth that becomes more rapid in proportion to the growing total number or size.

The function \(f(x) = e^x\) is a quintessential exponential function. In the context of our exercise:
\[ y = x^x e^{-x} e^C \]

• Here, \(e^{-x}\) represents an exponential decay, modifying the overall behavior of the solution.
• \(x^x\) showcases an unusual form, as the base and exponent are the same variable.

Exponential functions frequently appear in differential equations due to their capacity to model growth and decay processes accurately. In the solutions of such equations, they help encapsulate the dynamic behavior of the system being modeled.