Chapter 8: Problem 2

Identify each of the differential equations as type (for example, separable, linear first order, linear second order, etc.), and then solve it. $$x(\ln y) y^{\prime}-y \ln x=0$$

Short Answer

Expert verified

The differential equation is separable and the solution is $y = e^{\pm \sqrt{(\ln x)^2 + C'}}$.

Step by step solution

Rewrite the Differential Equation

First, let's rewrite the given differential equation in standard form. The equation is: \[x(\ln y) y^{\prime}-y \ln x = 0\]Rewrite it to make it more manageable:\[x(\ln y) \frac{dy}{dx}-y \ln x = 0\]

Identify the Type of Differential Equation

This differential equation can be identified as separable. A separable differential equation can be written as: \[N(y) dy = M(x) dx\]Let's transform the equation into this form.

Separate the Variables

Separate the variables to make it easier to solve:\[x(\ln y) \frac{dy}{dx} = y \ln x\]Divide both sides by $xy$:\[\frac{\ln y}{y} dy = \frac{\ln x}{x} dx\]

Integrate Both Sides

Integrate both sides of the equation:\[\int \frac{\ln y}{y} dy = \int \frac{\ln x}{x} dx\]Use integration by parts for both integrals.

Solve the Integrals

For the integral on the left:Set $u = \ln y$ and $dv = \frac{1}{y} dy$, so $du = \frac{1}{y} dy$ and $v = \ln x$:\[\int \ln y \cdot \frac{1}{y} dy = u v - \int v du\]Simplify and solve it similar for the right side.

Combine Results

After solving both integrals, combine the results to get:\[\frac{1}{2} (\ln y)^2 = \frac{1}{2} (\ln x)^2 + C\]

Simplify the Solution

Multiply the equation by 2 to simplify:\[(\ln y)^2 = (\ln x)^2 + C'\] Where $C' = 2C$.

Solve for y

Take the exponential of both sides to isolate $y$:\[\ln y = \pm \sqrt{(\ln x)^2 + C'}\]Then solve for $y$:\[y = e^{\pm \sqrt{(\ln x)^2 + C'}}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integral Calculus

Integral calculus is a major part of calculus focused on integrals and their properties. It is the process of finding the area under a curve, which is essential in many disciplines such as physics and engineering.
There are two main types of integrals: definite and indefinite.

Definite integrals calculate the net area under a curve between two points.
Indefinite integrals, also known as antiderivatives, represent a family of functions whose derivative is the integrand.

In differential equations, we often deal with indefinite integrals as part of our solution process. Integrals can be computed using various techniques such as substitution, partial fractions, and integration by parts, which we'll discuss further.

Integration by Parts

Integration by parts is a technique derived from the product rule of differentiation. It is particularly useful when the integral involves a product of two functions. The formula is given by:
\[ \int u dv = uv - \int v du \]Where:

u = a function that becomes simpler when differentiated (du)
dv = a function that remains simple when integrated (v)

In solving integral calculus problems within differential equations, such as in Step 5 of our exercise, we apply this technique to manage complex integrals.
Here is an example from our exercise:

Set u = ln y and dv = 1/y dy
then du = 1/y dy and v = y

The integral becomes:
\[ \int \ln y \cdot \frac{1}{y} dy = (\ln y) y - \int (1) dy \]
This simplifies and helps in finding the final solution.

Variable Separation

Variable separation is a method used to solve simple differential equations. In these equations, we can separate the variables on opposite sides of the equation allowing us to integrate each side independently.

The general form of a separable differential equation is:
\[ N(y) dy = M(x) dx \]
In Step 3 of our exercise, we manipulated the equation:

From: \[ x(\ln y) \frac{dy}{dx} = y \ln x \]
To: \[ \frac{\ln y}{y} dy = \frac{\ln x}{x} dx \]

By separating the variables, we could integrate both sides:

\[ \int \frac{\ln y}{y} dy = \int \frac{\ln x}{x} dx \]

This method simplifies the problem, enabling us to find the function y in terms of x. The integral results lead us towards the solution of the differential equation.