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Chapter 2: Problem 36

Devise an appropriate substitution to solve $$ x y^{\prime}=y \ln (x y) $$

Short Answer

Expert verified
The solution is \( y = \frac{C'}{x} \), where \( C' \) is a constant.

Step by step solution

01

Rewrite the Differential Equation

Start by rewriting the given differential equation. The equation is given by \[ x y' = y \ln (xy) \]This can be rewritten by recognizing that \( y' = \frac{dy}{dx} \). So, \[ x \frac{dy}{dx} = y \ln(xy) \]
02

Introduce a Substitution

To simplify, we'll use the substitution \( z = xy \). From this, it follows that \( y = \frac{z}{x} \), and differentiating with respect to \( x \), we have:\[ y' = \frac{d}{dx}\left(\frac{z}{x}\right) = \frac{1}{x} \frac{dz}{dx} - \frac{z}{x^2} \] Now substituting \( y \) and \( y' \) back in the original equation, we find:\[ x \left( \frac{1}{x} \frac{dz}{dx} - \frac{z}{x^2} \right) = \frac{z}{x} \ln(z) \]
03

Simplify the Equation

After substitution, the equation becomes:\[ \frac{dz}{dx} - \frac{z}{x} = \ln(z) \frac{z}{x} \]Simplifying further gives:\[ \frac{dz}{dx} = \ln(z) \frac{z}{x} + \frac{z}{x} \]\[ \frac{dz}{dx} = \frac{z}{x}(1 + \ln(z)) \]
04

Solve the Separable Equation

The equation \( \frac{dz}{dx} = \frac{z}{x}(1 + \ln(z)) \) is separable. We can separate it as:\[ \frac{1}{z(1+\ln(z))} dz = \frac{1}{x} dx \]Now integrate both sides:\[ \int \frac{1}{z(1+\ln(z))} dz = \int \frac{1}{x} dx \]
05

Perform the Integrations

The integral on the right side is straightforward, yielding:\[ \ln|x| + C_1 \]For the left side, use a substitution \( u = \ln(z) \), which simplifies the integration to:\[ \int \frac{1}{(u+1)e^u} d(e^u) = u \]\[ = \ln(z) \]
06

Write the General Solution

Combine the integrals to find the general solution:\[ \ln(z) = \ln|x| + C \]Which implies that:\[ z = |x| e^C \]Recall that \( z = xy \), so:\[ xy = |x|e^C \]
07

Express the Solution in Terms of Original Variables

Convert back to the original variables to express the solution in terms of \( y \):\[ y = \frac{|x|e^C}{x} \]Assuming \( x > 0 \), to remove absolute values, the solution is:\[ y = e^C \frac{1}{x} \]
08

Simplify the Constant

Allow \( e^C \) to be a new constant, say \( C' \). Therefore, simplify the solution to:\[ y = \frac{C'}{x} \]

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Key Concepts

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

Substitution Method
The substitution method is a powerful technique used in differential equations to simplify complex expressions. In dealing with differential equations, a smart substitution can transform a challenging problem into a manageable one.
  • Choose a substitution that simplifies the equation.
  • Identify expressions that can be changed into a single variable.
  • Transform derivatives according to the substitution.
In our exercise, the substitution was cleverly chosen as \( z = xy \). With this substitution, the differential equation鈥檚 complexity is reduced significantly. This change of variables helps isolate terms and makes the upcoming steps more straightforward.
However, it鈥檚 crucial to remember to always transform back into the original variables once the differential equation is solved. This ensures that the final answer corresponds to the original problem.
Separable Equations
Separable equations are a subtype of ordinary differential equations that can be rearranged such that all terms involving one variable are on one side, and all terms involving the other are on the opposite side.
  • Identify if an equation can be separated by variables.
  • Separate the variables to different sides of the equation.
  • Integrate both sides separately.
In the given problem, once the equation is rewritten using the substitution, it becomes separable: \( \frac{1}{z(1+\ln(z))} dz = \frac{1}{x} dx \). By isolating \( dz \) and \( dx \) in this manner, we can employ integration techniques effectively to solve the equation.
This method is valuable for problems where initial conditions are provided or when finding a specific solution, facilitating straightforward integration.
Integration Techniques
Integration is a fundamental process in solving differential equations, and applying the right technique is essential.
  • Direct integration is used when the integral is straightforward.
  • Substitution in integration may be necessary when the denominator involves expressions like \( (u+1)e^u \).
  • Be mindful to integrate both sides of a separable equation.
In our problem, integration was necessary after separating variables. On the right side, \( \int \frac{1}{x} dx \) led to \( \ln|x| \).
The left side required clever substitution to simplify the integral involving \( z \). The key here was to transform variables within the integral to obtain a solvable form, leading to \( \ln(z) \). This step is critical to coming up with the general solution for the differential equation.
Ordinary Differential Equation Solution
Ordinary Differential Equations (ODEs) often need unique strategies for solutions. Understanding how to derive the general solution after integrations is crucial.
  • Combine integrated results to form the general solution.
  • Substitute back any free constants to express the solution entirely.
  • Simplify the solution to reflect initial variables and conditions.
In this exercise, after integrating both sides, the general solution \( \ln(z) = \ln|x| + C \) is deduced. To return to original terms, remember \( z = xy \), thereby allowing the transformation back to original variables.
Substitute and simplify to find \( y \) in terms of \( x \). It's pivotal to express solutions in clear, reduced forms, such as \( y = \frac{C'}{x} \). Understanding these steps helps derive meaningful results from ODEs while addressing all initial conditions and substitutions involved.

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Most popular questions from this chapter

Two chemicals \(A\) and \(B\) are combined to form a chemical \(C\). The rate, or velocity, of the reaction is proportional to the product of the instantaneous amounts of \(A\) and \(B\) not converted to chemical \(C\). Initially there are 40 grams of \(A\) and 50 grams of \(B\), and for each gram of \(B\), 2 grams of \(A\) is used. It is observed that 10 grams of \(C\) is formed in 5 minutes. How much is formed in 20 minutes? What is the limiting amount of \(C\) after a long time? How much of chemicals \(A\) and \(B\) remains after a long time?

The population of a town grows at a rate proportional to the population present at time \(t\). The initial population of 500 increases by \(15 \%\) in 10 years. What will the population be in 30 years? How fast is the population growing at \(t=30 ?\)

The differential equation $$ \frac{d y}{d x}=\frac{-x+\sqrt{x^{2}+y^{2}}}{y} $$ describes the shape of a plane curve \(C\) that will reflect all incoming light beams to the same point and could be a model for the mirror of a reflecting telescope, a satellite antenna, or a solar collector. See Problem 29 in Exercises \(1.3\). There are several ways of solving this \(\mathrm{DE}\). (a) Verify that the differential equation is homogeneous (see Section 2.5). Show that the substitution \(y=u x\) yields $$ \frac{u d u}{\sqrt{1+u^{2}}\left(1-\sqrt{\left.1+u^{2}\right)}\right.}=\frac{d x}{x}. $$ Use a CAS (or another judicious substitution) to integrate the left-hand side of the equation. Show that the curve \(C\) must be a parabola with focus at the origin and is symmetric with respect to the \(x\) -axis. (b) Show that the first differential equation can also be solved by means of the substitution \(u=x^{2}+y^{2}\).

Solve the given differential equation by using an appropriate substitution. $$ \frac{d y}{d x}=\tan ^{2}(x+y) $$

Solve the given initial-value problem by finding, as in Example 4, an appropriate integrating factor. $$ \left(x^{2}+y^{2}-5\right) d x=(y+x y) d y, \quad y(0)=1 $$
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