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Chapter 2: Problem 31

The method outlined in Problem 30 can be used for any homogeneous equation. That is, the substitution \(y=x v(x)\) transforms a homogeneous equation into a separable equation. The latter equation can be solved by direct integration, and then replacing \(v\) by \(y / x\) gives the solution to the original equation. In each of Problems 31 through 38 : $$ \begin{array}{l}{\text { (a) Show that the given equation is homogeneous. }} \\\ {\text { (b) Solve the differential equation. }} \\ {\text { (c) Draw a direction field and some integral curves. Are they symmetric with respect to }} \\ {\text { the origin? }}\end{array} $$ $$ \frac{d y}{d x}=\frac{x^{2}+x y+y^{2}}{x^{2}} $$

Short Answer

Expert verified
#Answer# (a) The given equation is indeed homogeneous. (b) The general solution for the given differential equation is $$ \frac{2}{\sqrt{3}}\arctan \frac{2( \frac{y}{x} + \frac{1}{2})}{\sqrt{3}} = \ln|x| + C $$ (c) Due to the homogeneous nature of the differential equation, the integral curves are symmetric with respect to the origin.

Step by step solution

01

(a) Show the equation is homogeneous

Rewrite the given equation to show it is homogeneous. A differential equation is said to be homogeneous if it can be written in the form: \(\frac{dy}{dx} = F\left(\frac{y}{x}\right)\). So, we need to rewrite the given equation as $$ \frac{dy}{dx} = \frac{y^2+xy+x^2}{x^2} $$ Now, if we divide the numerator and the denominator by \(x^2\), we obtain: $$ \frac{dy}{dx} = \frac{\frac{y^2}{x^2}+\frac{y}{x}+\frac{x^2}{x^2}}{\frac{x^2}{x^2}} = \frac{(\frac{y}{x})^2+\frac{y}{x}+1}{1} = F\left(\frac{y}{x}\right) $$ Thus, the given equation is indeed homogeneous. Now, we move to the next part to solve the equation.
02

(b) Solve the differential equation using substitution

To solve the given differential equation, we use the substitution \(y=xv(x)\). Taking the derivative of \(y\) with respect to \(x\): $$ \frac{dy}{dx} = v(x) + x \cdot\frac{dv}{dx} $$ Now, we substitute \(y = xv(x)\) and \(\frac{dy}{dx} = v(x) + x \cdot\frac{dv}{dx}\) into the original equation: $$ v(x) + x \cdot\frac{dv}{dx} = \frac{(xv(x))^2 + x(xv(x)) + x^2}{x^2} $$ Now, we can simplify the equation by canceling out terms: $$ \frac{dv}{dx} + \frac{v(x)}{x} = v(x)^2 + v(x) + 1 $$ Separating the variables, we get: $$ \frac{dv}{v^2 + v + 1} = \frac{dx}{x} $$ Now, we must integrate both sides of the equation: $$ \int \frac{dv}{v^2 + v + 1} = \int \frac{dx}{x} $$ Let's now solve the integrals.
03

Solve the integrals

We can solve the left-hand side integral by using the substitution \(v = u - \frac{1}{2}\): $$ u = v + \frac{1}{2} \Rightarrow dv = du $$ Now, the integral becomes: $$ \int \frac{du}{u^2 + (\frac{3}{4})} = \int \frac{dx}{x} $$ Using the tangent half-angle substitution for the left-hand side: $$ u = \frac{\sqrt{3}}{2} \tan\theta \Rightarrow du = \frac{\sqrt{3}}{2} \sec^2\theta d\theta $$ So, the equation now looks like: $$ \int \frac{2 du}{3} = \int \frac{dx}{x} $$ Integrate both sides: $$ \frac{2}{\sqrt{3}}\arctan \frac{2( v + \frac{1}{2})}{\sqrt{3}} = \ln|x| + C $$ Now, substitute back \(y = xv(x)\): $$ \frac{2}{\sqrt{3}}\arctan \frac{2( \frac{y}{x} + \frac{1}{2})}{\sqrt{3}} = \ln|x| + C $$ This is the general solution of the given differential equation.
04

(c) Direction field and symmetry of integral curves

As the general solution is implicit, it is not practical to draw the integral curves here. However, general properties of homogeneous equations still apply. Due to the homogeneous nature of the differential equation, we can see that if a point \((x_0, y_0)\) is a solution, then \((\lambda x_0, \lambda y_0)\) with any \(\lambda > 0\) must also be a solution. Therefore, the integral curves are in fact symmetric with respect to the origin, as all curves pass through it.

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Key Concepts

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

Separable Equations
Separable equations are a special class of differential equations where you can separate variables on either side of the equation. This technique allows for simplification of complex expressions. Given a differential equation in the form \(\frac{dy}{dx} = g(x)h(y)\), you can rearrange it as \(\frac{1}{h(y)}dy = g(x)dx\). At this point, you can integrate both sides of the equation. This makes separable equations relatively straightforward to solve once separated.

For example, in our exercise, after using substitution, we arrive at the separable form with \(v\) and \(x\): \(\frac{dv}{v^2 + v + 1} = \frac{dx}{x}\). The key here is recognizing that the structure of the equation permits this separation, making it easier to solve.
Substitution Method
The substitution method is a powerful tool when dealing with homogeneous equations. By introducing a substitution, we can transform a complicated expression into a simpler form. In the given exercise, we apply the substitution \(y = x v(x)\).

This particular substitution helps transform a homogeneous differential equation into a separable one. After substitution, the derivative \(\frac{dy}{dx}\) changes to \(v(x) + x \cdot\frac{dv}{dx}\). From here, we can proceed with separating the variables as explained.

The beauty of this method lies in its ability to reduce complexity, enabling easier integration. Always remember, the goal of substitution is to simplify the equation structure, making it easier to solve. Keep in mind that proper selection of the substitution is crucial to simplify the problem effectively.
Direction Field
A direction field, also known as a slope field, provides a visual representation of a differential equation. It consists of tiny line segments or arrows that indicate the slope of the solution at any given point. This is particularly useful for analyzing the behavior of differential equations even when explicit solutions are hard to find.

In the context of our problem, plotting the direction field helps illustrate the symmetry of integral curves. Since the equation is homogeneous, the direction field shows that the curves are symmetric with respect to the origin.
  • Homogeneous equations show scaling symmetry, meaning if \((x_0, y_0)\) is on a curve, then \((\lambda x_0, \lambda y_0)\) is also on the same curve for any non-zero \(\lambda\).
Observing this symmetry provides an important qualitative understanding of the solution's behavior.
Integration Techniques
Integration is a fundamental step in solving differential equations, especially after transforming them into a separable form. In the exercise, after successfully separating the variables, we deal with integrals of the form \(\int \frac{dv}{v^2 + v + 1}\) and \(\int \frac{dx}{x}\).

The integral on the right, \(\int \frac{dx}{x}\), is a straightforward natural logarithm, \(\ln|x|\). The more complex integral, \(\int \frac{dv}{v^2 + v + 1}\), requires the use of techniques like completing the square or making trigonometric substitutions to solve it.

Choosing the correct technique is key. For instance, completing the square for \(v^2 + v + 1\) aids in setting up a tangent half-angle substitution, simplifying the integral. Employing these techniques simplifies the expression, allowing us to find the solution to the original homogeneous equation.

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Most popular questions from this chapter

Consider the initial value problem \(y^{\prime}=y^{1 / 3}, y(0)=0\) from Example 3 in the text. (a) Is there a solution that passes through the point \((1,1) ?\) If so, find it. (b) Is there a solution that passes through the point \((2,1)\) ? If so, find it. (c) Consider all possible solutions of the given initial value problem. Determine the set of values that these solutions have at \(t=2\)

A tank contains 100 gallons of water and 50 oz of salt. Water containing a salt concentration of \(\frac{1}{4}\left(1+\frac{1}{2} \sin t\right)\) oz/gal flows into the tank at a rate of 2 gal/min, and the mixture in the tank flows out at the same rate. $$ \begin{array}{l}{\text { (a) Find the amount of salt in the tank at any time. }} \\ {\text { (b) Plot the solution for a time period long enough so that you see the ultimate behavior }} \\ {\text { of the graph. }} \\ {\text { (c) The long-time behavior of the solution is an oscillation about a certain constant level, }} \\ {\text { What is this level? What is the amplitude of the oscillation? }}\end{array} $$

A recent college graduate borrows 100,000 dollar at an interest rate of \(9 \%\)to purchase a condominium. Anticipating steady salary increases, the buyer expects to make payments at a monthly rate of 800 dollar (1+t / 120), where t is the number of months since the loan was made. $$ \begin{array}{l}{\text { (a) Assuming that this payment schedule can be maintained, when will the loan be fully }} \\ {\text { paid? }} \\ {\text { (b) Assuming the same payment show large a loan could be paid off in exactly }} \\ {\text { 20 years? }}\end{array} $$

(a) Solve the Gompertz equation $$ d y / d t=r y \ln (K / y) $$ subject to the initial condition \(y(0)=y_{0}\) (b) For the data given in Example 1 in the text \([ \leftr=0.71 \text { per year, } K=80.5 \times 10^{6} \mathrm{kg}\), \right. \(\left.y_{0} / K=0.25\right]\), use the Gompertz model to find the predicted value of \(y(2) .\) (c) For the same data as in part (b), use the Gompertz model to find the time \(\tau\) at which \(y(\tau)=0.75 K .\) Hint: You may wish to let \(u=\ln (y / K)\).

Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 liters of a dye solution with a concentration of 1 g/liter. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 liters/min, the well-stirred solution flowing out at the same rate. Find the time that will elapse before the concentration of dye in the tank reaches \(1 \%\) of its original value.
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