Question

Homogencous Equations. If the right side of the equation $d y / d x=f(x, y)$ can be expressed as a function of the ratio $y / x$ only, then the equation is said to be homogeneous. Such equations can always be transformed into separable equations by a change of the dependent variable. Problem 30 illustrates how to solve first order homogencous equations. Consider the equation $$ \frac{d y}{d x}=\frac{y-4 x}{x-y} $$ $$ \text { (a) Show that } \mathrm{Eq} \text { . (i) can be rewritten as } $$ $$ \begin{array}{l}{\text { thus Eq. (i) is homogeneous, }} \\ {\text { (b) Introduce a new dependent variable } v \text { so that } v=y / x, \text { or } y=x v(x) \text { . Express } d y / d x} \\ {\text { in terms of } x, v, \text { and } d v / d x \text { . }} \\ {\text { (c) Replace } y \text { and } d y / d x \text { in Eq. (ii) by the expressions from part (b) that involve } v \text { and }} \\ {d \tau / d x . \text { Show that the resulting differential equation is }}\end{array} $$ $$ v+x \frac{d v}{d x}=\frac{v-4}{1-v} $$ $$ \text { or } $$ $$ x \frac{d v}{d x}=\frac{v^{2}-4}{1-v} $$ $$ \begin{array}{l}{\text { Observe that } \mathrm{Eq} \text { (iii) is separable. }} \\ {\text { (d) Solve } \mathrm{Eq} \text { (iii) for } v \text { in terms oy replaing } v \text { by } y / x \text { in the solution in part (d). }} \\ {\text { (t) Find the solution of Eq. (i) by replaing } v \text { by } y / x \text { in the solution in part (d). }} \\ {\text { (h) Draw a direction field and some intregan curves for } \mathrm{Eq} \text { (i) Recall that the right side }}\end{array} $$ $$ \begin{array}{l}{\text { of Eq. ( 1) actually depends only on the ratio } y / x . \text { This means that integral curves have the }} \\ {\text { same slope at all points on any given straight line through the origin, although the slope }} \\ {\text { changes from one line to another. Therefore the direction field and the integral curves are }} \\ {\text { symmetric with respect to the origin. Is this symmetry property evident from your plot? }}\end{array} $$

Name: Problem 30, Chapter 2: Elementary Differential Equations and Boundary Value Problem
Uploaded: 2021-07-22T01:51:34-08:00
Duration: 10 min 1 s
Channel: Aidan Mcnabb

   Homogencous Equations. If the right side of the equation $d y / d x=f(x, y)$ can be expressed as a function of the ratio $y / x$ only, then the equation is said to be homogeneous. Such equations can always be transformed into separable equations by a change of the dependent variable. Problem 30 illustrates how to solve first order homogencous equations.
Consider the equation
$$
\frac{d y}{d x}=\frac{y-4 x}{x-y}
$$
$$
\text { (a) Show that } \mathrm{Eq} \text { . (i) can be rewritten as }
$$
$$
\begin{array}{l}{\text { thus Eq. (i) is homogeneous, }} \\ {\text { (b) Introduce a new dependent variable } v \text { so that } v=y / x, \text { or } y=x v(x) \text { . Express } d y / d x} \\ {\text { in terms of } x, v, \text { and } d v / d x \text { . }} \\ {\text { (c) Replace } y \text { and } d y / d x \text { in Eq. (ii) by the expressions from part (b) that involve } v \text { and }} \\ {d \tau / d x . \text { Show that the resulting differential equation is }}\end{array}
$$
$$
v+x \frac{d v}{d x}=\frac{v-4}{1-v}
$$
$$
\text { or }
$$
$$
x \frac{d v}{d x}=\frac{v^{2}-4}{1-v}
$$
$$
\begin{array}{l}{\text { Observe that } \mathrm{Eq} \text { (iii) is separable. }} \\ {\text { (d) Solve } \mathrm{Eq} \text { (iii) for } v \text { in terms oy replaing } v \text { by } y / x \text { in the solution in part (d). }} \\ {\text { (t) Find the solution of Eq. (i) by replaing } v \text { by } y / x \text { in the solution in part (d). }} \\ {\text { (h) Draw a direction field and some intregan curves for } \mathrm{Eq} \text { (i) Recall that the right side }}\end{array}
$$
$$
\begin{array}{l}{\text { of Eq. ( 1) actually depends only on the ratio } y / x . \text { This means that integral curves have the }} \\ {\text { same slope at all points on any given straight line through the origin, although the slope }} \\ {\text { changes from one line to another. Therefore the direction field and the integral curves are }} \\ {\text { symmetric with respect to the origin. Is this symmetry property evident from your plot? }}\end{array}
$$

Elementary Differential Equations and Boundary Value Problem

William E. Boyce,… 7th Edition

Chapter 2, Problem 30

Instant Answer

Solved by Expert Aidan Mcnabb

07/22/2021

Step 1

We can rewrite this equation in terms of $y/x$ by factoring out an $x$ from both the numerator and the denominator. This gives us $\frac{d y}{d x}=\frac{y/x-4}{1-y/x}$. Show more…

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Thanks for your feedback!

Homogencous Equations. If the right side of the equation $d y / d x=f(x, y)$ can be expressed as a function of the ratio $y / x$ only, then the equation is said to be homogeneous. Such equations can always be transformed into separable equations by a change of the dependent variable. Problem 30 illustrates how to solve first order homogencous equations. Consider the equation $$ \frac{d y}{d x}=\frac{y-4 x}{x-y} $$ $$ \text { (a) Show that } \mathrm{Eq} \text { . (i) can be rewritten as } $$ $$ \begin{array}{l}{\text { thus Eq. (i) is homogeneous, }} \\ {\text { (b) Introduce a new dependent variable } v \text { so that } v=y / x, \text { or } y=x v(x) \text { . Express } d y / d x} \\ {\text { in terms of } x, v, \text { and } d v / d x \text { . }} \\ {\text { (c) Replace } y \text { and } d y / d x \text { in Eq. (ii) by the expressions from part (b) that involve } v \text { and }} \\ {d \tau / d x . \text { Show that the resulting differential equation is }}\end{array} $$ $$ v+x \frac{d v}{d x}=\frac{v-4}{1-v} $$ $$ \text { or } $$ $$ x \frac{d v}{d x}=\frac{v^{2}-4}{1-v} $$ $$ \begin{array}{l}{\text { Observe that } \mathrm{Eq} \text { (iii) is separable. }} \\ {\text { (d) Solve } \mathrm{Eq} \text { (iii) for } v \text { in terms oy replaing } v \text { by } y / x \text { in the solution in part (d). }} \\ {\text { (t) Find the solution of Eq. (i) by replaing } v \text { by } y / x \text { in the solution in part (d). }} \\ {\text { (h) Draw a direction field and some intregan curves for } \mathrm{Eq} \text { (i) Recall that the right side }}\end{array} $$ $$ \begin{array}{l}{\text { of Eq. ( 1) actually depends only on the ratio } y / x . \text { This means that integral curves have the }} \\ {\text { same slope at all points on any given straight line through the origin, although the slope }} \\ {\text { changes from one line to another. Therefore the direction field and the integral curves are }} \\ {\text { symmetric with respect to the origin. Is this symmetry property evident from your plot? }}\end{array} $$

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Key Concepts

Direction Fields and Symmetry of Integral Curves

Direction fields provide a visual representation of the slopes of the solution curves of a differential equation at various points in the plane. In the case of homogeneous differential equations, since the expression for the slope depends only on the ratio y/x, the direction field exhibits symmetry with respect to the origin. This means that along any straight line through the origin, the slopes are constant, which is evident in the symmetric patterns of both the direction field and the integral curves.

Separable Differential Equations

A separable differential equation is one where the variables can be isolated on different sides of the equation, typically in the form g(v) dv = h(x) dx. Once a homogeneous differential equation has been transformed using the substitution v = y/x, it frequently becomes separable, allowing the integration of each side separately. This method is a fundamental technique for solving first-order differential equations.

Homogeneous Differential Equations

A first-order differential equation is called homogeneous if the function on the right-hand side can be written as a function of the ratio y/x only. This means that the function is invariant under scaling of the independent and dependent variables, and it often exhibits symmetry properties. Homogeneous equations can be identified by noting that the degrees of x and y in every term are the same, which allows the substitution of y/x to simplify the equation.

Change of Variables (Substitution v = y/x)

The substitution v = y/x is a common method used to handle homogeneous differential equations. By expressing y in terms of x and v (as y = xv), the derivative dy/dx can be rewritten using the product rule, converting the original equation into one that involves v, x, and the derivative dv/dx. This substitution reduces the complexity of the equation by transforming it into an equation that often separates the variables.

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Transcript

00:01 Hello everyone, welcome back to another differential equations problem.

00:03 This is a long one, so let's get right into it.

00:05 Here we have dydx is equal to y minus 4x over x minus y.

00:11 Now the first part of this problem, part a, asks to rewrite this equation.

00:16 So all the variables are y over x instead of y and x separately.

00:21 So what i noticed is that we can factor out an x from both the top and the bottom.

00:27 I'm going to do that here.

00:28 So we'll take x and times y over x, remember this is just the numerator of this up here, and subtract 4.

00:37 If we multiply this x out and distribute it, we'll get the numerator in the last problem.

00:43 If we do the same thing in the denominator, we'll get 1 minus 5 over x.

00:50 And if we distribute this x out, we'll get the same as the initial problem.

00:57 And now that we have an x in the numerator and denominator, we can just cancel it out.

01:02 Great, so that's part a.

01:05 Part b is to replace y of x with a variable v.

01:10 This is just a temporary variable.

01:13 But we're also going to try to find dvdx, so the derivative of v with respect to x.

01:21 All right, so let's just start off with the easy part, replacing y over x with v.

01:27 So we'll have v minus 4 over 1 minus v.

01:32 Now we have to find dvdx.

01:37 This is a little bit more challenging, but if we realize that v is in terms of x, such that y equals x times v of x, this is given in the problem statement, then we can just do the product rule of derivatives and get what dvd x is.

01:58 So we'll have do y dx equal to x terms of the derivative of v.

02:06 With respect x and the same thing but flipped plus v the derivative of x with respect x not just one great part c says to conflate these two values so we see that this is actually dy dx itself since that's the original problem and this is also do idx right there so they're really equal all right in fact if we do some simple we can find that dv dx times x is equal to v squared minus 4 over 1 minus v just through algebraic manipulation all right let's move on to part d part d says to solve our differential equation explicitly getting v in terms of x without this dv d x term all right so to do this we have to realize that this is a separable differential equation then we can get the d x is on the right and the dv is on the left.

03:14 So i wanted to do that.

03:16 We'll have dv times 1 minus v all over v squared minus 4.

03:25 Then on the right side, we'll just be dx over x.

03:31 Great.

03:32 And let's integrate the right side.

03:36 The left side will have to do some partial fractions on, but it won't be too bad.

03:40 I'm just going to do these partial fractions quickly, skip all the labor intensive steps, because that's not what this problem is really about.

03:47 But it will be dv times this quantity of negative three -fourths times my bad negative three over four times v minus two the v plus two minus one -fourth and the v minus two is in the denominator recognize that this is just separating these two right here great and this is going to be equal to dx over x let's integrate both sides this should be simple because these are all logarithms so we will have negative those factors is negative 1 4th and we will have the natural log three times the natural log of v plus 2 and add this to the natural log of v minus 2 great and that will be equal to the natural log of x.

05:10 Perfect.

05:15 The next step is to, actually let's multiply this negative 4 over to the right side and do the power rule.

05:22 At the same time, let's combine these two terms.

05:26 So we will have the natural log of the quantity of v plus 2 cubed...

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