Exercise 9.4 · Q9: Homogeneous: $y\,dx + x\log\frac{y}{x}\,dy - 2x\,dy = 0$

Basics you need

Solution

1. Write the equation as $\frac{dy}{dx} = F(x, y)$

   Collect the $dy$ terms, $x\left(2 - \log\frac{y}{x}\right)dy = y\,dx$⁠, and divide:

   $$\frac{dy}{dx} = \frac{y}{2x - x\log\frac{y}{x}} = F(x, y).\quad\ldots(1)$$
   The log needs $\frac{y}{x} > 0$ (and so $x \ne 0$⁠, $y \ne 0$⁠). Dividing by $2 - \log\frac{y}{x}$ loses no solution: where $\log\frac{y}{x} = 2$⁠, that is on the line $y = e^2x$⁠, the given equation reads $y\,dx = 0$ with $y \ne 0$⁠, which forces $dx = 0$⁠, a vertical tangent. So no solution $y(x)$ with a finite slope passes through such a point.
   Section 9.4.2: a homogeneous equation has the form $\frac{dy}{dx} = F(x, y)$⁠. Lesson: homogeneous equations

2. Show that it is homogeneous: $F(\lambda x, \lambda y) = F(x, y)$

   For any $\lambda \ne 0$⁠, $\frac{\lambda y}{\lambda x} = \frac{y}{x}$⁠, so the log does not change:

   $$\begin{aligned} F(\lambda x, \lambda y) &= \frac{\lambda y}{2\lambda x - \lambda x\log\frac{y}{x}} \\ &= \frac{\lambda y}{\lambda\left(2x - x\log\frac{y}{x}\right)} \\ &= \lambda^0\cdot F(x, y). \end{aligned}$$

   So $F$ is homogeneous of degree zero, and (1) is a homogeneous differential equation.

   Top and bottom each pick up one factor $\lambda$⁠. Dividing top and bottom by $x$⁠: $F = \frac{y/x}{2 - \log(y/x)}$⁠, a function of $\frac{y}{x}$ alone.
   Section 9.4.2: definition of a homogeneous function of degree $n$⁠. Basics: scale x and y together

3. Substitute $y = vx$

   Put $y = vx$ ($v > 0$⁠). Then $\frac{dy}{dx} = v + x\frac{dv}{dx}$⁠, and (1) becomes

   $$v + x\frac{dv}{dx} = \frac{vx}{2x - x\log v} = \frac{v}{2 - \log v}.$$
   Cancel $x \ne 0$⁠: the right side is a function of $v$ only.
   Section 9.4.2, equations (2) and (3). Lesson: put y = vx

4. Simplify to $x\frac{dv}{dx} = \ldots$

   $$\begin{aligned} x\frac{dv}{dx} &= \frac{v}{2 - \log v} - v \\ &= \frac{v - v(2 - \log v)}{2 - \log v} \\ &= \frac{v - 2v + v\log v}{2 - \log v} \\ &= \frac{v(\log v - 1)}{2 - \log v}. \end{aligned}$$
   Subtract $v$ over the common denominator $2 - \log v$⁠, then take out the factor $v$⁠.
   Section 9.4.2, equation (4). Lesson: homogeneous equations

5. Separate $v$ and $x$⁠, and see what is set aside

   $$\frac{2 - \log v}{v(\log v - 1)}\,dv = \frac{dx}{x}.$$
   Dividing by $v(\log v - 1)$ needs $\log v \ne 1$ ($v > 0$ already). The excluded constant $v = e$ makes $v(\log v - 1) = 0$⁠, so the line $y = ex$ is a solution. (Check in the given equation: $dy = e\,dx$ and $\log\frac{y}{x} = 1$⁠, so $ex\,dx + x\cdot e\,dx - 2x\cdot e\,dx = 0$⁠.) Step 7 shows it comes back with $C = 0$⁠.
   Section 9.4.2, equation (5); separating as in Section 9.4.1, which needs the divisor to be nonzero, so its zeros are checked separately. Basics: dy = y′ dx

6. Put $t = \log v$ and integrate

   With $t = \log v$⁠, $dt = \frac{dv}{v}$⁠, the left side becomes $\frac{2 - t}{t - 1}\,dt$⁠. Divide first:

   $$\frac{2 - t}{t - 1} = \frac{-(t - 1) + 1}{t - 1} = -1 + \frac{1}{t - 1}.$$

   Integrating both sides, with the constant written as $\log|C|$ ($C \ne 0$⁠):

   $$\begin{aligned} &\log|t - 1| - t \\ &= \log|x| + \log|C|.\quad\ldots(2) \end{aligned}$$

   Check by differentiating the left side with respect to $t$⁠: $\frac{1}{t - 1} - 1 = \frac{2 - t}{t - 1}$⁠, the integrand.

   The substitution $t = \log v$ is allowed because $v > 0$⁠, and $\frac{dv}{v}$ is exactly the factor $\frac1v$ already present. Top and bottom have the same degree in $t$⁠, so we divide before integrating.
   Integration by substitution, Section 7.3.1; Section 7.2, formula (xii). Basics: every method ends with an integral

7. Put $t = \log v$ and $v = \frac{y}{x}$ back

   Move $t$ across and combine the logs: $\log|t - 1| = t + \log|Cx|$⁠. With $t = \log v$⁠:

   $$\begin{aligned} \log|\log v - 1| &= \log v + \log|Cx| \\ &= \log|Cxv| = \log|Cy|. \end{aligned}$$

   So $\log v - 1 = \pm Cy$⁠; naming the constant $C$ again and putting $v = \frac{y}{x}$⁠:

   $$Cy = \log\left(\frac{y}{x}\right) - 1.$$
   $xv = y$⁠. So far $C \ne 0$⁠; the value $C = 0$ gives $\log\frac{y}{x} = 1$⁠, the line $y = ex$ set aside in Step 5. So with $C$ any real number, nothing is lost. The answer holds for $x \ne 0$⁠, $\frac{y}{x} > 0$⁠. (The book's key writes $cy = \log\left|\frac{y}{x}\right| - 1$⁠; since $\frac{y}{x} > 0$ wherever the equation is defined, the bars change nothing.)
   $\log a + \log b = \log ab$⁠; $\log|A| = \log|B| \Rightarrow A = \pm B$⁠. Basics: one constant for order 1

8. Check: differentiate the answer back

   For $C \ne 0$ write the answer as $\log y - \log x - 1 = Cy$ (on $x, y > 0$⁠; the case $x, y < 0$ is the same with $\log(-y) - \log(-x)$⁠). Differentiating with respect to $x$⁠:

   $$\frac{y'}{y} - \frac1x = Cy'.$$

   Put $C = \frac{\log(y/x) - 1}{y}$ and multiply by $xy$⁠:

   $$\begin{aligned} xy' - y &= x\left(\log\frac{y}{x} - 1\right)y', \\ \left(2x - x\log\frac{y}{x}\right)y' &= y, \end{aligned}$$

   which is $y\,dx + x\log\frac{y}{x}\,dy - 2x\,dy = 0$⁠.

   We recover the equation for every $C \ne 0$⁠, and $C = 0$ was checked in Step 5, so the relation is the general solution.
   Implicit differentiation (Chapter 5); $\frac{d}{dx}\log|u| = \frac{u'}{u}$ (Section 7.2, formula (xii), with the chain rule). Basics: a rule for slopes

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