Exercise 9.4 · Q6: Homogeneous: $x\,dy - y\,dx = \sqrt{x^2 + y^2}\,dx$

Basics you need

Solution

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

   Move $y\,dx$ to the right and divide by $x\,dx$⁠, for $x \ne 0$⁠:

   $$\begin{aligned} \frac{dy}{dx} &= \frac{y + \sqrt{x^2 + y^2}}{x} \\ &= F(x, y).\quad\ldots(1) \end{aligned}$$
   We need $x \ne 0$ for $v = \frac{y}{x}$⁠, so we solve on $x > 0$ and on $x < 0$⁠. Dividing by $x$ does not lose solutions: Step 8 shows that the answer even joins up across $x = 0$⁠.
   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

   Replace $x$ by $\lambda x$ and $y$ by $\lambda y$⁠. Since $\sqrt{\lambda^2(x^2 + y^2)} = |\lambda|\sqrt{x^2 + y^2}$⁠, for $\lambda > 0$⁠:

   $$\begin{aligned} F(\lambda x, \lambda y) &= \frac{\lambda y + \lambda\sqrt{x^2 + y^2}}{\lambda x} \\ &= \lambda^0\cdot F(x, y). \end{aligned}$$

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

   A careful remark: for $\lambda < 0$ the root gives $-\lambda\sqrt{x^2 + y^2}$⁠, and $F(\lambda x, \lambda y) = \frac{y - \sqrt{x^2 + y^2}}{x}$⁠, which is not $F(x, y)$⁠. What really matters is the form $F = g\left(\frac{y}{x}\right)$⁠, and that holds on each side of the $y$-axis: for $x > 0$⁠, $F = \frac{y}{x} + \sqrt{1 + \frac{y^2}{x^2}}$⁠; for $x < 0$⁠, $F = \frac{y}{x} - \sqrt{1 + \frac{y^2}{x^2}}$⁠. So on each half-plane the equation is homogeneous in the sense of Section 9.4.2, and its method applies there.
   Section 9.4.2: definition of a homogeneous function, and $F(x, y) = g\left(\frac{y}{x}\right)$⁠. Basics: scale x and y together

3. Substitute $y = vx$ (first for $x > 0$⁠)

   Put $y = vx$⁠, so $\frac{dy}{dx} = v + x\frac{dv}{dx}$⁠. For $x > 0$⁠, $\sqrt{x^2 + v^2x^2} = x\sqrt{1 + v^2}$⁠, and (1) becomes

   $$\begin{aligned} v + x\frac{dv}{dx} &= \frac{vx + x\sqrt{1 + v^2}}{x} \\ &= v + \sqrt{1 + v^2}. \end{aligned}$$
   Taking $x$ out of the root as $x$ itself is right only because $x > 0$ here. The case $x < 0$ follows in Step 8.
   Section 9.4.2, equations (2) and (3). Lesson: put y = vx

4. Simplify to $x\frac{dv}{dx} = \sqrt{1 + v^2}$

   $$x\frac{dv}{dx} = \sqrt{1 + v^2}.$$
   Subtract $v$ from both sides; the $v$ terms cancel.
   Section 9.4.2, equation (4). Lesson: homogeneous equations

5. Separate $v$ and $x$

   $$\frac{dv}{\sqrt{1 + v^2}} = \frac{dx}{x}.$$
   Divide by $\sqrt{1 + v^2}$ and by $x$⁠. Since $\sqrt{1 + v^2} \ge 1$ is never zero, this division sets no solution aside.
   Section 9.4.2, equation (5). Basics: dy = y′ dx

6. Integrate, with the constant as $\log|C|$

   By Section 7.4, formula (6) with $a = 1$⁠:

   $$\begin{aligned} &\log\left|v + \sqrt{1 + v^2}\right| \\ &= \log|x| + \log|C|.\quad\ldots(2) \end{aligned}$$

   Check by differentiating the left side with respect to $v$⁠:

   $$\begin{aligned} &\frac{1 + \frac{v}{\sqrt{1 + v^2}}}{v + \sqrt{1 + v^2}} \\ &= \frac{\sqrt{1 + v^2} + v}{\sqrt{1 + v^2}\left(v + \sqrt{1 + v^2}\right)} \\ &= \frac{1}{\sqrt{1 + v^2}}. \end{aligned}$$
   The constant is written as $\log|C|$ ($C \ne 0$⁠) so that the logs combine. Note $v + \sqrt{1 + v^2} > 0$ for every $v$⁠, because $\sqrt{1 + v^2} > |v|$⁠.
   Section 7.4, formula (6); Section 7.2, formula (xii). Basics: every method ends with an integral

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

   From (2), $v + \sqrt{1 + v^2} = |Cx|$⁠; for $x > 0$ this is $Cx$ with $C > 0$⁠. With $v = \frac{y}{x}$ and $x\sqrt{1 + v^2} = \sqrt{x^2 + y^2}$ (as $x > 0$⁠):

   $$\begin{aligned} \frac{y + \sqrt{x^2 + y^2}}{x} &= Cx,\ \text{so} \\ y + \sqrt{x^2 + y^2} &= Cx^2. \end{aligned}$$
   Multiply by $x$⁠. Since $y + \sqrt{x^2 + y^2} > 0$ for $x \ne 0$⁠, the constant $C$ is positive.
   Section 9.4.2: replace $v$ by $\frac{y}{x}$⁠. Basics: one constant for order 1

8. The half-plane $x < 0$⁠, a check, and the curves

   For $x < 0$, $\sqrt{x^2 + y^2} = -x\sqrt{1 + v^2}$⁠, so Step 4 becomes $x\frac{dv}{dx} = -\sqrt{1 + v^2}$⁠, and integrating, $v + \sqrt{1 + v^2} = \frac{K}{|x|} = -\frac{K}{x}$ with $K > 0$⁠. Putting $v = \frac{y}{x}$ gives $\sqrt{x^2 + y^2} - y = K$⁠. Since $\left(\sqrt{x^2 + y^2} + y\right)\left(\sqrt{x^2 + y^2} - y\right) = x^2$⁠, this is $y + \sqrt{x^2 + y^2} = \frac{x^2}{K}$⁠: the same answer with $C = \frac1K$⁠.

   Explicit form. By the same product, $y + \sqrt{x^2 + y^2} = Cx^2$ means $\sqrt{x^2 + y^2} - y = \frac1C$⁠. Subtracting the two,

   $$y = \frac12\left(Cx^2 - \frac1C\right),$$

   an upward parabola for each $C > 0$⁠. Then $y' = Cx$⁠, and $\sqrt{x^2 + y^2} = Cx^2 - y = \frac12\left(Cx^2 + \frac1C\right)$⁠, so

   $$y + \sqrt{x^2 + y^2} = Cx^2 = x\cdot Cx = xy',$$

   which is the given equation $x\,dy - y\,dx = \sqrt{x^2 + y^2}\,dx$⁠. It holds for every real $x$⁠, even $x = 0$⁠, where it reads $0 = y + |y|$ with $y = -\frac{1}{2C} < 0$⁠.

   Both half-planes give one family, so $y + \sqrt{x^2 + y^2} = Cx^2$ is the general solution, and the check confirms it. (A parabola $y = \frac12\left(Cx^2 - \frac1C\right)$ is the set of points whose distance $\sqrt{x^2 + y^2}$ from $O$ equals their distance $y + \frac1C$ from the line $y = -\frac1C$⁠: its focus is $O$⁠.)
   $(a + b)(a - b) = a^2 - b^2$⁠; differentiation (Chapter 5). Basics: a rule for slopes

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