Exercise 9.4 · Q11: Particular solution of $(x + y)\,dy + (x - y)\,dx = 0$⁠, $y(1) = 1$

Basics you need

Solution

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

   Move $(x - y)\,dx$ across and divide by $(x + y)\,dx$⁠:

   $$\begin{aligned} \frac{dy}{dx} &= \frac{-(x - y)}{x + y} = \frac{y - x}{x + y} \\ &= F(x, y).\quad\ldots(1) \end{aligned}$$
   We need $x \ne 0$ anyway, because $v = \frac{y}{x}$ will be used. Dividing by $x + y$ loses no solution: at a point of the line $y = -x$ (with $x \ne 0$⁠) the given equation reads $0\cdot dy + 2x\,dx = 0$⁠, 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: $F(\lambda x, \lambda y) = F(x, y)$

   Replace $x$ by $\lambda x$ and $y$ by $\lambda y$⁠, for any $\lambda \ne 0$⁠:

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

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

   Top and bottom both have degree 1, so one factor $\lambda$ comes out of each and cancels.
   Section 9.4.2: homogeneous function of degree $n$⁠. Basics: scale x and y together

3. Substitute $y = vx$

   Put $y = vx$⁠; then $\frac{dy}{dx} = v + x\frac{dv}{dx}$⁠, and (1) becomes

   $$v + x\frac{dv}{dx} = \frac{vx - x}{x + vx} = \frac{v - 1}{1 + v}.$$
   Dividing top and bottom by $x \ne 0$ leaves a function of $v$ only, as homogeneity guarantees.
   Section 9.4.2, equations (2) and (3). Lesson: put y = vx

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

   $$\begin{aligned} x\frac{dv}{dx} &= \frac{v - 1}{1 + v} - v \\ &= \frac{v - 1 - v - v^2}{1 + v} \\ &= -\frac{1 + v^2}{1 + v}. \end{aligned}$$
   Write $v$ over the common denominator $1 + v$⁠: $v = \frac{v + v^2}{1 + v}$⁠. The $v$ terms cancel and $-1 - v^2 = -(1 + v^2)$⁠.
   Section 9.4.2, equation (4): $x\frac{dv}{dx} = g(v) - v$⁠. Lesson: homogeneous equations

5. Separate $v$ and $x$

   $$\frac{1 + v}{1 + v^2}\,dv = -\frac{dx}{x}.$$
   Multiply by $\frac{1 + v}{1 + v^2}$ and by $\frac{dx}{x}$⁠. Dividing by $1 + v^2$ loses no solution, because $1 + v^2 \ge 1$ is never 0 (and $1 + v \ne 0$ was settled in Step 1).
   Section 9.4.2, equation (5); separating as in Section 9.4.1. Basics: dy = y′ dx

6. Integrate both sides

   Split the fraction: $\frac{1 + v}{1 + v^2} = \frac{1}{1 + v^2} + \frac{v}{1 + v^2}$⁠. For the second piece put $t = 1 + v^2$⁠, $dt = 2v\,dv$⁠, so $\int\frac{v\,dv}{1 + v^2} = \frac12\int\frac{dt}{t} = \frac12\log(1 + v^2)$⁠. With one constant $C_1$⁠:

   $$\begin{aligned} \tan^{-1}v &+ \tfrac{1}{2}\log(1 + v^2) \\ &= -\log|x| + C_1.\quad\ldots(2) \end{aligned}$$

   Check: differentiating the left side of (2) with respect to $v$⁠,

   $$\begin{aligned} &\frac{d}{dv}\left(\tan^{-1}v + \tfrac12\log(1 + v^2)\right) \\ &= \frac{1}{1 + v^2} + \frac{v}{1 + v^2}, \end{aligned}$$

   the integrand.

   No modulus is needed in $\log(1 + v^2)$⁠, because $1 + v^2 > 0$⁠. The equation has order 1 (and degree 1, as a polynomial in $\frac{dy}{dx}$⁠), so one arbitrary constant is enough.
   Section 7.2, formulas (x) and (xii); substitution, Section 7.3.1. Basics: every method ends with an integral

7. Put $v = \frac{y}{x}$ back: the general solution

   Multiply (2) by 2, move $2\log|x| = \log x^2$ to the left, and write $C = 2C_1$⁠:

   $$2\tan^{-1}v + \log\left[(1 + v^2)x^2\right] = C.$$

   Since $(1 + v^2)x^2 = x^2 + y^2$⁠,

   $$\log(x^2 + y^2) + 2\tan^{-1}\frac{y}{x} = C.\quad\ldots(3)$$
   $\log(1 + v^2) + \log x^2 = \log\left[(1 + v^2)x^2\right]$⁠, and $x^2\left(1 + \frac{y^2}{x^2}\right) = x^2 + y^2$⁠. $C = 2C_1$ is again one arbitrary constant. This is the general solution, for $x \ne 0$ and $x + y \ne 0$⁠.
   Section 9.4.2: replace $v$ to get the general solution. Lesson: Examples 10 to 13

8. Use the condition: $y = 1$ when $x = 1$

   Putting $x = 1$⁠, $y = 1$ in (3):

   $$\begin{aligned} C &= \log 2 + 2\tan^{-1}1 \\ &= \log 2 + 2\cdot\frac{\pi}{4} = \frac{\pi}{2} + \log 2. \end{aligned}$$

   So the particular solution is

   $$\begin{aligned} &\log(x^2 + y^2) + 2\tan^{-1}\frac{y}{x} \\ &= \frac{\pi}{2} + \log 2. \end{aligned}$$
   $1^2 + 1^2 = 2$⁠, and $\tan^{-1}1 = \frac{\pi}{4}$ (principal value, in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$⁠). Of all the curves (3), only this one passes through $(1, 1)$⁠.
   Section 9.3: a particular solution from the general one. Lesson: general and particular solutions

9. Check by substituting back, and see the curve

   Check: differentiate (3) with respect to $x$⁠. Since $\frac{d}{dx}\tan^{-1}\frac{y}{x} = \frac{1}{1 + y^2/x^2}\cdot\frac{xy' - y}{x^2} = \frac{xy' - y}{x^2 + y^2}$⁠,

   $$\begin{aligned} \frac{2x + 2yy'}{x^2 + y^2} + \frac{2(xy' - y)}{x^2 + y^2} &= 0 \\ x + yy' + xy' - y &= 0 \\ (x + y)y' + (x - y) &= 0, \end{aligned}$$

   which is the given equation, for every $C$⁠. The point: $\log 2 + 2\cdot\frac{\pi}{4} = \frac{\pi}{2} + \log 2$⁠, so $(1, 1)$ lies on the particular solution.

   Where it lives. The method needs $x \ne 0$⁠, and the given point has $x = 1$⁠, so we work on $x > 0$⁠. There, with $x = r\cos\theta$⁠, $y = r\sin\theta$ and $\theta = \tan^{-1}\frac{y}{x}$⁠, the answer reads $\log r^2 + 2\theta = \frac{\pi}{2} + \log 2$⁠, that is

   $$r = \sqrt2\,e^{\frac{\pi}{4} - \theta}:$$

   a spiral. As $\theta$ runs from $\frac{\pi}{2}$ down to $-\frac{\pi}{4}$⁠, $x$ runs from 0 up to $e^{\pi/2} \approx 4.81$⁠. At $\theta = -\frac{\pi}{4}$ the curve meets the line $y = -x$ at $\left(e^{\pi/2}, -e^{\pi/2}\right)$ with a vertical tangent, as Step 1 predicts. So the particular solution is a function $y(x)$ on $0 < x < e^{\pi/2}$⁠.

   Getting the equation back proves the answer. The polar form shows what the implicit answer hides: the curve is part of a spiral around the origin, and as a function of $x$ it stops where it meets $y = -x$⁠.
   Implicit differentiation and the chain rule for $\tan^{-1}\frac{y}{x}$ (Chapter 5). Basics: a rule for slopes

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