Exercise 9.4 · Q1: Homogeneous: $(x^2 + xy)\,dy = (x^2 + y^2)\,dx$

Basics you need

Solution

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

   For $x^2 + xy = x(x + y) \ne 0$⁠, divide by $x^2 + xy$ and by $dx$⁠:

   $$\frac{dy}{dx} = \frac{x^2 + y^2}{x^2 + xy} = F(x, y).\quad\ldots(1)$$
   The homogeneity test is stated for $\frac{dy}{dx} = F(x, y)$⁠. 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 $0\cdot dy = 2x^2\,dx$⁠, 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 $\lambda \ne 0$⁠:

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

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

   Top and bottom both have degree 2, so the factor $\lambda^2$ cancels. Equivalently, dividing top and bottom by $x^2$ gives $F = \frac{1 + (y/x)^2}{1 + y/x}$⁠, 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 $y = vx$⁠, where $v$ is a function of $x$⁠. By the product rule $\frac{dy}{dx} = v + x\frac{dv}{dx}$⁠, and (1) becomes

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

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

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

5. Separate $v$ and $x$

   $$\frac{1 + v}{1 - v}\,dv = \frac{dx}{x}.$$
   Multiply by $\frac{1 + v}{1 - v}$ and by $\frac{dx}{x}$⁠. This needs $x \ne 0$ and $v \ne 1$ ($v \ne -1$ is already assumed, since $x + y \ne 0$⁠). Dividing by $1 - v$ does set one solution aside: the constant $v = 1$⁠, that is $y = x$⁠, makes both sides of $x\frac{dv}{dx} = \frac{1 - v}{1 + v}$ zero, so $y = x$ is a solution (indeed on $y = x$ the given equation reads $2x^2\,dy = 2x^2\,dx$⁠). We will see in Step 7 that the final answer still contains it, 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. Integrate both sides

   The $v$-fraction has top and bottom of the same degree, so divide first:

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

   Then, with $\int\frac{dv}{1 - v} = -\log|1 - v|$ (put $t = 1 - v$⁠, $dt = -dv$⁠) and one constant $C_1$⁠,

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

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

   $$\begin{aligned} &\frac{d}{dv}\big(-v - 2\log|1 - v|\big) \\ &= -1 + \frac{2}{1 - v} = \frac{1 + v}{1 - v}. \end{aligned}$$
   Since $\frac{d}{dv}\log|1 - v| = \frac{-1}{1 - v}$⁠, the derivative of $-2\log|1 - v|$ is $+\frac{2}{1 - v}$⁠, and the check returns the integrand. A first order equation needs one arbitrary constant, written once on the right.
   Section 7.2, $\int\frac{dx}{x} = \log|x| + C$⁠; integration by substitution (Section 7.3.1). Basics: every method ends with an integral

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

   Move the logs together, using $2\log|1 - v| = \log(1 - v)^2$⁠:

   $$\begin{aligned} \log\left|x(1 - v)^2\right| &= -v - C_1, \\ \left|x(1 - v)^2\right| &= e^{-C_1}e^{-v}. \end{aligned}$$

   Now $x(1 - v)^2 = x\cdot\frac{(x - y)^2}{x^2} = \frac{(x - y)^2}{x}$⁠. Removing the modulus puts a sign $\pm$ in front of $e^{-C_1}$⁠; call $C = \pm e^{-C_1}$⁠. Then $\frac{(x - y)^2}{x} = Ce^{-\frac{y}{x}}$⁠, that is

   $$(x - y)^2 = Cxe^{-\frac{y}{x}}.$$
   $e^{-C_1}$ takes every positive value, and with the sign every nonzero value. The value $C = 0$ gives $(x - y)^2 = 0$⁠, the line $y = x$⁠: exactly the solution set aside in Step 5. So with $C$ any real number, the one relation contains every solution, and nothing was lost. It holds for $x \ne 0$ and $y \ne -x$⁠.
   $\log|A| = B$ means $|A| = e^B$⁠. Basics: one constant for order 1

8. Check: differentiate the answer back

   For $C \ne 0$ and $y \ne x$⁠, write the answer as $Ce^{-\frac{y}{x}} = \frac{(x - y)^2}{x}$⁠. Differentiate $(x - y)^2 = Cxe^{-\frac{y}{x}}$ with respect to $x$⁠; since $\frac{d}{dx}\frac{y}{x} = \frac{xy' - y}{x^2}$⁠,

   $$\begin{aligned} &2(x - y)(1 - y') \\ &= Ce^{-\frac{y}{x}}\left(1 - \frac{xy' - y}{x}\right) \\ &= \frac{(x - y)^2}{x}\cdot\frac{x + y - xy'}{x}. \end{aligned}$$

   Divide by $x - y$ and multiply by $x^2$⁠:

   $$\begin{aligned} 2x^2(1 - y') &= (x - y)(x + y - xy') \\ 2x^2 - 2x^2y' &= x^2 - y^2 - (x^2 - xy)y' \\ (x^2 + xy)\,y' &= x^2 + y^2. \end{aligned}$$
   This is the given equation, for every $C \ne 0$⁠; and $C = 0$ gives $y = x$⁠, which we checked in Step 5. So the relation is the general solution.
   Implicit differentiation with the product and chain rules (Chapter 5). Basics: a rule for slopes

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