Exercise 9.4 · Q2: Homogeneous: $y' = \frac{x + y}{x}$

Basics you need

Solution

1. Name the right side $F(x, y)$

   $$\frac{dy}{dx} = \frac{x + y}{x} = F(x, y).\quad\ldots(1)$$
   The equation is already in the form $\frac{dy}{dx} = F(x, y)$⁠, so nothing is divided and nothing is lost. The only restriction is the one the question itself carries: $x \ne 0$⁠, because $x$ is in the denominator. So the solutions live on $x > 0$ or on $x < 0$⁠.
   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: $F(\lambda x, \lambda y) = F(x, y)$

   For any $\lambda \ne 0$⁠:

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

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

   Top and bottom both have degree 1, so $\lambda$ cancels. In the other form, $F = 1 + \frac{y}{x}$⁠, 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 $y = vx$⁠, with $v$ a function of $x$⁠. Then $\frac{dy}{dx} = v + x\frac{dv}{dx}$⁠, and (1) becomes

   $$v + x\frac{dv}{dx} = \frac{x + vx}{x} = 1 + v.$$
   The product rule differentiates $v\cdot x$⁠. Dividing $x + vx$ by $x \ne 0$ leaves $1 + v$⁠, a function of $v$ only.
   Section 9.4.2, equations (2) and (3). Lesson: put y = vx

4. Simplify to $x\frac{dv}{dx} = 1$

   $$x\frac{dv}{dx} = (1 + v) - v = 1.$$
   Subtract $v$ from both sides. Here $g(v) - v = 1$ is never zero, so the next division (by $g(v) - v$⁠) cannot lose any solution.
   Section 9.4.2, equation (4): $x\frac{dv}{dx} = g(v) - v$⁠. Lesson: homogeneous equations

5. Separate and integrate

   $$dv = \frac{dx}{x},\quad\text{so}\quad \int dv = \int\frac{dx}{x},$$ $$v = \log|x| + C.\quad\ldots(2)$$
   Divide by $x \ne 0$ and multiply by $dx$⁠. Both integrals are standard; one arbitrary constant $C$ is enough for a first order equation. Check: $\frac{d}{dx}\log|x| = \frac1x$ on both $x > 0$ and $x < 0$⁠.
   Section 9.4.2, equations (5) and (6); $\int\frac{dx}{x} = \log|x| + C$⁠. Basics: every method ends with an integral

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

   $$\begin{aligned} \frac{y}{x} &= \log|x| + C,\ \text{so} \\ y &= x\log|x| + Cx. \end{aligned}$$
   Multiply by $x$⁠. This is an explicit formula for $y$⁠, valid on $x > 0$ and on $x < 0$⁠; $C$ is any real number.
   Section 9.4.2: replace $v$ by $\frac{y}{x}$ in (6). Basics: one constant for order 1

7. Check: substitute the answer back

   For $y = x\log|x| + Cx$⁠, the product rule gives

   $$\begin{aligned} y' &= \log|x| + x\cdot\frac1x + C \\ &= \log|x| + 1 + C, \end{aligned}$$

   and the right side of the equation is

   $$\begin{aligned} \frac{x + y}{x} &= \frac{x + x\log|x| + Cx}{x} \\ &= 1 + \log|x| + C. \end{aligned}$$
   The two agree for every $x \ne 0$ and every $C$⁠, so $y = x\log|x| + Cx$ is the general solution.
   Product rule (Chapter 5, Section 5.3); $\frac{d}{dx}\log|x| = \frac1x$ (Section 7.2, formula (xii)). Basics: y′ is the slope

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