Example 4: General solution of $\frac{dy}{dx} = \frac{x+1}{2-y}$

Basics you need

Solution

1. Recognise the form: variables separable

   $$\begin{aligned} \frac{dy}{dx} &= \frac{x + 1}{2 - y} = (x + 1)\cdot\frac{1}{2 - y} \\ &= g(x)\,h(y), \end{aligned}$$

   with $g(x) = x + 1$ and $h(y) = \frac{1}{2 - y}$⁠. Since $y \ne 2$⁠, $h(y)$ is defined and is never 0.

   The method of Section 9.4.1 needs the right side to split as (a function of $x$ only) $\times$ (a function of $y$ only), and needs $h(y) \ne 0$ so that we may divide by it. Both hold here.
   Section 9.4.1: $\frac{dy}{dx} = h(y)\,g(x)$ is the variables separable type. Lesson: separate, then integrate

2. Separate the variables

   Multiply both sides by $(2 - y)\,dx$⁠:

   $$(2 - y)\,dy = (x + 1)\,dx.$$
   Now the left side contains only $y$ (with $dy$⁠) and the right side only $x$ (with $dx$⁠). Multiplying by $2 - y$ is allowed because $2 - y \ne 0$⁠. In the language of Section 9.4.1 this is $\frac{1}{h(y)}\,dy = g(x)\,dx$ with $\frac{1}{h(y)} = 2 - y$⁠.
   Separating: $y$ with $dy$⁠, $x$ with $dx$⁠. Basics: dy = y′ dx

3. Integrate both sides, with one constant

   $$\int(2 - y)\,dy = \int(x + 1)\,dx$$ $$2y - \frac{y^2}{2} = \frac{x^2}{2} + x + C_1.$$

   Check each integral by differentiating: $\frac{d}{dy}\left(2y - \frac{y^2}{2}\right) = 2 - y$ and $\frac{d}{dx}\left(\frac{x^2}{2} + x\right) = x + 1$⁠.

   Each side is integrated with respect to its own variable by the power rule. Each integration produces a constant, but two constants $k_1$⁠, $k_2$ on opposite sides combine into a single one, $C_1 = k_2 - k_1$⁠. A first order equation needs exactly one arbitrary constant.
   Power rule, Section 7.2; one constant for order 1. Basics: order n gives n constants

4. Tidy up and rename the constant

   Multiply both sides by $-2$⁠:

   $$-4y + y^2 = -x^2 - 2x - 2C_1.$$

   Move every term to the left:

   $$x^2 + y^2 + 2x - 4y + 2C_1 = 0.$$

   Write $C = 2C_1$⁠. Then the general solution is

   $$x^2 + y^2 + 2x - 4y + C = 0.$$
   Multiplying by $-2$ clears the fractions. Since $C_1$ can be any real number, so can $2C_1$⁠; naming it $C$ keeps the answer clean without losing any solution. There is still exactly one arbitrary constant, as order 1 requires.
   A fixed multiple of an arbitrary constant is again an arbitrary constant. Basics: order n gives n constants

5. Check: put the answer back into the equation

   Differentiate $x^2 + y^2 + 2x - 4y + C = 0$ with respect to $x$⁠, treating $y$ as a function of $x$⁠:

   $$\begin{aligned} 2x + 2y\frac{dy}{dx} + 2 - 4\frac{dy}{dx} &= 0 \\ (2y - 4)\frac{dy}{dx} &= -(2x + 2) \\ \frac{dy}{dx} &= \frac{x + 1}{2 - y}. \end{aligned}$$
   This is exactly the given equation, for every value of $C$⁠, so the relation really is a solution. The last line divides by $2y - 4 = -2(2 - y)$⁠, which is not 0 because $y \ne 2$⁠. The constant $C$ disappears on differentiation, as an arbitrary constant must.
   Implicit differentiation (Section 5.3.2). Basics: y′ is the slope

6. What the solution curves look like

   Complete the squares in $x$ and in $y$⁠:

   $$(x + 1)^2 + (y - 2)^2 = 5 - C.$$

   For $C < 5$ this is a circle with centre $(-1, 2)$ and radius $\sqrt{5 - C}$⁠. (For $C = 5$ it is the single point $(-1, 2)$⁠, which has $y = 2$⁠; for $C > 5$ there are no real points.) The line $y = 2$ is excluded, so each circle splits into an upper and a lower half, and each half is a solution curve. At the two points where a circle meets $y = 2$⁠, its tangent is vertical: there the slope $\frac{x + 1}{2 - y}$ would be infinite.

   The picture shows what "general solution" means: one curve for each value of the constant, and every curve follows the slope marks. Here different values of $C$ give circles of different sizes about the same centre.
   A differential equation is a rule for slopes at every point. Basics: dy/dx = F(x, y) draws an arrow at every point

Every fact used, and where it comes from