Example 6: Particular solution of $\frac{dy}{dx} = -4xy^2$⁠, $y(0) = 1$

Basics you need

Solution

1. Recognise the form, and set $y = 0$ aside

   $$\frac{dy}{dx} = (-4x)\cdot y^2 = g(x)\,h(y),$$

   with $g(x) = -4x$ and $h(y) = y^2$⁠. Here $h(y) = 0$ when $y = 0$⁠. The constant function $y = 0$ does satisfy the equation (both sides are 0), but it does not satisfy $y = 1$ when $x = 0$⁠. So we look for a solution with $y \ne 0$⁠, and may divide by $y^2$⁠.

   Dividing by $h(y)$ is only allowed where $h(y) \ne 0$⁠. Checking the zero of $h$ separately makes sure no solution is lost; here the lost one, $y = 0$⁠, is not the one we want.
   Section 9.4.1: separation needs $h(y) \ne 0$⁠. Lesson: separate, then integrate

2. Separate the variables

   $$\frac{dy}{y^2} = -4x\,dx.$$
   Divide both sides by $y^2$ (allowed, $y \ne 0$⁠) and multiply by $dx$⁠. The left side now has only $y$⁠, the right side only $x$⁠.
   Separating: $y$ with $dy$⁠, $x$ with $dx$⁠. Basics: dy = y′ dx

3. Integrate both sides, with one constant

   $$\begin{aligned} \int y^{-2}\,dy &= -4\int x\,dx \\ -\frac{1}{y} &= -2x^2 + C. \end{aligned}$$

   Check each integral by differentiating: $\frac{d}{dy}\left(-\frac{1}{y}\right) = \frac{1}{y^2}$ and $\frac{d}{dx}(-2x^2) = -4x$⁠.

   The power rule with $n = -2$ gives $\frac{y^{-1}}{-1} = -\frac1y$⁠; with $n = 1$ it gives $\frac{x^2}{2}$⁠, and $-4 \cdot \frac{x^2}{2} = -2x^2$⁠. The two constants of integration combine into one $C$⁠.
   Section 7.2, formula (i). Basics: every method ends with an integral

4. Solve for $y$⁠: the general solution

   Multiply by $-1$⁠: $\frac{1}{y} = 2x^2 - C$⁠. Taking reciprocals,

   $$y = \frac{1}{2x^2 - C}\quad\ldots(1)$$

   for the $x$ with $2x^2 \ne C$⁠.

   This is the general solution (for $y \ne 0$⁠): one curve for each value of $C$⁠. Taking the reciprocal is allowed because $\frac1y$ is never 0.
   Section 9.3: the solution with an arbitrary constant is the general solution. Lesson: general and particular solutions

5. Use the condition to find $C$

   Put $x = 0$ and $y = 1$ in (1):

   $$\begin{aligned} 1 &= \frac{1}{2\cdot 0 - C} = \frac{1}{-C} \\ \Rightarrow\ -C &= 1 \;\Rightarrow\; C = -1. \end{aligned}$$

   Substituting $C = -1$ in (1):

   $$y = \frac{1}{2x^2 + 1}.$$
   The condition says the curve passes through $(0, 1)$⁠. Exactly one value of $C$ makes (1) pass through that point. With it, no arbitrary constant is left: this is the particular solution.
   Section 9.3: a particular solution comes from giving the constant a value. Basics: one constant, one condition

6. Check: substitute back, and see the curve

   For $y = \frac{1}{2x^2 + 1} = (2x^2 + 1)^{-1}$⁠, the chain rule gives

   $$\begin{aligned} \frac{dy}{dx} &= -\frac{4x}{(2x^2 + 1)^2} \\ &= -4x\cdot\left(\frac{1}{2x^2 + 1}\right)^2 = -4xy^2. \end{aligned}$$

   And at $x = 0$⁠, $y = \frac{1}{0 + 1} = 1$⁠. Both the equation and the condition hold. Since $2x^2 + 1 \ge 1$⁠, the solution is defined for every real $x$⁠, satisfies $0 < y \le 1$⁠, and so is never 0, so the division by $y^2$ in Step 2 was valid all along the curve.

   Substituting back is the definition of a solution (Section 9.3), so this check is complete proof. The picture shows the one curve picked out by the point $(0, 1)$ among its neighbours; every curve follows the slope marks.
   A solution makes L.H.S. = R.H.S. Basics: dy/dx = F(x, y) draws an arrow at every point

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