Example 7: Curve through $(1, 1)$ with $x\,dy = (2x^2+1)\,dx$

Basics you need

Solution

1. Divide by $x$⁠: the variables separate

   Since $x \ne 0$⁠, divide both sides of $x\,dy = (2x^2 + 1)\,dx$ by $x$⁠:

   $$dy = \frac{2x^2 + 1}{x}\,dx = \left(2x + \frac{1}{x}\right)dx.$$
   The left side has only $dy$ and the right side only $x$ and $dx$⁠: this is the separated form with $h(y) = 1$⁠. Splitting the fraction term by term, $\frac{2x^2}{x} = 2x$ and $\frac{1}{x}$⁠, turns it into two standard integrals.
   Section 9.4.1: separate, then integrate. Basics: dy = y′ dx

2. Integrate both sides, with one constant

   $$\int dy = \int\left(2x + \frac{1}{x}\right)dx$$ $$y = x^2 + \log|x| + C\quad\ldots(1)$$

   Check by differentiating: $\frac{d}{dx}\left(x^2 + \log|x|\right) = 2x + \frac{1}{x}$ for $x \ne 0$⁠.

   $\int 2x\,dx = x^2$ by the power rule, and $\int\frac1x\,dx = \log|x|$⁠; the absolute value makes the formula correct for negative $x$ too. One constant $C$ for this first order equation. Equation (1) is the general solution: a family of curves, each a vertical shift of the others.
   Section 7.2, formulas (i) and (xii). Basics: every method ends with an integral

3. Use the point $(1, 1)$ to find $C$

   Put $x = 1$⁠, $y = 1$ in (1):

   $$\begin{aligned} 1 &= 1^2 + \log|1| + C = 1 + 0 + C \\ \Rightarrow\ C &= 0. \end{aligned}$$

   So the curve is $y = x^2 + \log|x|$⁠.

   Among all the curves of the family, exactly one passes through $(1, 1)$⁠. Substituting the point gives one equation for the one unknown $C$⁠.
   $\log 1 = 0$⁠; particular solution, Section 9.3. Basics: one constant, one condition

4. Which branch is the curve through $(1, 1)$⁠?

   The equation is not defined at $x = 0$⁠, so a solution curve cannot cross the $y$-axis. The curve through $(1, 1)$ therefore lives on $x > 0$⁠, where $|x| = x$⁠:

   $$y = x^2 + \log x,\quad x > 0.$$

   The formula $y = x^2 + \log|x|$ also has a branch for $x < 0$⁠. That branch is a separate solution curve that does not pass through $(1, 1)$⁠; the condition at $(1, 1)$ says nothing about it.

   The book's answer $y = x^2 + \log|x|$ is right, and on the curve through $(1, 1)$ it reads $y = x^2 + \log x$⁠. Noticing that $x = 0$ splits the plane explains why the absolute value is harmless here.
   A solution curve follows the slope marks and cannot pass where the equation is undefined. Basics: dy/dx = F(x, y) draws an arrow at every point

5. Check: substitute back

   For $y = x^2 + \log|x|$ ($x \ne 0$⁠): $\frac{dy}{dx} = 2x + \frac1x$⁠, so

   $$x\frac{dy}{dx} = x\left(2x + \frac1x\right) = 2x^2 + 1,$$

   which is $x\,dy = (2x^2 + 1)\,dx$⁠. And at $x = 1$⁠, $y = 1 + \log 1 = 1$⁠.

   Both the equation and the point are satisfied, so this is the required curve.
   $\frac{d}{dx}\log|x| = \frac1x$ (Section 7.2, formula (xii)). Basics: y′ is the slope

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