Exercise 9.3 · Q11: Particular solution of $(x^3 + x^2 + x + 1)\frac{dy}{dx} = 2x^2 + x$⁠, $y(0) = 1$

Basics you need

Solution

1. Factor the cubic and separate

   Group the terms:

   $$\begin{aligned} x^3 + x^2 + x + 1 &= x^2(x + 1) + (x + 1) \\ &= (x + 1)(x^2 + 1). \end{aligned}$$

   For $x \ne -1$ this is nonzero ($x^2 + 1 \ge 1$⁠), so divide by it and multiply by $dx$⁠:

   $$dy = \frac{2x^2 + x}{(x + 1)(x^2 + 1)}\,dx.$$
   The equation is $\frac{dy}{dx} = g(x)\cdot 1$⁠, separable with $h(y) = 1$⁠, so nothing in $y$ is divided by. The division by the cubic needs $x \ne -1$⁠; the condition is given at $x = 0$⁠, so the solution lives on the interval $x > -1$⁠.
   Section 9.4.1: separate, then integrate. Lesson: separate, then integrate

2. Partial fractions

   $x^2 + 1$ has no real factors, so take

   $$\frac{2x^2 + x}{(x + 1)(x^2 + 1)} = \frac{A}{x + 1} + \frac{Bx + C}{x^2 + 1}.$$

   Multiplying by $(x + 1)(x^2 + 1)$⁠:

   $$\begin{aligned} 2x^2 + x &= A(x^2 + 1) \\ &\quad + (Bx + C)(x + 1). \end{aligned}$$

   Put $x = -1$⁠: $2 - 1 = 2A$⁠, so $A = \frac{1}{2}$⁠. Compare coefficients of $x^2$⁠: $2 = A + B$⁠, so $B = \frac{3}{2}$⁠. Compare constants: $0 = A + C$⁠, so $C = -\frac{1}{2}$⁠. Coefficients of $x$ check it: $B + C = \frac{3}{2} - \frac{1}{2} = 1$⁠.

   The numerator has degree 2, less than the degree 3 of the denominator, so no division is needed first. Putting $x = -1$ kills the $B$⁠, $C$ bracket; two coefficient comparisons give the rest, and the third (the $x$ terms) confirms them. The letter $C$ is used here for a coefficient, so the constant of integration is called $c$⁠.
   Section 7.5, partial fractions. Basics: every method ends with an integral

3. Integrate each piece, with one constant

   $$\begin{aligned} y &= \frac{1}{2}\int\frac{dx}{x + 1} + \frac{3}{2}\int\frac{x\,dx}{x^2 + 1} \\ &\quad - \frac{1}{2}\int\frac{dx}{x^2 + 1} \\ &= \frac{1}{2}\log|x + 1| + \frac{3}{4}\log(x^2 + 1) \\ &\quad - \frac{1}{2}\tan^{-1}x + c. \end{aligned}$$

   In the middle integral $t = x^2 + 1$⁠, $dt = 2x\,dx$⁠, so $\int\frac{x\,dx}{x^2 + 1} = \frac{1}{2}\log(x^2 + 1)$⁠, and $\frac{3}{2}\cdot\frac{1}{2} = \frac{3}{4}$⁠.

   Check by differentiating:

   $$\begin{aligned} &\frac{1}{2(x + 1)} + \frac{3x}{2(x^2 + 1)} - \frac{1}{2(x^2 + 1)} \\ &= \frac{2x^2 + x}{(x + 1)(x^2 + 1)}. \end{aligned}$$
   Each piece is a standard form: a log for the linear factor, "derivative over function" for $\frac{x}{x^2 + 1}$⁠, and $\tan^{-1}$ for $\frac{1}{x^2 + 1}$⁠. Since $x^2 + 1 > 0$⁠, its log needs no modulus. All constants combine into one $c$⁠: this is the general solution.
   Section 7.2, formulas (x), (xii); Section 7.3.1. Basics: every method ends with an integral

4. Use the condition to find $c$

   Put $x = 0$⁠, $y = 1$⁠:

   $$\begin{aligned} 1 &= \frac{1}{2}\log 1 + \frac{3}{4}\log 1 \\ &\quad - \frac{1}{2}\tan^{-1}0 + c = c. \end{aligned}$$

   So $c = 1$⁠.

   $\log 1 = 0$ and $\tan^{-1}0 = 0$⁠, so every term except $c$ vanishes. One condition fixes the one constant; nothing arbitrary is left.
   Section 9.3: a particular solution gives the constant a value. Basics: one constant, one condition

5. The textbook's form, and where it holds

   Combine the logs: $\frac{1}{2}\log|x + 1| = \frac{1}{4}\log(x + 1)^2$ and $\frac{3}{4}\log(x^2 + 1) = \frac{1}{4}\log(x^2 + 1)^3$⁠. So

   $$\begin{aligned} y &= \frac{1}{4}\log\left[(x + 1)^2(x^2 + 1)^3\right] \\ &\quad - \frac{1}{2}\tan^{-1}x + 1, \end{aligned}$$

   for $x > -1$⁠.

   $k\log a = \log a^k$ and $\log a + \log b = \log ab$⁠. Writing $(x + 1)^2 = |x + 1|^2$ makes the modulus unnecessary. The interval is $x > -1$⁠: the solution must be defined on an interval containing $x = 0$⁠, and it cannot cross $x = -1$⁠, where the equation breaks down and $y \to -\infty$⁠.
   Laws of logarithms. Lesson: general and particular solutions

6. Check: substitute back

   Differentiating the answer gives, by Step 3, $\frac{dy}{dx} = \frac{2x^2 + x}{(x + 1)(x^2 + 1)}$⁠, and multiplying by $(x + 1)(x^2 + 1) = x^3 + x^2 + x + 1$ gives the equation. At $x = 0$⁠: $y = \frac{1}{4}\log 1 - 0 + 1 = 1$⁠.

   Both the equation and the condition hold, so this is the particular solution asked for (Section 9.3).
   A solution makes L.H.S. = R.H.S. Basics: y′ is the slope

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