Exercise 9.5 · Q6: Linear: $x\frac{dy}{dx} + 2y = x^2\log x$

Basics you need

Solution

1. Divide by $x$ to reach the standard form

   $$\frac{dy}{dx} + \frac{2}{x}\,y = x\log x.$$
   Because of $\log x$⁠, the equation only makes sense for $x > 0$⁠; there $x \ne 0$⁠, so dividing by $x$ is allowed, and $\frac{x^2\log x}{x} = x\log x$⁠.
   Section 9.4.3, step (i). Lesson: linear equations

2. Read off $P$ and $Q$

   $P = \frac{2}{x}$ and $Q = x\log x$⁠.

   Both depend on $x$ only, and $y$ appears to the first power, so the equation is linear.
   Section 9.4.3: $P$ and $Q$ are constants or functions of $x$ only. Lesson: linear equations

3. The integrating factor: $x^2$

   $$\begin{aligned} \text{I.F.} &= e^{\int\frac{2}{x}\,dx} = e^{2\log|x|} \\ &= e^{2\log x} = e^{\log x^2} = x^2. \end{aligned}$$
   For $x > 0$⁠, $|x| = x$⁠; then $2\log x = \log x^2$ and $e^{\log t} = t$⁠. (Even for negative $x$ the result would be $x^2$⁠, since $|x|^2 = x^2$⁠.)
   Section 9.4.3, step (ii); Section 7.2, formula (xii). Basics: exponentials

4. Multiply and integrate, with one constant

   $$\begin{aligned} \frac{d}{dx}\left(x^2y\right) &= x^2\cdot x\log x = x^3\log x, \\ x^2y &= \int x^3\log x\,dx + C. \end{aligned}$$
   $x^2\frac{dy}{dx} + 2x\,y = \frac{d}{dx}(x^2y)$ by the product rule. Integrating gives $y\,(\text{I.F.}) = \int(Q\times\text{I.F.})\,dx + C$⁠, with the one constant $C$⁠.
   Section 9.4.3, step (iii); product rule (Chapter 5, Section 5.3). Basics: (y · g)′ = g(y′ + Py)

5. $\int x^3\log x\,dx$ by parts

   Take $\log x$ as the first function and $x^3$ as the second, so $\int x^3\,dx = \frac{x^4}{4}$⁠:

   $$\begin{aligned} &\int x^3\log x\,dx \\ &= \log x\cdot\frac{x^4}{4} - \int\frac{1}{x}\cdot\frac{x^4}{4}\,dx \\ &= \frac{x^4}{4}\log x - \frac{x^4}{16} \\ &= \frac{x^4}{16}(4\log x - 1). \end{aligned}$$
   $\frac{1}{x}\cdot\frac{x^4}{4} = \frac{x^3}{4}$⁠, whose integral is $\frac{x^4}{16}$⁠. Taking $\frac{x^4}{16}$ out as a common factor gives the tidy form. Check: the derivative of $\frac{x^4}{4}\log x - \frac{x^4}{16}$ is $x^3\log x + \frac{x^3}{4} - \frac{x^3}{4}$ $= x^3\log x$⁠.
   Section 7.6, integration by parts; Section 7.2, formula (i). Basics: every method ends with an integral

6. Solve for $y$

   $$\begin{aligned} x^2y &= \frac{x^4}{16}(4\log x - 1) + C \\ y &= \frac{x^2}{16}(4\log x - 1) + Cx^{-2}. \end{aligned}$$
   Divide by $x^2 > 0$⁠; $\frac{C}{x^2} = Cx^{-2}$⁠.
   Section 9.4.3: the general solution. Lesson: Exercise 9.5 player

7. Check by substituting back, and the key's $\log|x|$

   From $y = \frac{x^2}{4}\log x - \frac{x^2}{16} + Cx^{-2}$⁠: $\frac{dy}{dx} = \frac{x}{2}\log x + \frac{x}{4} - \frac{x}{8} - 2Cx^{-3}$⁠, so

   $$\begin{aligned} x\frac{dy}{dx} &= \frac{x^2}{2}\log x + \frac{x^2}{8} - 2Cx^{-2}, \\ 2y &= \frac{x^2}{2}\log x - \frac{x^2}{8} + 2Cx^{-2}, \end{aligned}$$

   and adding, $x\frac{dy}{dx} + 2y = x^2\log x$⁠, for every $C$⁠.

   The derivative of $\frac{x^2}{4}\log x$ is $\frac{x}{2}\log x + \frac{x^2}{4}\cdot\frac{1}{x}$ by the product rule. The answer key writes $\log|x|$ in place of $\log x$⁠; since the equation itself needs $x > 0$⁠, where $\log|x| = \log x$⁠, the two answers are the same.
   A solution makes L.H.S. = R.H.S. (Section 9.3). Lesson: general and particular solutions

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