Example 15: Linear: $x\frac{dy}{dx} + 2y = x^2$

Basics you need

Solution

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

   $$\frac{dy}{dx} + \frac{2}{x}\,y = x.$$
   The standard linear form needs $\frac{dy}{dx}$ with coefficient 1. Dividing by $x$ is allowed because the question says $x \ne 0$⁠.
   Section 9.4.3, step (i). Lesson: linear equations

2. Read off $P$ and $Q$

   $P = \frac{2}{x}$ and $Q = x$⁠.

   Both depend on $x$ only, and $y$ appears to the first power, so the equation is linear. (It is not separable: $\frac{dy}{dx} = x - \frac{2y}{x}$ is not a product of a function of $x$ and a function of $y$⁠. It is not homogeneous either, since $x^2$ on the right has a different degree from the other terms.)
   Section 9.4.3: $P$ and $Q$ are constants or functions of $x$ only. Lesson: linear equations

3. The integrating factor: $x^2$

   $$\begin{aligned} \text{I.F.} &= e^{\int\frac{2}{x}\,dx} = e^{2\log|x|} \\ &= e^{\log x^2} = x^2. \end{aligned}$$
   $2\log|x| = \log|x|^2 = \log x^2$⁠, and $e^{\log t} = t$⁠. The book writes $e^{2\log x}$⁠; with $|x|$ the result is the same $x^2$ for negative $x$ too, because $|x|^2 = x^2$⁠. No constant is needed in $\int P\,dx$⁠: it would only multiply the I.F. by a nonzero constant.
   Section 9.4.3, step (ii); $e^{\log f(x)} = f(x)$⁠. Basics: exponentials

4. Multiply by $x^2$⁠: the left side is one derivative

   $$\begin{aligned} x^2\frac{dy}{dx} + 2x\,y &= x^3 \\ \frac{d}{dx}\left(x^2y\right) &= x^3. \end{aligned}$$
   By the product rule, $\frac{d}{dx}(x^2y) = x^2\frac{dy}{dx} + 2x\,y$⁠, exactly the left side. (Notice this is just the original equation multiplied by $x$⁠.)
   Product rule (Chapter 5, Section 5.3). Basics: (y · g)′ = g(y′ + Py)

5. Integrate, with one constant

   $$\begin{aligned} y\cdot x^2 &= \int x\cdot x^2\,dx + C \\ &= \int x^3\,dx + C \\ &= \frac{x^4}{4} + C. \end{aligned}$$
   This is $y\,(\text{I.F.}) = \int(Q\times\text{I.F.})\,dx + C$ with $Q\times\text{I.F.} = x\cdot x^2$⁠. The $C$ here is the one arbitrary constant of the general solution.
   Section 9.4.3, step (iii); Section 7.2, formula (i). Basics: one constant for order 1

6. Solve for $y$

   $$y = \frac{x^2}{4} + \frac{C}{x^2} = \frac{x^2}{4} + Cx^{-2}.$$
   Divide by $x^2 \ne 0$⁠. The solution lives on an interval where $x \ne 0$⁠: on $x > 0$ or on $x < 0$⁠, since the equation itself is not defined at $x = 0$⁠.
   Section 9.4.3: the general solution. Lesson: Examples 14 to 18

7. Check by substituting back

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

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

   for every $C$⁠.

   The $C$ terms cancel, as they must: $Cx^{-2}$ alone solves $x\frac{dy}{dx} + 2y = 0$⁠, and $\frac{x^2}{4}$ supplies the $x^2$⁠.
   A solution makes L.H.S. = R.H.S. (Section 9.3). Lesson: general and particular solutions

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