Chapter 9 · Differential Equations

Exercise 9.5 · Q3: Linear: $\frac{dy}{dx} + \frac{y}{x} = x^2$

Exercise 9.5Linear differential equations · Section 9.4.3

The question

For each of the differential equations given in Exercises 1 to 12, find the general solution:

3. $\dfrac{dy}{dx} + \dfrac{y}{x} = x^2$

Exam answer

This is of the form $\frac{dy}{dx} + Py = Q$ ⁠, where $P = \frac{1}{x}$ and $Q = x^2$ $(x \ne 0)$ ⁠.

\text{I.F.} = e^{\int\frac{1}{x}dx} = e^{\log x} = x

So the solution is given by

\begin{aligned} y\cdot x &= \int x^2\cdot x\,dx + C \\ &= \int x^3\,dx + C \end{aligned}

$\therefore$ $xy = \dfrac{x^4}{4} + C$ is the general solution.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

The equation is already in the standard linear form, with $P = \frac{1}{x}$ ⁠. Its integrating factor is $e^{\int\frac{dx}{x}} = x$ (up to sign), and multiplying by $x$ turns the left side into $\frac{d}{dx}(xy)$ ⁠, by the product rule. So $xy$ is just an antiderivative of $x^3$ ⁠.

Given, to find, and the plan

Given: $\frac{dy}{dx} + \frac{y}{x} = x^2$ ⁠; the term $\frac{y}{x}$ needs $x \ne 0$ ⁠.
To find: The general solution (one arbitrary constant).
Plan: (1) Read off $P$ and $Q$ ⁠. (2) I.F. $= e^{\int P\,dx}$ ⁠, with care over $\log|x|$ ⁠. (3) Multiply and see the product rule. (4) Integrate, with one constant. (5) Solve for $y$ ⁠. (6) Check by substitution.

Basics you need

1First order linear equations

$\frac{dy}{dx} + Py = Q$ ⁠, where $P$ and $Q$ are constants or functions of $x$ only. Its solution is $y\,(\text{I.F.}) = \int(Q\times\text{I.F.})\,dx + C$ ⁠, where $\text{I.F.} = e^{\int P\,dx}$ (Section 9.4.3). Lesson: linear equations, the integrating factor

2Logs and exponentials

$\int\frac{dx}{x} = \log|x|$ (Section 7.2, formula (xii)), and $e^{\log t} = t$ for $t > 0$ ⁠. Basics: exponentials

3A constant multiple of an I.F. is an I.F.

If $g$ is an integrating factor, so is $kg$ for any constant $k \ne 0$ ⁠: the $k$ cancels from both sides. Basics: (y · g)′ = g(y′ + Py)

4Power rule

$\int x^3\,dx = \frac{x^4}{4} + C$ (Section 7.2, formula (i)). Basics: every method ends with an integral

Solution

Recognise the linear form: $P = \frac{1}{x}$ ⁠, $Q = x^2$
$\frac{dy}{dx} + \frac{1}{x}\,y = x^2.$
The coefficient of $\frac{dy}{dx}$ is 1, $y$ appears to the first power, and $P = \frac{1}{x}$ ⁠, $Q = x^2$ depend on $x$ only. Because of $\frac{1}{x}$ ⁠, we work on an interval with $x \ne 0$ ⁠: either $x > 0$ or $x < 0$ ⁠.

Section 9.4.3, step (i). Lesson: linear equations
The integrating factor: $x$
$\text{I.F.} = e^{\int\frac{1}{x}\,dx} = e^{\log|x|} = |x|.$
We take $\text{I.F.} = x$ ⁠, as the book does.

For $x > 0$ ⁠, $|x| = x$ ⁠. For $x < 0$ ⁠, $|x| = -x$ ⁠, which is $-1$ times $x$ ⁠; a nonzero constant multiple of an integrating factor is again one (the constant cancels from both sides). So $x$ works on both sides of 0. The book writes $e^{\log x} = x$ ⁠.

Section 9.4.3, step (ii); Section 7.2, formula (xii). Basics: exponentials
Multiply by $x$ ⁠: the left side is one derivative
$\begin{aligned} x\frac{dy}{dx} + y &= x^3 \\ \frac{d}{dx}(xy) &= x^3. \end{aligned}$
By the product rule, $\frac{d}{dx}(xy) = x\frac{dy}{dx} + y\cdot 1$ ⁠, exactly the left side.

Product rule (Chapter 5, Section 5.3). Basics: (y · g)′ = g(y′ + Py)
Integrate both sides, with one constant
$\begin{aligned} y\cdot x &= \int x^2\cdot x\,dx + C = \int x^3\,dx + C \\ xy &= \frac{x^4}{4} + C. \end{aligned}$
This is $y\,(\text{I.F.}) = \int(Q\times\text{I.F.})\,dx + C$ ⁠, with $Q\times\text{I.F.} = x^2\cdot x = x^3$ ⁠. The $C$ is the one arbitrary constant.

Section 9.4.3, step (iii); Section 7.2, formula (i). Basics: one constant for order 1
Solve for $y$ (optional)
$y = \frac{x^3}{4} + \frac{C}{x}\quad(x \ne 0).$
Divide by $x \ne 0$ ⁠. The book leaves the answer as $xy = \frac{x^4}{4} + C$ ⁠; both say the same thing.

Section 9.4.3: the general solution. Lesson: Exercise 9.5 player
Check by substituting back

From $y = \frac{x^3}{4} + \frac{C}{x}$ ⁠: $\frac{dy}{dx} = \frac{3x^2}{4} - \frac{C}{x^2}$ ⁠, and $\frac{y}{x} = \frac{x^2}{4} + \frac{C}{x^2}$ ⁠. Adding,
$\frac{dy}{dx} + \frac{y}{x} = \frac{3x^2}{4} + \frac{x^2}{4} = x^2,$
for every $C$ ⁠.

The $\frac{C}{x^2}$ terms cancel, and $\frac{3}{4} + \frac{1}{4} = 1$ ⁠. So the family satisfies the equation on $x > 0$ and on $x < 0$ ⁠.

A solution makes L.H.S. = R.H.S. (Section 9.3). Lesson: general and particular solutions

Result

The general solution is $xy = \frac{x^4}{4} + C$ ⁠, that is $y = \frac{x^3}{4} + \frac{C}{x}$ ⁠, where $C$ is an arbitrary constant ( $x \ne 0$ ⁠). (Textbook answer: $xy = \frac{x^4}{4} + C$ ⁠.)

Check it with numbers

Take $C = 1$ and $x = 2$ ⁠: $y = 2 + \frac12 = 2.5$ ⁠. The equation says $\frac{dy}{dx} = x^2 - \frac{y}{x} = 4 - 1.25 = 2.75$ ⁠.

The curve itself: $y(2.001) \approx 2.5027516252$ and $y(1.999) \approx 2.4972516248$ ⁠, so

\begin{aligned} &\frac{2.5027516252 - 2.4972516248}{0.002} \\ &\approx 2.75000. \end{aligned}

Common mistakes

Taking the I.F. as $e^{x}$ or $\log x$ ⁠; it is $e^{\int P\,dx} = e^{\log|x|} = |x|$ ⁠, and $x$ may be used.
Integrating $Q$ instead of $Q\times\text{I.F.}$ ⁠: the integral is $\int x^2\cdot x\,dx$ ⁠, not $\int x^2\,dx$ ⁠.
Dividing only $\frac{x^4}{4}$ by $x$ when solving for $y$ ⁠: the answer is $\frac{x^3}{4} + \frac{C}{x}$ ⁠.

What to take away

$P = \frac{k}{x}$ gives the I.F. $x^k$ (here $k = 1$ ⁠).
$e^{\log|x|} = |x|$ ⁠, and $x$ can replace $|x|$ because a constant factor $\pm1$ does not matter in an I.F.
The general solution may be left in the form $y\,(\text{I.F.}) = \ldots + C$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
Linear equation; I.F. $= e^{\int P\,dx}$ ⁠; $y\,(\text{I.F.}) = \int(Q\times\text{I.F.})\,dx + C$	Steps 1, 2, 4	Section 9.4.3, steps (i) to (iii) (Lesson)
$\int\frac{dx}{x} = \log\|x\|$ ⁠; $e^{\log t} = t$	Step 2	Section 7.2, formula (xii); Chapter 5, Section 5.4 (Basics)
Product rule	Step 3	Chapter 5, Section 5.3 (Basics)
$\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$	Step 4	Section 7.2, formula (i) (Basics)
Substituting back into the equation	Step 6	Section 9.3 (Lesson)

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Exercise 9.5 · Q3: Linear: dydx+yx=x2\frac{dy}{dx} + \frac{y}{x} = x^2dxdy​+xy​=x2

Basics you need

Solution

Recognise the linear form: P=1xP = \frac{1}{x}P=x1​⁠, Q=x2Q = x^2Q=x2

The integrating factor: xxx

Multiply by xxx⁠: the left side is one derivative

Integrate both sides, with one constant

Solve for yyy (optional)

Check by substituting back

Every fact used, and where it comes from

Exercise 9.5 · Q3: Linear: $\frac{dy}{dx} + \frac{y}{x} = x^2$

Recognise the linear form: $P = \frac{1}{x}$ ⁠, $Q = x^2$

The integrating factor: $x$

Multiply by $x$ ⁠: the left side is one derivative

Solve for $y$ (optional)