Differential-Equations · Unit 3 · Video 1 · Interactive Practice

First Order Linear: Standard Form and the Sign of p

IKey Formulas

Formula	Name	What you need
$a(x)\,y' + b(x)\,y = c(x)$	First-order linear equation — linear in $y$ and $y'$	$y$ and $y'$ alone, first power, each times a function of $x$ ; $a$ , $b$ , $c$ otherwise arbitrary
$a\,y_1 + b\,y_2 = c$	The high-school analogy: $y'$ plays $y_1$ , $y$ plays $y_2$	Nothing — it is the same shape, one equation in two variables
$y' + p(x)\,y = q(x)$ , $p = \dfrac{b}{a}$ , $q = \dfrac{c}{a}$	Standard linear form	Divide through by $a(x)$ , where $a(x) \neq 0$
$y' + p\,y = 0 \;\Longrightarrow\; y = C e^{-px}$	Homogeneous test case, $p$ constant	$q \equiv 0$ ; the sign of $p$ then decides decay or growth

Key Insight: Keep the $p$ term on the left. The other standard form, $y' = -p(x)y + q(x)$ , carries $-p$ as the coefficient of $y$ , and one wrong sign turns $e^{-px}$ into $e^{+px}$ — decay into blow-up, with no later step able to repair it.

IIFreeze x: Straight Line or Not

Hold $x$ fixed: the pairs $(y, y')$ the equation allows form a straight line exactly when it is linear.

Equation x frozen at 1.00

IIIDividing Down to Standard Form

One division by $a(x)$ separates the general linear equation from its standard form.

Equation Step 1

1 · the equation as given

x\,y' + 3y = x^2 \qquad (x > 0)

$a(x) = x$ , $b(x) = 3$ , $c(x) = x^2$ — the coefficient of $y'$ is not $1$ .

2 · divide every term by $a(x) = x$

\frac{x\,y'}{x} + \frac{3y}{x} = \frac{x^2}{x} \;\Longrightarrow\; y' + \frac{3}{x}\,y = x

3 · read off $p$ and $q$

p(x) = \frac{b}{a} = \frac{3}{x}, \qquad q(x) = \frac{c}{a} = x

4 · check the sign of $p$

$p(x) = 3/x$ is positive on $x > 0$ , and $q \not\equiv 0$ , so the equation is not homogeneous. Its homogeneous partner $y' + \tfrac{3}{x}y = 0$ has decaying solutions.

1 · the equation as given

2y' + 6y = 8

$a = 2$ , $b = 6$ , $c = 8$ — all three constant.

2 · divide every term by $a = 2$

y' + 3y = 4

3 · read off $p$ and $q$

p = \frac{6}{2} = 3, \qquad q = \frac{8}{2} = 4

4 · check the sign of $p$

$p = 3 > 0$ and constant, so the homogeneous partner $y' + 3y = 0$ is solved by $e^{-3x}$ : every solution of it dies out.

1 · the equation as given

y' = 3y - x

Already solved for $y'$ — this is the other standard form, with everything else on the right.

2 · bring the $y$ term to the left

y' - 3y = -x

$a = 1$ , so no division is needed; only the move matters.

3 · read off $p$ and $q$

p(x) = -3, \qquad q(x) = -x

4 · check the sign of $p$

$p = -3$ , not $+3$ : the $3$ visible in $y' = 3y - x$ is $-p$ . Here $p < 0$ , and the homogeneous partner $y' - 3y = 0$ grows like $e^{3x}$ .

💡 Dividing is legal only where $a(x) \neq 0$ : the standard form of $x\,y' + 3y = x^2$ lives on $x > 0$ or on $x < 0$ , never across $x = 0$ .

IVOne Sign, Opposite Behavior

With $p$ constant and nothing on the right, the sign in front of $py$ is the whole difference.

Equation p 0.8 x 1.40

VQuiz Questions

Problem 1 · Which One Is Linear

Given: four first-order equations — find the one that is linear in $y$ and $y'$ , then decide whether it is homogeneous.

Which equation is linear?

Is that equation homogeneous?

✅ Correct! In

y' + (\cos x)y = x^3

the unknown appears alone and differentiated once, each multiplied by a function of

x

; the coefficients may be as wild as they like. Homogeneous is decided by the right-hand side alone, and

x^3 \neq 0

❌ Not quite.

y^2

is the unknown multiplied by itself — the one product that breaks linearity, and the reason this lecture-one equation has no elementary solution.

❌ Not quite.

y\,y'

multiplies the unknown by its own derivative. Linear means linear in the pair

(y, y')

jointly, so a product of the two is out.

❌ Not quite.

\sin y

puts the unknown inside a function. Only

y

and

y'

themselves, times functions of

x

, are allowed.

❌ Not quite. Test each equation on

y

and

y'

only: they must appear alone, to the first power, never multiplied together and never inside a function.

❌ Check the definition. Homogeneous means the right-hand side

c(x)

is identically zero — nothing to do with the coefficients on the left.

Show solution

Treat $y'$ as a variable in its own right and ask whether the equation is linear in the pair $(y, y')$ :

$y' = x - y^2$ : the unknown is squared — not linear.
$y\,y' + y = x$ : the unknown multiplies its own derivative — not linear.
$y' + (\cos x)\,y = x^3$ : matches $a(x)y' + b(x)y = c(x)$ with $a = 1$ , $b = \cos x$ , $c = x^3$ — linear.
$y' + \sin y = 0$ : the unknown sits inside a sine — not linear.

It is already in standard form, so $p(x) = \cos x$ and $q(x) = x^3$ .

Homogeneous? The word means $c(x) \equiv 0$ , exactly as for $a y_1 + b y_2 = c$ in algebra. Here $c(x) = x^3$ , which is zero only at $x = 0$ , not identically — so the equation is not homogeneous.

Problem 2 · Divide by a(x)

Given: $x\,y' - 3y = x^5$ on $x > 0$ — write it in standard linear form $y' + p(x)y = q(x)$ and read off the two coefficients.

What is $p(x)$ ?

What is $q(x)$ ?

✅ Correct! Dividing by

a(x) = x

gives

y' - \frac{3}{x}y = x^4

, so

p(x) = -\frac{3}{x}

and

q(x) = x^4

— the sign of

p

is the one it carries on the left.

❌ Half a step. Every term must be divided by

a(x) = x

, the

y

term included:

-3y

becomes

-\frac{3}{x}y

❌ Close, but check the sign.

p

is read off the left-hand side of

y' + p(x)y = q(x)

. You have read the coefficient from

y' = \frac{3}{x}y + x^4

, the other standard form, whose coefficient is

-p

❌ Half a step. The right-hand side is divided by

a(x) = x

as well:

x^5/x = x^4

❌ Not quite. Divide the whole equation by

a(x) = x

, then read the coefficient of

y

exactly as it stands on the left.

❌ Not quite.

q

is the right-hand side after the division:

x^5 \div x

, not a derivative of anything.

Show solution

Step 1 — identify $a$ , $b$ , $c$ : $a(x) = x$ , $b(x) = -3$ , $c(x) = x^5$ .

Step 2 — divide every term by $a(x) = x$ (legal on $x > 0$ ):

\frac{x\,y'}{x} - \frac{3y}{x} = \frac{x^5}{x} \qquad\Longrightarrow\qquad y' - \frac{3}{x}\,y = x^4

Step 3 — compare with $y' + p(x)y = q(x)$ :

p(x) = \frac{b}{a} = -\frac{3}{x}, \qquad q(x) = \frac{c}{a} = x^4

Check: multiplying $y' - \frac{3}{x}y = x^4$ back by $x$ returns $x y' - 3y = x^5$ .

Written the other way, $y' = \frac{3}{x}y + x^4$ — and the $+\frac{3}{x}$ there is $-p$ , not $p$ . That is exactly the confusion the standard form is chosen to avoid.

Problem 3 · What the Sign of p Does

Given: the homogeneous equation $y' + \tfrac{1}{2}y = 0$ — find its solutions and their behavior as $x \to \infty$ .

✅ Correct! Differentiating

e^{-x/2}

returns

-\tfrac{1}{2}

times the function — precisely

-p\,y

— and

-p < 0

makes every solution decay.

❌ Close, but check the sign.

y = Ce^{x/2}

gives

y' = +\tfrac{1}{2}y

, which solves

y' - \tfrac{1}{2}y = 0

— the sign-flipped equation, with the opposite behavior.

❌ Not quite.

y = Ce^{-2x}

has

y' = -2y

, so it solves

y' + 2y = 0

. The exponent is

-p\,x

itself, not the reciprocal of

p

❌ Not quite. That is what you would get from

y' = -\tfrac{1}{2}

; here the right-hand side is

0

and the

-\tfrac{1}{2}

multiplies

y

❌ Not quite. Substitute each candidate into

y' + \tfrac{1}{2}y

and keep the one that returns

0

Show solution

The equation is the constant- $p$ test case with $p = \tfrac{1}{2}$ and $q = 0$ . Try $y = Ce^{rx}$ :

y' + \tfrac{1}{2}y = \left(r + \tfrac{1}{2}\right)Ce^{rx} = 0 \qquad\Longrightarrow\qquad r = -\tfrac{1}{2}

y = Ce^{-x/2}

Check by differentiating: $\frac{d}{dx}e^{-x/2} = -\tfrac{1}{2}e^{-x/2}$ , which is $-p$ times the function — exactly what $y' = -py$ demands.

Since $-p = -\tfrac{1}{2} < 0$ , every solution decays to $0$ ; at $x = 6$ it has already fallen to $C e^{-3} \approx 0.050\,C$ .

Flip the sign of $p$ and the equation $y' - \tfrac{1}{2}y = 0$ is solved by $Ce^{+x/2}$ : the same $|p|$ , unbounded growth. One sign, opposite fates.

Problem 4 · The Other Standard Form

Given: a textbook writes the equation as $y' = 5y - e^{2x}$ — put it into the standard linear form used in this course and identify $p$ and $q$ .

What is $p(x)$ ?

What is $q(x)$ ?

✅ Correct! Moving the

y

term across gives

y' - 5y = -e^{2x}

, so

p = -5

and

q = -e^{2x}

. The

5

printed in the textbook's version is

-p

❌ Close, but check the sign.

5

is the coefficient of

y

on the right. Carried to the left it becomes

-5y

, and

p

is read from the left:

p = -5

❌ Not quite.

p

is a function of

x

alone — it is the coefficient multiplying

y

, never a term containing

y

❌ Close, but check the sign. Only the

y

term moves;

-e^{2x}

stays on the right exactly as it is, so

q = -e^{2x}

❌ Not quite. Rewrite as

y' + p(x)y = q(x)

first: the

y

term has to sit on the left before

p

can be read.

❌ Not quite.

q

is whatever is left on the right once the

y

term has moved — no differentiating, no sign change.

Show solution

$y' = 5y - e^{2x}$ is written in the general first-order standard form ( $y'$ equals everything else). The linear standard form wants the $y$ term on the left:

y' - 5y = -e^{2x}

Compare with $y' + p(x)y = q(x)$ :

p(x) = -5, \qquad q(x) = -e^{2x}

Half the books write this equation as $y' = r(x)y + q(x)$ with $r = -p$ ; we never do, because the sign of $p$ is what the whole method hangs on.

What the sign predicts: $p = -5 < 0$ , so the homogeneous partner $y' - 5y = 0$ grows like $e^{5x}$ — and indeed the general solution is $y = Ce^{5x} + \tfrac{1}{3}e^{2x}$ . Reading $p = +5$ instead would have predicted a decaying $e^{-5x}$ : the wrong behavior from the first line on.

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