Differential-Equations · Unit 3 · Video 4 · Interactive Practice

Integrating Factors: Two Mistakes and a Trick Question

IKey Formulas

Formula	Name	What it demands
$y' + p(x)\,y = q(x)$	Standard linear form	The coefficient of $y'$ divided out to $1$ first
$u = e^{\int p\,dx}$	Integrating factor	The $p$ read off after that division, minus sign included
$(u\,y)' = u\,q$	Multiply both sides	The right-hand side carries $u$ too
$\int \dfrac{f'}{f}\,dx = \ln\lvert f\rvert$	Log integral	Numerator = derivative of denominator

Key Insight: For $(1 + \cos x)\,y' - (\sin x)\,y = 2x$ the method returns $u = 1 + \cos x$ — the coefficient already standing in front of $y'$ — so multiplying through hands back the equation you started with. The round trip is the method working: it certifies that the left side was $\big[(1 + \cos x)\,y\big]'$ all along.

IIWhat Makes $u$ an Integrating Factor

A factor works exactly when $u' = u\,p$ — and a $p$ read off without standard form fails that test.

Candidate factor Probe x 1.00

Any constant multiple works too: $u = 5(1 + \cos x)$ satisfies the same test $u' = u\,p$ and cancels out of the final answer — which is why $e^{\int p\,dx}$ never needs a $+C$ .

IIIThe Round Trip

Four steps on $(1 + \cos x)\,y' - (\sin x)\,y = 2x$ return the equation itself; two more finish it.

Step 1

Step 1 — Standard form

y' - \frac{\sin x}{1 + \cos x}\,y = \frac{2x}{1 + \cos x}, \qquad p = -\frac{\sin x}{1 + \cos x}

Divide by the coefficient of

y'

. The

-\sin x

standing in the given equation is not

p

, and the division needs

1 + \cos x \neq 0

— true on

(-\pi, \pi)

, the interval carrying

x = 0

Step 2 — Integrating factor

\int p\,dx = \int \frac{-\sin x}{1 + \cos x}\,dx = \ln(1 + \cos x)

u = e^{\ln(1 + \cos x)} = 1 + \cos x

The numerator is the derivative of the denominator, so the integral is a logarithm:

\dfrac{d}{dx}\ln(1 + \cos x) = \dfrac{-\sin x}{1 + \cos x}\ \checkmark

Step 3 — Multiply both sides

(1 + \cos x)\,y' - (\sin x)\,y = 2x

The given equation, letter for letter — because

u\,q = (1 + \cos x)\cdot\dfrac{2x}{1 + \cos x} = 2x

. Its left side is now certified to be a derivative:

\big[(1 + \cos x)\,y\big]' = (1 + \cos x)\,y' + (-\sin x)\,y = 2x\ \checkmark

Step 4 — Integrate

(1 + \cos x)\,y = x^2 + C \qquad \Longrightarrow \qquad y = \frac{x^2 + C}{1 + \cos x}

One parabola

x^2 + C

for each constant, every one of them divided by the same

1 + \cos x

Step 5 — The initial condition $y(0) = 1$

1 = \frac{0^2 + C}{1 + \cos 0} = \frac{C}{2} \qquad \Longrightarrow \qquad C = 2

y = \frac{x^2 + 2}{1 + \cos x}

💡 One product-rule check on the original equation already shows its left side to be $\big[(1 + \cos x)\,y\big]'$ — the clever route writes that line at once and skips steps 1 to 3 entirely.

IVForgetting the Right Side

Both curves satisfy $y(0) = 1$ ; only one of them satisfies the equation that was asked.

Reach x 0.70

Substituting the dashed curve into the given equation returns $\dfrac{2x}{1 + \cos x}$ where $2x$ was required — a leftover of $-\dfrac{2x\cos x}{1 + \cos x}$ , which vanishes only at $x = 0$ and $x = \pm\tfrac{\pi}{2}$ .

VQuiz Questions

Problem 1 · Read Off $p$

Given: $(1 + \cos x)\,y' - (\sin x)\,y = 2x$ — find the $p$ that belongs in $u = e^{\int p\,dx}$ .

✅ Correct! Standard form first: dividing by

1 + \cos x

gives

y' - \frac{\sin x}{1+\cos x}\,y = \frac{2x}{1+\cos x}

, and

p

is that whole coefficient of

y

, minus sign included. It leads to

u = 1 + \cos x

❌ That is mistake 1.

-\sin x

is the coefficient of

y

before the equation is in standard form. Reading it as

p

gives

u = e^{\int -\sin x\,dx} = e^{\cos x}

, and everything downstream is wrong.

❌ The sign belongs to $p$ . The standard form has

-\frac{\sin x}{1+\cos x}\,y

on the left, so

p

is negative here; dropping the minus gives

u = \frac{1}{1+\cos x}

instead of

1 + \cos x

❌ That is $q$ . After dividing,

\frac{2x}{1+\cos x}

is the right-hand side.

p

is what multiplies

y

on the left.

❌ Not quite. Divide every term by the coefficient of

y'

first, then read the coefficient of

y

together with its sign.

Show solution

The formula $u = e^{\int p\,dx}$ is stated for the standard form $y' + p(x)\,y = q(x)$ , in which the coefficient of $y'$ is $1$ . The given equation is not there yet, so divide through by $1 + \cos x$ :

y' - \frac{\sin x}{1 + \cos x}\,y = \frac{2x}{1 + \cos x}

Matching against $y' + p\,y = q$ :

p = -\frac{\sin x}{1 + \cos x}, \qquad q = \frac{2x}{1 + \cos x}

The payoff is immediate: the numerator of $-p$ is the derivative of its denominator, so

\int p\,dx = \ln(1 + \cos x), \qquad u = e^{\ln(1 + \cos x)} = 1 + \cos x

Skipping the division and taking $p = -\sin x$ would have produced $u = e^{\cos x}$ — a function that fails the test $u' = u\,p$ and collapses nothing.

Problem 2 · What Step 3 Produces

Given: the standard form $y' - \frac{\sin x}{1+\cos x}\,y = \frac{2x}{1+\cos x}$ and its integrating factor $u = 1 + \cos x$ — which line does multiplying through by $u$ produce?

✅ Correct!

u\,q = (1+\cos x)\cdot\frac{2x}{1+\cos x} = 2x

, and the left side collapses to

\big[(1+\cos x)\,y\big]'

— which is the original equation, now written as a single derivative.

❌ That is mistake 2. The right side was left unmultiplied. Every term gets

u

(1+\cos x)\cdot\frac{2x}{1+\cos x} = 2x

. Integrating your version solves a different equation.

❌ $u$ was applied twice on the right. The standard form already carries

\frac{2x}{1+\cos x}

; multiplying that by

u

gives

2x

, not

2x(1+\cos x)

❌ The $y$ term cannot be dropped. Multiplying gives

(1+\cos x)\,y' - (\sin x)\,y

, and it is precisely that pair — not

(1+\cos x)\,y'

alone — that is the derivative of

(1+\cos x)\,y

❌ Not quite. Multiply every term of the standard form by

u = 1 + \cos x

and watch the denominators cancel.

Show solution

Multiply each of the three terms by $u = 1 + \cos x$ :

the $y'$ term becomes $(1+\cos x)\,y'$ ;
the $y$ term becomes $(1+\cos x)\cdot\left(-\dfrac{\sin x}{1+\cos x}\right)y = -(\sin x)\,y$ ;
the right side becomes $(1+\cos x)\cdot\dfrac{2x}{1+\cos x} = 2x$ .

(1 + \cos x)\,y' - (\sin x)\,y = 2x

That is the equation the problem started from — the round trip. Its value is the guarantee that comes with it: the left side is now known to be a derivative, and the product rule confirms which one,

\big[(1+\cos x)\,y\big]' = (1+\cos x)\,y' + (1+\cos x)'\,y = (1+\cos x)\,y' - (\sin x)\,y

so the equation is $\big[(1+\cos x)\,y\big]' = 2x$ , ready to integrate.

Leaving $\frac{2x}{1+\cos x}$ on the right instead would integrate to $2x\tan\frac{x}{2} + 4\ln\cos\frac{x}{2}$ rather than to $x^2$ — a valid solution of an equation nobody asked about.

Problem 3 · The Same Trick, New Equation

Given: $(x^2 + 1)\,y' + 2x\,y = 4x$ with $y(0) = 3$ — find the integrating factor and the solution.

What is the integrating factor?

What is the solution?

✅ Correct! Standard form gives

p = \frac{2x}{x^2+1}

, whose integral is

\ln(x^2+1)

, so

u = x^2 + 1

— the coefficient that was already there. Then

\big[(x^2+1)y\big]' = 4x

(x^2+1)y = 2x^2 + C

, and

y(0) = C = 3

❌ That is the skipped-standard-form factor.

e^{x^2}

comes from reading

p = 2x

off the given equation; the true

p

\frac{2x}{x^2+1}

, giving

u = e^{\ln(x^2+1)} = x^2 + 1

❌ Check the exponent. Divide by

x^2 + 1

first, then integrate

p = \frac{2x}{x^2+1}

— numerator = derivative of denominator, so the integral is a logarithm.

❌ One division short. Integration gives

(x^2+1)\,y = 2x^2 + C

; you still have to divide by

u = x^2+1

to isolate

y

❌ Right family, wrong constant.

\frac{2x^2+1}{x^2+1}

does solve the equation, but it starts at

y(0) = 1

; the condition

y(0) = 3

forces

C = 3

❌ Check the integration.

\int 4x\,dx = 2x^2

, not

x^2

. This candidate hits

y(0) = 3

but leaves a remainder of

-2x

in the equation.

❌ Not quite. Integrate

\big[(x^2+1)\,y\big]' = 4x

, divide by

x^2+1

, then use

y(0) = 3

to pin the constant.

Show solution

Step 1 — standard form. Divide by the coefficient of $y'$ :

y' + \frac{2x}{x^2+1}\,y = \frac{4x}{x^2+1}, \qquad p = \frac{2x}{x^2+1}

Step 2 — integrating factor. The numerator is the derivative of the denominator:

\int p\,dx = \ln(x^2+1), \qquad u = e^{\ln(x^2+1)} = x^2 + 1

Step 3 — multiply both sides. Both denominators cancel and the original equation comes back:

(x^2+1)\,y' + 2x\,y = 4x \qquad \Longrightarrow \qquad \big[(x^2+1)\,y\big]' = 4x

Step 4 — integrate, then apply the initial condition.

(x^2+1)\,y = 2x^2 + C, \qquad y = \frac{2x^2 + C}{x^2+1}

y(0) = \frac{C}{1} = 3 \qquad \Longrightarrow \qquad C = 3, \qquad y = \frac{2x^2 + 3}{x^2 + 1}

Check: $(x^2+1)\,y = 2x^2 + 3$ , and differentiating that product gives $(x^2+1)\,y' + 2x\,y = 4x\ \checkmark$

Problem 4 · Spot the Derivative

Given: $a(x)\,y' + b(x)\,y$ is already $\big[a(x)\,y\big]'$ exactly when $b = a'$ — which equation can therefore be integrated at once, with no integrating factor at all?

✅ Correct!

\big[x^3 y\big]' = x^3 y' + 3x^2 y

, so the equation reads

\big[x^3 y\big]' = e^x

and one integration finishes it:

x^3 y = e^x + C

❌ Short by a factor of $3$ .

\big[x^3y\big]' = x^3y' + 3x^2y

, not

x^3y' + x^2y

. This equation still yields to the method — standard form gives

p = \frac{1}{x}

and

u = x

— but not for free.

❌ Watch the sign.

\big[(\cos x)\,y\big]' = (\cos x)\,y' - (\sin x)\,y

. With

+\sin x

the left side is that derivative's near miss, not the derivative.

❌ Check $a'$ .

\big[e^{2x}y\big]' = e^{2x}y' + 2e^{2x}y

, and

2e^{2x} \neq e^{x}

❌ Not quite. Differentiate the coefficient of

y'

and compare it with the coefficient of

y

; only when they agree is the left side a completed product rule.

Show solution

The product rule gives $\big[a(x)\,y\big]' = a(x)\,y' + a'(x)\,y$ , so the test is simply whether the coefficient of $y$ equals the derivative of the coefficient of $y'$ :

$a = x^3$ : $a' = 3x^2$ , and the coefficient of $y$ is $3x^2\ \checkmark$
$a = x^3$ with $b = x^2$ : $a' = 3x^2 \neq x^2$
$a = \cos x$ : $a' = -\sin x \neq +\sin x$
$a = e^{2x}$ : $a' = 2e^{2x} \neq e^{x}$

Only the first passes, so it collapses immediately:

\big[x^3 y\big]' = e^x \qquad \Longrightarrow \qquad x^3 y = e^x + C

y = \frac{e^x + C}{x^3}

The three failures are not unsolvable — each one still has an integrating factor. Running the four steps on the second, for instance, gives $p = \frac{1}{x}$ , then $u = x$ , then $\big[x\,y\big]' = \frac{e^x}{x^2}$ . The method never needs you to spot anything; spotting only saves work.

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Integrating Factors: Two Mistakes and a Trick Question

IKey Formulas

IIWhat Makes uuu an Integrating Factor

IIIThe Round Trip

IVForgetting the Right Side

VQuiz Questions

IIWhat Makes $u$ an Integrating Factor