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

The Integrating Factor and the Four-Step Method

IKey Formulas

Formula	Name	What it is for
$y' + p(x)\,y = q(x)$	Standard linear form	$p$ is read off this form and no other
$(u\,y)' = u\,y' + u'\,y$	Product rule	The identity the whole method rests on
$u' = p\,u \;\Longrightarrow\; u = e^{\int p(x)\,dx}$	Integrating factor	No arbitrary constant — one $u$ is enough
$(u\,y)' = q\,u$	The line to always write	Integrate both sides, then solve for $y$

Key Insight: After multiplying by $u$ the left side is $u\,y' + p\,u\,y$ , and the product rule already supplies the first term of $(u\,y)'$ for free — so the entire method costs exactly one condition, $u' = p\,u$ .

IIWhich Multiplier Collapses the Left Side?

Multiplying $y' - \tfrac{1}{x}y = x^2$ by $u$ folds the left side into $(u y)'$ only where $u' = p\,u$ .

Candidate Test point x 2

Both $1/x$ and $5/x$ pass, because adding a constant to $\int p\,dx$ only multiplies $u$ by $e^{C}$ — which is why the formula carries no $+C$ .

IIIThe Four Steps on $x\,y' - y = x^3$

One well-chosen factor turns an equation whose variables do not separate into a single integration.

Step 1

Step 1 — Standard linear form

x\,y' - y = x^3 \quad\Longrightarrow\quad y' - \frac{1}{x}\,y = x^2

The coefficient of

y'

must be

1

, so divide by

x

; only now can

p

be read off:

p = -\dfrac{1}{x}

\ q = x^2

Step 2 — The integrating factor

u = e^{\int -\frac{1}{x}\,dx} = e^{-\ln x} = e^{\ln (x^{-1})} = \frac{1}{x}

Never leave it as

e^{-\ln x}

: since

e^{\ln x} = x

, the factor is simply

1/x

Step 3 — Multiply BOTH sides

\frac{1}{x}\,y' - \frac{1}{x^2}\,y = x^2 \cdot \frac{1}{x} = x

The right-hand side gets the factor too:

q\,u = x

Step 4 — Write the derivative line

\left(\frac{1}{x}\,y\right)' = x

Product-rule check:

\dfrac{1}{x}y' + y\left(\dfrac{1}{x}\right)' = \dfrac{1}{x}y' - \dfrac{1}{x^2}y \ \checkmark

Step 5 — Integrate

\frac{1}{x}\,y = \frac{x^2}{2} + c \quad\Longrightarrow\quad y = \frac{x^3}{2} + c\,x

Check:

x\left(\tfrac{3}{2}x^2 + c\right) - \left(\tfrac{1}{2}x^3 + cx\right) = x^3 \ \checkmark

IVThe Constant and the Initial Condition

Every curve here solves $x\,y' - y = x^3$ ; a single initial value selects one of them.

Initial value y(1) 1.5

The $c\,x$ term is the general solution of the homogeneous equation $x\,y' - y = 0$ , which is why it survives untouched while $x^3/2$ carries the right-hand side.

VQuiz Questions

Problem 1 · Build the Factor

Given: $y' + \dfrac{2}{x}\,y = x$ for $x > 0$ — find the integrating factor $u$ .

✅ Correct!

e^{2\ln x} = e^{\ln x^2} = x^2

, and the condition holds:

(x^2)' = 2x = \dfrac{2}{x}\cdot x^2

❌ Check the integral. The exponent is

\int p\,dx

, and

\int \dfrac{2}{x}\,dx = 2\ln x

, not

2x

❌ One step short.

2\ln x

\int p\,dx

; the integrating factor is

e

raised to that.

❌ That is the reciprocal.

x^{-2} = e^{-\int p\,dx}

, and multiplying by it gives

(x^{-2}y)' = x^{-2}y' - \dfrac{2}{x}x^{-2}y

— a minus sign where the equation has a plus.

Show solution

The equation is already in standard form, so $p = \dfrac{2}{x}$ can be read off directly.

\int p\,dx = \int \frac{2}{x}\,dx = 2\ln x

u = e^{2\ln x} = e^{\ln x^{2}} = x^{2}

Verify the defining condition $u' = p\,u$ :

(x^{2})' = 2x, \qquad \frac{2}{x}\cdot x^{2} = 2x \ \checkmark

Multiplying through: $\left(x^{2}y\right)' = x^{3}$ , so $x^{2}y = \dfrac{x^{4}}{4} + C$ and $y = \dfrac{x^{2}}{4} + \dfrac{C}{x^{2}}$ .

Problem 2 · Standard Form First

Given: $x\,y' + 3y = x^{5}$ for $x > 0$ — find the integrating factor $u$ .

✅ Correct! Dividing by

x

first gives

p = \dfrac{3}{x}

, so

u = e^{3\ln x} = x^{3}

, and

(x^3)' = 3x^2 = \dfrac{3}{x}\cdot x^3

❌ Close, but the equation was not in standard form. The coefficient of

y'

must be

1

: dividing by

x

turns the

3

into

p = 3/x

, so the exponent is

3\ln x

, not

3x

❌ Careful with the exponential.

e^{3\ln x} = e^{\ln x^{3}} = x^{3}

; the

3

multiplies the logarithm inside the exponent, so it becomes a power, not a factor.

❌ Sign of $p$ .

x^{-3} = e^{-3\ln x}

comes from

p = -3/x

; here the

3y

term is positive.

Show solution

Step 1 — standard form. Divide through by $x$ (the coefficient of $y'$ must be $1$ ):

y' + \frac{3}{x}\,y = x^{4}, \qquad p = \frac{3}{x}, \quad q = x^{4}

Step 2 — integrating factor.

u = e^{\int \frac{3}{x}dx} = e^{3\ln x} = e^{\ln x^{3}} = x^{3}

Steps 3 and 4 — multiply both sides and integrate:

\left(x^{3}y\right)' = x^{3}\cdot x^{4} = x^{7} \quad\Longrightarrow\quad x^{3}y = \frac{x^{8}}{8} + C

y = \frac{x^{5}}{8} + \frac{C}{x^{3}}

Reading $p = 3$ off the original equation would give $e^{3x}$ , and $\left(e^{3x}y\right)' = e^{3x}y' + 3e^{3x}y$ never matches the $3y/x$ that the equation actually has.

Problem 3 · Carry It Through

Given: $y' + \dfrac{1}{x}\,y = 4x$ for $x > 0$ , whose integrating factor is $u = x$ — find the derivative line and the general solution.

After multiplying both sides by $u = x$ :

General solution:

✅ Correct!

xy = \dfrac{4}{3}x^{3} + C

, and dividing by

x

carries the constant with it:

y = \dfrac{4}{3}x^{2} + \dfrac{C}{x}

❌ The right-hand side gets the factor too.

q\,u = 4x\cdot x = 4x^{2}

— multiplying only the left side is the most common slip in this method.

❌ Check the collapse. With

u = x

the left side is

x\,y' + y

, which is exactly

(x\,y)'

, and the right side is

4x\cdot x

❌ Divide the whole line by $u$ . From

xy = \frac{4}{3}x^{3} + C

, the constant becomes

C/x

, not

C

— check it:

\frac{4}{3}x^{2} + C

gives

y' + y/x = 4x + C/x

❌ Not quite. Solve

xy = \frac{4}{3}x^{3} + C

for

y

by dividing every term by

x

Show solution

Step 3 — multiply both sides by $u = x$ :

x\,y' + y = 4x^{2}

Step 4 — the left side is one derivative:

(x\,y)' = 4x^{2} \qquad \text{(check: } (xy)' = x y' + y \text{)}

Integrate:

x\,y = \frac{4x^{3}}{3} + C \quad\Longrightarrow\quad y = \frac{4}{3}x^{2} + \frac{C}{x}

Verify: $y' = \dfrac{8}{3}x - \dfrac{C}{x^{2}}$ , so

y' + \frac{y}{x} = \frac{8}{3}x - \frac{C}{x^{2}} + \frac{4}{3}x + \frac{C}{x^{2}} = 4x \ \checkmark

Problem 4 · Negative $p$ , With an Initial Condition

Given: $y' - \dfrac{2}{x}\,y = x^{2}$ for $x > 0$ with $y(1) = 3$ — find $y(x)$ . Every option below satisfies $y(1) = 3$ ; only one satisfies the equation.

✅ Correct!

u = x^{-2}

makes

q\,u = 1

, so

x^{-2}y = x + C

and

y = x^{3} + Cx^{2}

; the condition

y(1) = 3

gives

C = 2

❌ That is the $u = x^{2}$ path. Here

p = -2/x

, so

\int p\,dx = -2\ln x

and

u = e^{-2\ln x} = x^{-2}

; using

x^{2}

is using

e^{-\int p\,dx}

, the reciprocal factor.

❌ Substitute and check. Compute

y' - \dfrac{2}{x}y

for your choice — it must come out to exactly

x^{2}

, and the homogeneous part must be a multiple of

x^{2}

Show solution

Step 1. Already standard: $p = -\dfrac{2}{x}$ , $q = x^{2}$ .

Step 2. $\displaystyle \int p\,dx = -2\ln x$ , so

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

Step 3. Multiply both sides — including the right-hand side:

\frac{1}{x^{2}}y' - \frac{2}{x^{3}}y = x^{2}\cdot\frac{1}{x^{2}} = 1

Step 4. Write the derivative line and integrate:

\left(\frac{y}{x^{2}}\right)' = 1 \quad\Longrightarrow\quad \frac{y}{x^{2}} = x + C \quad\Longrightarrow\quad y = x^{3} + Cx^{2}

Initial condition: $y(1) = 1 + C = 3$ , so $C = 2$ and $y = x^{3} + 2x^{2}$ .

Verify: $y' = 3x^{2} + 4x$ and $\dfrac{2}{x}y = 2x^{2} + 4x$ , so $y' - \dfrac{2}{x}y = x^{2} \ \checkmark$

Solved: 0 / 4

The Integrating Factor and the Four-Step Method

IKey Formulas

IIWhich Multiplier Collapses the Left Side?

IIIThe Four Steps on x y′−y=x3x\,y' - y = x^3xy′−y=x3

IVThe Constant and the Initial Condition

VQuiz Questions

IIIThe Four Steps on $x\,y' - y = x^3$