MHT-CET Maths · Differential Equations

Linear Differential Equations — the Integrating Factor

A first-order linear ODE has the shape dy/dx + P(x)y = Q(x). Multiply by the integrating factor IF = e to the power of the integral of P, and the left side collapses into d/dx(y times IF) — integrate once and you are done.

All Differential Equations notes MHT-CET Maths strategy

Why this matters

This is the workhorse subtopic and the densest HARD pool in the chapter — 24 PYQs, most of them HARD. Nearly every question is one skill: force the equation into standard form, read off P and Q, build the integrating factor, and integrate. The recurring MHT-CET traps live entirely here: reading P before the equation is in standard form, missing that some equations are only linear in x (swap the roles of x and y), and failing to spot a Bernoulli equation that becomes linear after one substitution.

Concept 1 of 8

Recognizing the Standard Linear Form

Intuition

Before you can use any of this, the equation must be written so that dy/dx sits alone with coefficient 1, the plain-y term is on the same side, and everything free of y is on the right. That shape — dy/dx plus P(x)y equals Q(x) — is what makes the whole integrating-factor machine run. Get the equation into it FIRST; only then read P and Q.

Definition

A first-order ODE is linear when it can be written in the standard form

\dfrac{dy}{dx} + P(x)\,y = Q(x),

where

P

and

Q

depend on

x

only (not on

y

). To reach it:

Divide through by whatever multiplies $\dfrac{dy}{dx}$ so its coefficient becomes $1$ .
Collect every term containing $y$ on the left; the rest becomes $Q(x)$ on the right.
$P(x)$ is then the coefficient of $y$ , read off only after the coefficient of $\dfrac{dy}{dx}$ is $1$ .

Standard linear form

\dfrac{dy}{dx} + P(x)\,y = Q(x)

P(x)coefficient of y — read AFTER dividing so dy/dx has coefficient 1
Q(x)everything with no y, on the right

Worked example

Put

x\dfrac{dy}{dx} - 2y = x^3

into standard linear form and identify

P(x)

and

Q(x)

Divide every term by $x$ so $\dfrac{dy}{dx}$ has coefficient $1$ : $\dfrac{dy}{dx} - \dfrac{2}{x}y = x^2$ .
Compare with $\dfrac{dy}{dx} + P(x)y = Q(x)$ .
Read off: $P(x) = -\dfrac{2}{x}$ , $Q(x) = x^2$ .

Answer:

P(x) = -\dfrac{2}{x}

Q(x) = x^2

Practice this conceptself-check · 4 quick reps

Try it yourself

Write

(1+x^2)\dfrac{dy}{dx} + 2xy = 4x^2

in standard form and give

P(x)

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Standard form of $2\dfrac{dy}{dx} + 6y = e^x$ : give $P$ .
2.
For $x\dfrac{dy}{dx} + y = \log x$ , what is $P(x)$ ?
3.
For $\cos x\dfrac{dy}{dx} - y\sin x = 6x$ , what is $P(x)$ ?
4.
Is $\dfrac{dy}{dx} + xy^2 = x$ linear?

Read $P$ only AFTER making the $\dfrac{dy}{dx}$ coefficient $1$

\cos x\,\dfrac{dy}{dx} - y\sin x = 6x

, the naive read

P = -\sin x

is wrong. Divide by

\cos x

first:

\dfrac{dy}{dx} - y\tan x = 6x\sec x

, so

P = -\tan x

. Reading

P

before normalizing the leading coefficient is the single most common mistake here.

A $y^2$ , $\sqrt{y}$ , or $1/y$ means it is NOT linear (yet)

Linear means

y

appears only to the first power. Terms like

y^2\sec x

y^4\cos x

are NON-linear — those are Bernoulli equations that first need a substitution before an integrating factor applies.

Concept 2 of 8

The Integrating Factor and the Solution Formula

Intuition

Once the equation is in standard form, multiply the whole thing by the integrating factor IF = e to the integral of P. The magic: the left side becomes an exact derivative, d/dx of (y times IF). So integrating both sides just once gives y times IF equals the integral of Q times IF. Two formulas carry the entire subtopic.

Definition

For the standard linear ODE $\dfrac{dy}{dx} + P(x)y = Q(x)$ :

The integrating factor is $\text{IF} = e^{\int P\,dx}$ .
Multiplying by IF turns the left side into a perfect derivative: $\dfrac{d}{dx}\big(y\cdot\text{IF}\big) = Q\cdot\text{IF}$ .
Integrating once gives the solution formula

y\cdot\text{IF} = \int Q\cdot\text{IF}\,dx + c.

The constant

c

is fixed later by any initial condition. Everything in this subtopic is: normalize, compute IF, integrate

Q\cdot\text{IF}

Integrating factor and general solution

\text{IF} = e^{\int P(x)\,dx}, \qquad y\cdot\text{IF} = \int Q(x)\cdot\text{IF}\,dx + c

IFthe integrating factor e to the integral of P
cthe single arbitrary constant, fixed by an initial condition

Worked example

Solve

\dfrac{dy}{dx} + 2y = e^{-2x}

Already standard: $P = 2$ , $Q = e^{-2x}$ .
$\text{IF} = e^{\int 2\,dx} = e^{2x}$ .
Solution formula: $y\,e^{2x} = \int e^{-2x}\cdot e^{2x}\,dx + c = \int 1\,dx + c = x + c$ .
Divide by $e^{2x}$ : $y = (x+c)e^{-2x}$ .

Answer:

y = (x + c)e^{-2x}

Practice this conceptself-check · 4 quick reps

Try it yourself

y + \dfrac{d}{dx}(xy) = x(\sin x + \log x)

, reduce it to a linear ODE, find the integrating factor, and give the solution.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
IF for $\dfrac{dy}{dx} + 3y = x$ .
2.
IF for $\dfrac{dy}{dx} + \dfrac{y}{x} = x$ .
3.
After multiplying by IF, the left side equals?
4.
Solve $\dfrac{dy}{dx} + y = 0$ .

From the bank · past-year question

Example 2Differential EquationsHARD

y+\frac{d}{dx}(xy) =x(\sin x+ \log x)

then

[Q118 · 22 April Shift I · 2025]

The left side is $\dfrac{d}{dx}(y\cdot\text{IF})$ — do not re-differentiate the product

After multiplying by IF, the entire left side is ALREADY the derivative of

y\cdot\text{IF}

. Integrating both sides simply un-does it, giving

y\cdot\text{IF} = \int Q\cdot\text{IF}\,dx + c

. Students who try to apply the product rule again are re-doing work the integrating factor already handled.

One arbitrary constant only, added at the integration step

The solution formula produces exactly ONE constant

c

, not one per side. Add it when you integrate

\int Q\cdot\text{IF}\,dx

; an initial condition then pins its value.

Concept 3 of 8

Simple Integrating Factors

Intuition

Most exam questions have a friendly P whose integral you can do in your head, so the IF comes out as a clean power, a clean exponential, or a trig factor. The three you meet constantly: P = n/x gives IF = x to the n; a constant P gives an exponential; and P = -tan x gives IF = cos x. Recognize the pattern and the IF is instant.

Definition

Common integrating factors worth recognizing at a glance:

$P = \dfrac{1}{x} \Rightarrow \text{IF} = e^{\log x} = x$ ; more generally $P = \dfrac{n}{x} \Rightarrow \text{IF} = x^{n}$ .
$P = \dfrac{2x}{1+x^2} \Rightarrow \text{IF} = e^{\log(1+x^2)} = 1+x^2$ (and likewise $\dfrac{3x^2}{1+x^3} \Rightarrow 1+x^3$ ).
$P = \text{constant } k \Rightarrow \text{IF} = e^{kx}$ .
$P = -\tan x \Rightarrow \text{IF} = e^{\log\cos x} = \cos x$ ; $P = \cot x \Rightarrow \text{IF} = \sin x$ .

In every case the pattern is: $\int P\,dx$ is a logarithm, so the IF is what that logarithm is a log OF.

Common integrating factors

P=\dfrac{n}{x}\Rightarrow\text{IF}=x^{n}, \quad P=k\Rightarrow\text{IF}=e^{kx}, \quad P=-\tan x\Rightarrow\text{IF}=\cos x

Worked example

Solve

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

with

y(1)=1

, and find

y\!\left(\tfrac12\right)

Standard form: $\dfrac{dy}{dx} + \dfrac{2}{x}y = x$ , so $P = \dfrac{2}{x}$ .
$\text{IF} = e^{\int \frac{2}{x}dx} = x^2$ .
$y x^2 = \int x\cdot x^2\,dx + c = \dfrac{x^4}{4} + c$ . At $x=1,\,y=1$ : $1 = \tfrac14 + c \Rightarrow c = \tfrac34$ .
At $x = \tfrac12$ : $y\cdot\tfrac14 = \tfrac{1}{64} + \tfrac34 \Rightarrow y = \tfrac{49}{16}$ .

Answer:

y\!\left(\tfrac12\right) = \dfrac{49}{16}

Practice this conceptself-check · 4 quick reps

Try it yourself

Find the general solution of

\dfrac{dy}{dx} - y = x - 5

, then the curve through

(0,2)

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
IF for $\dfrac{dy}{dx} + \dfrac{y}{x} = \sin x$ .
2.
IF for $(1+x^3)\dfrac{dy}{dx}$ -form with $P = \dfrac{3x^2}{1+x^3}$ .
3.
IF for $\sin x\dfrac{dy}{dx} + y\cos x = 4x$ .
4.
IF for $\dfrac{dy}{dx} + y = e^{-x}$ .

From the bank · past-year question

Example 3Differential EquationsMODERATE

y=y(x)

is the solution of the differential equation

x\frac{dy}{dx}+2y=x^2

satisfying

y(1)=1

, then the value of

y\!\left(\frac{1}{2}\right)

[Q116 · 9th May Shift 1 · 2023]

$e^{\log f(x)} = f(x)$ — simplify the exponential of a log

When

\int P\,dx = \log(1+x^2)

, the IF is

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

, NOT

e^{1+x^2}

. Any time

\int P\,dx

turns out to be a logarithm, the IF is simply the thing inside that log. Forgetting to cancel

e

and

\log

leaves an unusable IF.

Watch the sign of $P$ in the exponential

P = -\tan x

gives

\int P\,dx = \log\cos x

, so

\text{IF} = \cos x

. A dropped minus sign would give

\sec x

and the wrong solution. Track the sign of

P

all the way into the IF.

Drill 9 more on simple integrating factors

Concept 4 of 8

Tricky Integrating Factors

Intuition

Some P's need real work before the IF appears: the integral might be a log-of-a-log, might combine an exponential with a power, or might need partial fractions. The method is identical — IF = e to the integral of P — but the integral itself is the challenge. Do that integral carefully and the rest is routine.

Definition

Harder integrating factors seen in HARD questions:

Log-of-a-log: $P = \dfrac{1}{x\log x}$ gives $\int P\,dx = \log(\log x)$ , so $\text{IF} = \log x$ .
Exponential times a power: $P = -\dfrac{x}{1+x} = -1 + \dfrac{1}{1+x}$ gives $\int P\,dx = -x + \log(1+x)$ , so $\text{IF} = e^{-x}(1+x)$ .
Combine-then-cancel: $P = 1 - \dfrac{1}{x}$ gives $\int P\,dx = x - \log x$ , so $\text{IF} = \dfrac{e^{x}}{x}$ .
Partial fractions: $P = -\dfrac{2}{x} + \dfrac{1}{x-1}$ gives $\int P\,dx = -2\log x + \log(x-1)$ , so $\text{IF} = \dfrac{x-1}{x^2}$ .

Split $P$ into standard pieces, integrate each, then exponentiate.

A tricky IF built by partial fractions

P = -\dfrac{2}{x} + \dfrac{1}{x-1} \;\Rightarrow\; \int P\,dx = \log\dfrac{x-1}{x^2} \;\Rightarrow\; \text{IF} = \dfrac{x-1}{x^2}

Worked example

Find the integrating factor of

x\log x\,\dfrac{dy}{dx} + y = 2x\log x

Divide by $x\log x$ : $\dfrac{dy}{dx} + \dfrac{1}{x\log x}\,y = 2$ , so $P = \dfrac{1}{x\log x}$ .
Substitute $u = \log x$ , $du = \dfrac{dx}{x}$ : $\int \dfrac{1}{x\log x}\,dx = \int \dfrac{du}{u} = \log(\log x)$ .
So $\text{IF} = e^{\log(\log x)} = \log x$ .

Answer:

\text{IF} = \log x

Practice this conceptself-check · 4 quick reps

Try it yourself

Solve

(1+x)\dfrac{dy}{dx} - xy = 1 - x

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
IF when $P = \dfrac{1}{x\log x}$ .
2.
IF when $P = 1 - \dfrac{1}{x}$ .
3.
IF when $P = -\dfrac{x}{1+x}$ .
4.
First step when $P$ is a rational function?

From the bank · past-year question

Example 4Differential EquationsHARD

Let

y = y(x)

be the solution of the differential equation

x\log x\,\frac{dy}{dx} + y = 2x\log x\; (x \geq 1)

, then

y(e)

is equal to

[Q124 · 16th May Shift 1 · 2023]

Split $P$ before integrating a rational coefficient

For

P = -\dfrac{x}{1+x}

, do polynomial/partial-fraction division first:

-\dfrac{x}{1+x} = -1 + \dfrac{1}{1+x}

. Integrating the un-split form is where students stall. The same trick handles

\dfrac{2-x}{x(x-1)}

via partial fractions before the IF appears.

Do not stop at $\int P\,dx$ — exponentiate it

The IF is

e^{\int P\,dx}

, so after finding

\int P\,dx = -x + \log(1+x)

you still must exponentiate to

e^{-x}(1+x)

. Using the raw integral as the IF is a common slip on the harder coefficients.

Drill 3 more on tricky integrating factors

Concept 5 of 8

Linear in x — Swap the Roles of x and y

Intuition

Some equations are hopeless as dy/dx but become perfectly linear when you flip them to dx/dy. If y appears in awkward places but x appears only to the first power, treat x as the unknown function of y: write dx/dy + P(y)x = Q(y), and use exactly the same integrating factor machine with y as the variable.

Definition

An ODE is **linear in $x$ ** if it fits

\dfrac{dx}{dy} + P(y)\,x = Q(y),

with

P, Q

functions of

y

only. Then

\text{IF} = e^{\int P(y)\,dy}

and

x\cdot\text{IF} = \int Q(y)\cdot\text{IF}\,dy + c

. Signals to flip: the equation has

y\,dx

and

x\,dy

terms, or a coefficient of

\dfrac{dy}{dx}

that is a messy function of

y

. Rewriting

\dfrac{dy}{dx}

1/\dfrac{dx}{dy}

exposes the linear-in-

x

shape.

Linear in x (reciprocal form)

\dfrac{dx}{dy} + P(y)\,x = Q(y), \qquad \text{IF} = e^{\int P(y)\,dy}

Worked example

Solve

y\,dx - (x + 3y^2)\,dy = 0

, the curve through

(1,1)

Rewrite as $\dfrac{dx}{dy} = \dfrac{x}{y} + 3y$ , i.e. $\dfrac{dx}{dy} - \dfrac{1}{y}x = 3y$ — linear in $x$ , $P(y) = -\dfrac{1}{y}$ .
$\text{IF} = e^{\int -\frac{1}{y}dy} = e^{-\log y} = \dfrac{1}{y}$ .
$x\cdot\dfrac{1}{y} = \int 3y\cdot\dfrac{1}{y}\,dy + c = 3y + c$ , so $x = 3y^2 + cy$ . Through $(1,1)$ : $1 = 3 + c \Rightarrow c = -2$ .
The curve is $x = 3y^2 - 2y$ .

Answer:

x = 3y^2 - 2y

(so it passes through

\left(-\tfrac13, \tfrac13\right)

Practice this conceptself-check · 4 quick reps

Try it yourself

Solve

(1 + y^2) + (x - e^{\tan^{-1}y})\dfrac{dy}{dx} = 0

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
When is flipping to $\dfrac{dx}{dy}$ worth it?
2.
IF of $\dfrac{dx}{dy} - \dfrac{1}{y}x = 3y$ .
3.
IF of $\dfrac{dx}{dy} + \dfrac{x}{1+y^2} = \cdots$ .
4.
In linear-in- $x$ , the variable of integration is?

From the bank · past-year question

Example 5Differential EquationsHARD

The solution of

\left( 1 +y^{2} \right)+\left( x-e^{\tan - 1y} \right)\frac{dy}{\text{ }dx}= 0

[Q126 · 20 April Shift II · 2025]

If $y$ is tangled, check whether $x$ is linear before giving up

y\,dx - (x+3y^2)\,dy = 0

never separates and is not linear in

y

— but dividing by

dy

shows

x

appears only to the first power, so it is linear in

x

. Flipping to

\dfrac{dx}{dy}

is the move; forcing

\dfrac{dy}{dx}

leads nowhere.

After flipping, integrate with respect to $y$ , not $x$

Everything shifts:

P

and

Q

are functions of

y

, the IF is

e^{\int P(y)\,dy}

, and the solution formula integrates

Q\cdot\text{IF}

over

y

. Slipping back to

dx

mid-solution is a classic error.

Drill 1 more on linear in x — swap the roles of x and y

Concept 6 of 8

Bernoulli Equations — Substitute to Linearize

Intuition

A Bernoulli equation has a lone power of y on the right: dy/dx + P y = Q y to the n. It is not linear as written, but one substitution fixes it. Divide through by y to the n, then let v = y to the (1 minus n) — the equation becomes linear in v, and you finish with the ordinary integrating factor.

Definition

A Bernoulli equation is $\dfrac{dy}{dx} + P(x)y = Q(x)y^{n}$ with $n \neq 0, 1$ . To solve:

**Divide by $y^{n}$ :** $y^{-n}\dfrac{dy}{dx} + P\,y^{1-n} = Q$ .
Substitute $v = y^{1-n}$ , so $\dfrac{dv}{dx} = (1-n)y^{-n}\dfrac{dy}{dx}$ .
The equation becomes **linear in $v$ :** $\dfrac{dv}{dx} + (1-n)P\,v = (1-n)Q$ — now use $\text{IF} = e^{\int (1-n)P\,dx}$ .

Special common case $n = 2$ : $v = y^{-1} = 1/y$ .

Bernoulli substitution

\dfrac{dy}{dx} + Py = Qy^{n} \;\xrightarrow{\;v = y^{1-n}\;}\; \dfrac{dv}{dx} + (1-n)Pv = (1-n)Q

nthe power on the right-hand y; must not be 0 or 1
vthe new unknown y to the power (1 minus n)

Worked example

Solve

\dfrac{dy}{dx} = y\tan x - y^2\sec x

Rewrite: $\dfrac{dy}{dx} - y\tan x = -y^2\sec x$ — Bernoulli with $n = 2$ .
Divide by $y^2$ and let $v = y^{-1}$ , so $\dfrac{dv}{dx} = -y^{-2}\dfrac{dy}{dx}$ : the equation becomes $\dfrac{dv}{dx} - v\tan x = \sec x$ .
$\text{IF} = e^{\int -\tan x\,dx} = \cos x$ . Then $\dfrac{d}{dx}(v\cos x) = \cos x\sec x = 1$ , so $v\cos x = x + c$ .
With $v = 1/y$ : $\dfrac{\cos x}{y} = x + c$ , i.e. $\cos x = y(x + c)$ . Matching the bank's tan-form solution: $\sec x = y(\tan x + c)$ .

Answer:

\sec x = y(\tan x + c)

Practice this conceptself-check · 4 quick reps

Try it yourself

Solve

\dfrac{dy}{dx} + \dfrac{y}{x} = x\,y^2

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
For $\dfrac{dy}{dx} + Py = Qy^2$ , substitute?
2.
For $n = 3$ , what is $v$ ?
3.
First step to linearize a Bernoulli equation?
4.
Is $\dfrac{dy}{dx} + xy = x^2 y^0$ Bernoulli?

From the bank · past-year question

Example 6Differential EquationsHARD

The solution of the equation

x^{2}y-x^{3}\cdot\frac{\text{ }dy}{\text{ }dx}=y^{4}\cos x

, where

y(0) = 1

, is

[Q123 · 25 April Shift I · 2025]

Divide by $y^{n}$ BEFORE substituting

You cannot substitute

v = y^{1-n}

usefully until the

y^{-n}\dfrac{dy}{dx}

term is exposed. Divide the whole equation by

y^{n}

first; only then does

\dfrac{dv}{dx}

appear cleanly. Skipping this step leaves an equation you cannot linearize.

Spot the lone $y^{n}$ — it is not a linear ODE

\dfrac{dy}{dx} = y\tan x - y^2\sec x

looks linear until you see the

y^2

. Treating it as linear (integrating factor straight away) is wrong. The

y^{n}

on the right is the tell: substitute first.

Drill 3 more on bernoulli equations — substitute to linearize

Concept 7 of 8

Exact Equations by d(·)-Grouping

Intuition

Sometimes the fastest route is not an integrating factor at all — it is recognizing that a clump of terms is itself the differential of a simple product or quotient. Group the terms into pieces like d(xy) or d(x/y), integrate each piece directly, and the answer falls out. This beats the linear machine when the grouping is obvious.

Definition

Recognize these exact differentials and integrate by grouping:

$x\,dy + y\,dx = d(xy)$ .
$\dfrac{x\,dy - y\,dx}{y^2} = d\!\left(\dfrac{x}{y}\right)$ , and $\dfrac{y\,dx - x\,dy}{x^2} = d\!\left(\dfrac{y}{x}\right)$ .
$x\,dx + y\,dy = \tfrac12\,d(x^2 + y^2)$ .
For products like $(1+xy)y\,dx + (1-xy)x\,dy = 0$ , divide by a factor such as $x^2y^2$ to expose $d\!\left(-\tfrac{1}{xy}\right)$ , $d(\log x)$ , and $d(\log y)$ .

Exact differentials to spot

x\,dy + y\,dx = d(xy), \qquad \dfrac{x\,dy - y\,dx}{y^2} = d\!\left(\dfrac{x}{y}\right)

Worked example

Solve

x\,dy = y(dx + y\,dy)

with

y(1) = 1

y > 0

, and find

y(-3)

Expand: $x\,dy = y\,dx + y^2\,dy \Rightarrow x\,dy - y\,dx = y^2\,dy \Rightarrow \dfrac{x\,dy - y\,dx}{y^2} = dy$ .
Recognize the left side: since $\dfrac{y\,dx - x\,dy}{y^2} = d\!\left(\dfrac{x}{y}\right)$ , the negative of it is $\dfrac{x\,dy - y\,dx}{y^2} = d\!\left(-\dfrac{x}{y}\right)$ . So $d\!\left(-\dfrac{x}{y}\right) = dy$ .
Integrate: $-\dfrac{x}{y} = y + c$ . At $(1,1)$ : $-1 = 1 + c \Rightarrow c = -2$ .
At $x = -3$ : $\dfrac{3}{y} = y - 2 \Rightarrow y^2 - 2y - 3 = 0 \Rightarrow y = 3$ (taking $y > 0$ ).

Answer:

y(-3) = 3

Practice this conceptself-check · 4 quick reps

Try it yourself

Solve

(1 + xy)y\,dx + (1 - xy)x\,dy = 0

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$x\,dy + y\,dx = ?$
2.
$\dfrac{x\,dy - y\,dx}{y^2} = ?$
3.
$x\,dx + y\,dy = ?$
4.
To expose a grouping in $(1+xy)y\,dx + \cdots$ , divide by?

From the bank · past-year question

Example 7Differential EquationsHARD

The solution of

(1+xy)y\,dx+(1-xy)x\,dy=0

[Q146 · 13th May Shift 2 · 2024]

Mind the sign and denominator of the quotient differentials

x\,dy + y\,dx = d(xy)

(a PLUS), but

\dfrac{x\,dy - y\,dx}{y^2} = d\!\left(\tfrac{x}{y}\right)

(a MINUS, over

y^2

). Swapping the sign or writing

x^2

in the denominator gives the wrong grouping and the wrong answer.

Try grouping before reaching for an integrating factor

When you see

x\,dy

and

y\,dx

sitting together, test for an exact differential first. Forcing the equation into standard linear form when a clean

d(xy)

d(x/y)

is staring at you wastes the whole solution.

Drill 1 more on exact equations by d(·)-grouping

Concept 8 of 8

Direct Integration and Reduction of Order

Intuition

The simplest first-order equations are the ones where dy/dx is already isolated as a function of x alone — just integrate. And a second-order equation with no y term (only y'' and x) drops an order: integrate once to get dy/dx, apply a condition, integrate again. No integrating factor needed at all.

Definition

Two direct routes:

Direct integration: if $\dfrac{dy}{dx} = f(x)$ , then $y = \int f(x)\,dx + c$ . Likewise $(x+2)\dfrac{dy}{dx} = x^2 + 4x - 9$ becomes $\dfrac{dy}{dx} = \dfrac{x^2+4x-9}{x+2}$ , integrate after dividing.
Reduction of order: for $x\dfrac{d^2y}{dx^2} = 1$ (no $y$ , no $\dfrac{dy}{dx}$ ), write it as $\dfrac{d^2y}{dx^2} = \dfrac{1}{x}$ ; integrate to $\dfrac{dy}{dx} = \log x + c_1$ , fix $c_1$ with the slope condition, then integrate again for $y$ .

Each integration introduces one constant — a second-order problem needs two conditions.

Reduction of order (integrate twice)

\dfrac{d^2y}{dx^2} = g(x) \;\Rightarrow\; \dfrac{dy}{dx} = \int g(x)\,dx + c_1 \;\Rightarrow\; y = \int\!\left(\int g\,dx\right)dx + c_1 x + c_2

Worked example

Solve

\dfrac{d^2y}{dx^2} = 12x

given

y = 5

and

\dfrac{dy}{dx} = 3

x = 0

The right side has no $y$ , so integrate twice. Integrate once: $\dfrac{dy}{dx} = 6x^2 + c_1$ .
Apply $\dfrac{dy}{dx} = 3$ at $x = 0$ : $3 = 0 + c_1 \Rightarrow c_1 = 3$ .
Integrate again: $y = 2x^3 + 3x + c_2$ . Apply $y = 5$ at $x = 0$ : $5 = c_2$ .

Answer:

y = 2x^3 + 3x + 5

Practice this conceptself-check · 4 quick reps

Try it yourself

Solve

\dfrac{dy}{dx} = 3x^2 - 4x + 1

, the curve through

(0, 5)

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Solve $\dfrac{dy}{dx} = \cos x$ .
2.
$\int \log x\,dx = ?$
3.
How many constants does a 2nd-order ODE solution carry?
4.
First step for $x\dfrac{d^2y}{dx^2} = 1$ ?

From the bank · past-year question

Example 8Differential EquationsMODERATE

The solution of the differential equation

x\frac{{\text{ }d}^{2}y}{\text{ }dx^{2}}= 1

x=y= 1

with

\frac{dy}{\text{ }dx}= 0

x= 1

, is

[Q126 · 19 April Shift I · 2025]

Apply the slope condition after the FIRST integration

For a second-order equation, use the

\dfrac{dy}{dx}

condition to fix

c_1

as soon as you have

\dfrac{dy}{dx}

— do not wait until the end. Fixing both constants only at the final step tangles the algebra and often gives the wrong

c_2

Divide out the leading factor before integrating

x\dfrac{d^2y}{dx^2} = 1

is NOT

\dfrac{d^2y}{dx^2} = 1

; first isolate

\dfrac{d^2y}{dx^2} = \dfrac{1}{x}

. Integrating the un-isolated form gives

y = x + \cdots

instead of the correct

x\log x

term.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (8)

Recognizing the Standard Linear Form
Standard linear form
$\dfrac{dy}{dx} + P(x)\,y = Q(x)$
The Integrating Factor and the Solution Formula
Integrating factor and general solution
$\text{IF} = e^{\int P(x)\,dx}, \qquad y\cdot\text{IF} = \int Q(x)\cdot\text{IF}\,dx + c$
Simple Integrating Factors
Common integrating factors
$P=\dfrac{n}{x}\Rightarrow\text{IF}=x^{n}, \quad P=k\Rightarrow\text{IF}=e^{kx}, \quad P=-\tan x\Rightarrow\text{IF}=\cos x$
Tricky Integrating Factors
A tricky IF built by partial fractions
$P = -\dfrac{2}{x} + \dfrac{1}{x-1} \;\Rightarrow\; \int P\,dx = \log\dfrac{x-1}{x^2} \;\Rightarrow\; \text{IF} = \dfrac{x-1}{x^2}$
Linear in x — Swap the Roles of x and y
Linear in x (reciprocal form)
$\dfrac{dx}{dy} + P(y)\,x = Q(y), \qquad \text{IF} = e^{\int P(y)\,dy}$
Bernoulli Equations — Substitute to Linearize
Bernoulli substitution
$\dfrac{dy}{dx} + Py = Qy^{n} \;\xrightarrow{\;v = y^{1-n}\;}\; \dfrac{dv}{dx} + (1-n)Pv = (1-n)Q$
Exact Equations by d(·)-Grouping
Exact differentials to spot
$x\,dy + y\,dx = d(xy), \qquad \dfrac{x\,dy - y\,dx}{y^2} = d\!\left(\dfrac{x}{y}\right)$
Direct Integration and Reduction of Order
Reduction of order (integrate twice)
$\dfrac{d^2y}{dx^2} = g(x) \;\Rightarrow\; \dfrac{dy}{dx} = \int g(x)\,dx + c_1 \;\Rightarrow\; y = \int\!\left(\int g\,dx\right)dx + c_1 x + c_2$

Watch out for (16)

Read $P$ only AFTER making the $\dfrac{dy}{dx}$ coefficient $1$
→ Recognizing the Standard Linear Form
A $y^2$ , $\sqrt{y}$ , or $1/y$ means it is NOT linear (yet)
→ Recognizing the Standard Linear Form
The left side is $\dfrac{d}{dx}(y\cdot\text{IF})$ — do not re-differentiate the product
→ The Integrating Factor and the Solution Formula
One arbitrary constant only, added at the integration step
→ The Integrating Factor and the Solution Formula
$e^{\log f(x)} = f(x)$ — simplify the exponential of a log
→ Simple Integrating Factors
Watch the sign of $P$ in the exponential
→ Simple Integrating Factors
Split $P$ before integrating a rational coefficient
→ Tricky Integrating Factors
Do not stop at $\int P\,dx$ — exponentiate it
→ Tricky Integrating Factors
If $y$ is tangled, check whether $x$ is linear before giving up
→ Linear in x — Swap the Roles of x and y
After flipping, integrate with respect to $y$ , not $x$
→ Linear in x — Swap the Roles of x and y
Divide by $y^{n}$ BEFORE substituting
→ Bernoulli Equations — Substitute to Linearize
Spot the lone $y^{n}$ — it is not a linear ODE
→ Bernoulli Equations — Substitute to Linearize
Mind the sign and denominator of the quotient differentials
→ Exact Equations by d(·)-Grouping
Try grouping before reaching for an integrating factor
→ Exact Equations by d(·)-Grouping
Apply the slope condition after the FIRST integration
→ Direct Integration and Reduction of Order
Divide out the leading factor before integrating
→ Direct Integration and Reduction of Order

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Differential EquationsMODERATE

The equation of the curve passing through origin and satisfying

\left( 1 +x^{2} \right)\frac{dy}{\text{ }dx}+ 2xy= 4x^{2}

[Q115 · 25 April Shift II · 2025]

Example 2Differential EquationsMODERATE

Integrating factor of the differential equation

\frac{dy}{dx} + y = 1 + \frac{y}{x}

[Q149 · 9th May Shift 2 · 2023]

Example 3Differential EquationsHARD

The curve satisfying the differential equation

y\,dx - (x+3y^2)\,dy = 0

and passing through the point

(1,1)

also passes through the point

[Q125 · 9th May Shift 1 · 2023]

Example 4Differential EquationsHARD

The general solution of the differential equation

\frac{dy}{dx} = y\tan x - y^2\sec x

[Q142 · Shift 1 · 2023]

Example 5Differential EquationsHARD

x\,dy=y(dx+y\,dy),\ y(1)=1,\ y(x)>0

, then

y(-3)

is:

[Q114 · 13th May Shift 2 · 2024]

Drill every past-year question on this subtopic

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