MHT-CET Maths · Formula sheet

Differential Equations formulas

41 formulas and 82 common traps for MHT-CET Maths Differential Equations, grouped by subtopic.

Order, Degree, Formation, and Verification

Differential Equation Terminology

The master link

\text{order of the ODE} \;=\; \text{number of independent arbitrary constants in its general solution}

orderorder of the highest derivative appearing
arbitrary constantsindependent free parameters in the solution family

Order = Order of the Highest Derivative Present

Order

\text{order} = \text{the order of the highest derivative appearing in the equation}

Degree = Power of the Highest Derivative After Clearing Radicals

Degree

\text{degree} = \text{power of the highest-order derivative, once the equation is polynomial in its derivatives}

When Degree Is Undefined (Derivative Inside a Transcendental)

Degree-undefined criterion

\text{degree undefined} \iff \text{a derivative sits inside a transcendental (}\log,\ \sin,\ \cos,\ e^{(\cdot)}\text{)}

Collapse Redundant Arbitrary Constants Before Counting Order

Constant-absorption identity

C_3\,e^{\,x + C_4} = \big(C_3 e^{C_4}\big)e^{x} = B\,e^{x}

$B$ the single surviving constant after absorbing $C_3, C_4$

Formation: n Independent Constants ⇒ Order-n Differential Equation

Formation order

n \text{ independent arbitrary constants} \;\Longrightarrow\; \text{differential equation of order } n

Forming the Differential Equation of a Curve Family

Elimination recipe

\text{differentiate } n \text{ times} \;\to\; \text{solve for the constants} \;\to\; \text{substitute back to eliminate them}

Forming the Differential Equation of Circles and Parabolas

Two workhorses

x^2 + y^2 = 2ax \;(\text{circle}) \qquad x^2 = 4ay \;(\text{parabola, axis } +Y)

$a$ the single geometric parameter to eliminate by one differentiation

Verifying a Solution and Identifying Its Family

Parametric derivative

\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} \qquad \dfrac{d^2y}{dx^2} = \dfrac{1}{dx/dt}\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)

$t$ the parameter — differentiate x and y with respect to it, then divide

Common traps

Order and degree are separate labels

Order is about WHICH derivative is highest; degree is about the POWER on it.

\big(\tfrac{d^2y}{dx^2}\big)^3 = x

is order 2 but degree 3. Don't conflate the two.

"Number of constants" means INDEPENDENT constants

Two constants that always merge into one (like

c_1 + c_3

) count as a single arbitrary constant. Collapse the family first, then count — the order equals the number of constants that survive.

A power on the top derivative is DEGREE, never order

\big(\tfrac{d^2y}{dx^2}\big)^{5}

reads as "order 2, degree 5", not "order 5". The exponent belongs to degree; the order only counts how many times you differentiated.

Clear radicals BEFORE you read the degree

The degree is NOT the fractional exponent you first see. For

\big(\tfrac{d^2y}{dx^2}\big)^{0.6} = y'

, raise to the 5th power to get

\big(\tfrac{d^2y}{dx^2}\big)^{3} = (y')^5

: degree

= 3

, not

0.6

. Make it polynomial first.

Raise to the LCM of the fractional exponents

With a

\sqrt{\;}

(power

\tfrac12

) and a

\sqrt[5]{\;}

(power

\tfrac15

), raise both sides to the 10th power in one shot — squaring alone leaves the 5th root, and 5th-powering alone leaves the square root.

Seeing a first power does NOT mean degree 1

For

\tfrac{d^2y}{dx^2} + \sin\!\big(\tfrac{dy}{dx}\big) = 0

, writing "degree 1" because

\tfrac{d^2y}{dx^2}

appears once is the trap. A derivative inside

\sin

\cos

\log

, or

e^{(\cdot)}

makes the degree UNDEFINED regardless of the visible power.

Order survives; only degree dies

"Degree undefined" never means "order undefined". Always still report the order — it is just the highest derivative present.

$e^{x + C}$ hides a constant, it does not add one

C_3 e^{x + C_4}

LOOKS like two constants but is just

B e^x

— one constant. Counting

C_4

separately over-states the order. Absorb every exponential shift before you count.

Only INDEPENDENT constants count

A sum like

C_1 + C_2

is a single free parameter. Two constants that can only ever appear as their sum contribute one to the order, not two.

Collapse constants BEFORE fixing the order

For all parabolas with axis parallel to Y,

(x-h)^2 = 4a(y-k)

has THREE independent constants

h, a, k

— so its ODE is order 3 (

\tfrac{d^3y}{dx^3} = 0

). Miscounting the constants sets the wrong order from the start.

A fixed point removes a constant

All lines through a fixed point have only the slope free (order 1), while all lines in the plane have slope AND intercept free (order 2). Read what is fixed before counting.

Eliminate the CONSTANT, not the known function

x^2 y = 4e^x + c

only

c

is arbitrary — the

e^x

is a fixed function that survives differentiation. Dropping

e^x

as if it were the constant gives the wrong equation. The correct ODE keeps the

4e^x

term.

Differentiate ONCE per constant — no more, no less

Ax^2 + By^2 = 1

has two constants, so it needs TWO differentiations to eliminate both (giving

xy y'' + x(y')^2 - yy' = 0

). Stopping after one differentiation leaves a constant behind.

Translate the geometry into the RIGHT free constants

"Centre on the X-axis and touching the Y-axis" fixes the centre as

(a,0)

with radius

|a|

— ONE free constant, giving an order-1 equation. Treating it as a general circle (two/three constants) inflates the order and the answer.

Mind the sign when substituting the eliminated constant

For circles through the origin centred on the X-axis, substituting

a = x + yy'

yields

y^2 = x^2 + 2xy\,y'

— a plus sign. Careless algebra flips it to

y^2 = x^2 - 2xy\,y'

, which is a different (wrong) option.

Convert parametric derivatives correctly

\dfrac{dy}{dx} \ne \dfrac{dy}{dt}

— you must divide by

\dfrac{dx}{dt}

. For

x = \sin t

\dfrac{dx}{dt} = \cos t

; skipping this factor is the most common error in find-

k

questions and gives the wrong constant.

Identify the conic from the SIMPLIFIED solution

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

only becomes

x^2 - y^2 = 1

(a hyperbola) AFTER applying the given point to fix

c

. Reading the conic type off the un-simplified, constant-carrying form is unreliable.

Variable-Separable Differential Equations

Learn this subtopic in the notes

The Separate-Then-Integrate Idea

Separable form and its solution

\dfrac{dy}{dx} = f(x)\,g(y) \;\Longrightarrow\; \int \dfrac{dy}{g(y)} = \int f(x)\,dx + c

$f(x)$ the x-only factor (integrated in x)
$g(y)$ the y-only factor (its reciprocal is integrated in y)
$c$ the single arbitrary constant of a first-order equation

Basic Separation and Integrating Both Sides

Standard integrals used after separating

\int \dfrac{dy}{y} = \log y + c,\qquad \int \dfrac{dx}{x^2} = -\dfrac{1}{x} + c

Applying an Initial Condition (Particular Solutions)

General → particular via the condition

y = \Phi(x, c),\qquad y(x_0) = y_0 \;\Rightarrow\; c = c_0 \;\Rightarrow\; y = \Phi(x, c_0)

Separables in Disguise — Logs and Exponential Right Sides

Exponentiate to separate; the eˣ(f + f′) trick

\log\!\Big(\tfrac{dy}{dx}\Big) = ax+by \;\Rightarrow\; e^{-by}\,dy = e^{ax}\,dx,\qquad \int e^x\big(f(x)+f'(x)\big)\,dx = e^x f(x) + c

Trigonometric-Product Separables

The log-integrals you reach for

\int \dfrac{\cos x}{1+\sin x}\,dx = \log(1+\sin x) + c,\qquad \int \tan y\,dy = -\log\cos y + c = \log\sec y + c

Rational Separables — arctan, arcsin, and Families of Circles

arctan and the circle-producing integral

\int \dfrac{dy}{1+y^2} = \tan^{-1}y + c,\qquad \int \dfrac{y\,dy}{\sqrt{k^2-y^2}} = -\sqrt{k^2-y^2} + c

Direct Integration — dy/dx = f(x) and Slope-of-Curve Problems

Pure x-side integration

\dfrac{dy}{dx} = f(x) \;\Longrightarrow\; y = \int f(x)\,dx + c

Common traps

One arbitrary constant, and add it at the integration step

Integrating both sides of a first-order equation gives ONE constant, not one per side.

\int e^y\,dy = \int e^x\,dx

e^y = e^x + c

, never

e^y + c_1 = e^x + c_2

. Dropping the constant, or writing two, is the classic separable-method slip.

You cannot divide by a factor that might be zero

To separate

\dfrac{dy}{dx} = f(x)\,g(y)

you divide by

g(y)

— but if

g(y)=0

for some

y=y_0

, that constant function

y=y_0

is a solution you would lose by dividing. Note any such

g(y)=0

branch before dividing.

Absorb the constant as $\log c$ , not $+c$ , when both sides are logs

When integration gives

\log y = 2\log x + \text{(const)}

, write the constant as

\log c

so the answer collapses to the clean family

y = cx^2

. Leaving it as

+c

blocks the tidy multiplicative form the options are written in.

$x\dfrac{dy}{dx} = 2y$ is a parabola family, not a linear one

It separates to

y = cx^2

— parabolas with vertex at the origin and axis along the Y-axis (since

x^2 = \tfrac1c\,y

). Reading it as

y = cx

(a line) or picking the X-axis parabola is the standard MHT-CET distractor pair.

Don't forget the $+c$ BEFORE applying the initial condition

The whole point of an IVP is to determine the constant from the condition — so you must carry

c

through the integration. Substituting the point before integrating, or dropping

c

, leaves you nothing to solve for and gives the wrong particular solution.

Watch the $\log$ → product conversion

\log(y+1) = -\log(2+\sin x) + c

becomes

(y+1)(2+\sin x) = k

(a PRODUCT, since the constant absorbs as

\log k

). Writing it as a sum, or keeping a stray minus outside, mis-fixes the constant and throws the final value.

Take logs / exponentials to unlock separation

An equation like

\log(dy/dx)=ax+by

looks non-separable until you exponentiate to

dy/dx = e^{ax}e^{by}

, which splits cleanly. Always test whether one rewrite makes the variables come apart before reaching for a heavier method.

Spot the $\int e^x(f+f')\,dx = e^x f$ pattern

On the x-side,

e^x(\sin x + \cos x) = e^x(f + f')

with

f = \sin x

, so its integral is

e^x\sin x

(NOT

e^x\cos x

). Missing this pattern — or picking

f=\cos x

— sends you to the wrong option; note that the y-side of

\frac{e^y(1+y\log y)}{y}

integrates to

e^y\log y

by the same trick with

f = \log y

Apply product-to-sum BEFORE trying to separate

\dfrac{dy}{dx} + \sin\tfrac{x+y}{2} = \sin\tfrac{x-y}{2}

does not separate as written. Convert the difference of sines:

\sin\tfrac{x-y}{2} - \sin\tfrac{x+y}{2} = -2\cos\tfrac{x}{2}\sin\tfrac{y}{2}

. Only then does

\csc\tfrac{y}{2}\,dy = -2\cos\tfrac{x}{2}\,dx

fall out, integrating to

\log\tan\tfrac{y}{4} = c - 2\sin\tfrac{x}{2}

Signs of the trig log-integrals

\int \dfrac{\sin y}{1+\cos y}\,dy = -\log(1+\cos y)

(a MINUS, because

\tfrac{d}{dy}(1+\cos y) = -\sin y

), while

\int \dfrac{\cos x}{1+\sin x}\,dx = +\log(1+\sin x)

. Dropping that minus turns the product answer

(1+\sin x)(1+\cos y)=c

into a wrong sum.

Write the arctan constant as $\tan^{-1}c$ , then use the subtraction formula

\tan^{-1}y = \tan^{-1}x + \tan^{-1}c

only collapses to

y-x = c(1+xy)

if you set the constant as

\tan^{-1}c

and apply

\tan^{-1}A - \tan^{-1}B = \tan^{-1}\tfrac{A-B}{1+AB}

. A bare

+c

leaves you stuck at

\tan^{-1}y - \tan^{-1}x = c

Identify the circle's centre-axis and radius carefully

For

\dfrac{dy}{dx} = \dfrac{\sqrt{1-y^2}}{y}

the solution

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

has FIXED radius 1 and centres on the X-axis. For

y\,dy = (a-x)\,dx

the radius is

\sqrt{a^2+2c}

— variable, centre

(a,0)

. Read which quantity is fixed vs variable before choosing the option.

Simplify the RHS before integrating

\dfrac{3e^{2x}+3e^{4x}}{e^x+e^{-x}}

looks like it needs a substitution, but it collapses to

3e^{3x}

after factoring — then

y = e^{3x}+c

in one line. Grinding the quotient without simplifying invites algebra errors.

Divide the polynomial before integrating a rational $f(x)$

For

\dfrac{dy}{dx} = \dfrac{x^2+4x-9}{x+2}

, do the division first:

(x+2) - \dfrac{13}{x+2}

. Integrating term-by-term gives the

\log|x+2|

piece cleanly; trying to integrate the raw quotient is where students lose the log term.

Homogeneous and Reducible Differential Equations

Learn this subtopic in the notes

Recognizing a Homogeneous Differential Equation

Homogeneity test

f(\lambda x, \lambda y) = \lambda^n f(x,y) \quad\Longleftrightarrow\quad \dfrac{dy}{dx} = g\!\left(\dfrac{y}{x}\right)

$n$ degree of homogeneity — for the DE, P and Q must share it
$g(y/x)$ the right side collapses to a function of the ratio alone

The y = vx Substitution

Homogeneous substitution

y = vx \;\Rightarrow\; \dfrac{dy}{dx} = v + x\dfrac{dv}{dx} \;\Rightarrow\; \dfrac{dv}{g(v) - v} = \dfrac{dx}{x}

$v$ the ratio y/x, itself a function of x
$v + x dv/dx$ the derivative dy/dx after the product rule — never just dv/dx

Worked Homogeneous Equations and Initial Conditions

Particular solution from an IC

\text{integrate} \;\to\; \text{back-substitute } v=\tfrac{y}{x} \;\to\; \text{plug the point to find } c

Homogeneous Curves Through a Point (Trig Ratio Slopes)

Trig-ratio homogeneous slopes

\dfrac{dy}{dx} = \dfrac{y}{x} + h\!\left(\dfrac{y}{x}\right) \;\xrightarrow{\,y=vx\,}\; \dfrac{dv}{h(v)} = \dfrac{dx}{x}

$h(y/x)$ trig function of the ratio; the bare y/x cancels

Log-Form Homogeneous Equations (v = y/x)

The log-form integral

\int\dfrac{dv}{v\,\log v} = \log(\log v) \;\Rightarrow\; \log\dfrac{y}{x} = c\,x

$t = log v$ substitution making $dt = dv/v$ , so the integrand becomes $dt/t$

Reducible to Separable via v = x + y (or v = ax + by)

Linear-argument substitution

v = x + y \;\Rightarrow\; \dfrac{dv}{dx} = 1 + \dfrac{dy}{dx} \;\Rightarrow\; \dfrac{dv}{1 + f(v)} = dx

$v = x + y$ collapses the repeated pair into one variable
$dv/dx = 1 + dy/dx$ the +1 from differentiating x — never omit it

Common traps

A stray constant breaks homogeneity

x + y + 1

is NOT homogeneous — the

+1

does not scale with

\lambda

. Likewise

\cos x + \sin y

fails because

x

and

y

enter separately, not through the ratio

y/x

. Only when EVERY term scales by the same power of

\lambda

is the function homogeneous.

Same degree top and bottom is the fast check

For a quotient

\dfrac{P}{Q}

, you rarely need the full scaling test — just confirm

P

and

Q

have the SAME degree.

\dfrac{x^2 + 2y^2}{xy}

: both degree 2, so homogeneous. If the degrees differ,

y = vx

will not clean it up.

dy/dx is v + x·dv/dx, not just dv/dx

The single most common homogeneous-substitution error is writing

\dfrac{dy}{dx} = \dfrac{dv}{dx}

. Since

y = vx

is a PRODUCT, the product rule gives

\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}

. Miss the

v

term and every subsequent step is wrong.

Substitute v = y/x back at the very end

After integrating in

v

and

x

, students often leave the answer in

v

. The options are always in

x

and

y

, so you MUST replace

v

with

\dfrac{y}{x}

in the final line — e.g.

\sin^{-1} v = \log x + c

becomes

\sin^{-1}\dfrac{y}{x} = \log x + c

Use the initial condition only after back-substituting

Plug the given point in

x, y

AFTER you have replaced

v

with

y/x

. Trying to apply

y(1)=0

while still in

v

(where

v = y/x

may be

0/1 = 0

but the equation is mid-integration) is where sign and constant errors creep in.

Track the sign of g(v) − v

The separable step needs

x\dfrac{dv}{dx} = g(v) - v

. For

(y^2 - x^2)\,dx = xy\,dy

g(v) - v = \dfrac{v^2 - 1}{v} - v = -\dfrac{1}{v}

— a MINUS sign that carries into the final answer as

+y^2

after moving terms. Dropping it flips the whole solution.

The bare y/x cancels — don't integrate it

v + x\dfrac{dv}{dx} = v + h(v)

, the

v

on both sides cancels. Students who keep it and try to integrate

v

as well produce an extra

\log

term. Only the trig part

h(v)

survives to the separable equation.

Feed the initial point to find c — always

These are 'curve through a point' questions, so the constant is fixed by the point, not left as

c

. Forgetting to substitute

(1, \pi/4)

etc. leaves you with a general-solution decoy rather than the specific curve the options list.

The integral is log(log v), not log v

\displaystyle\int\dfrac{dv}{v\log v}

uses

t = \log v

so it becomes

\int\dfrac{dt}{t} = \log(\log v)

. Stopping at

\log v

(as if the denominator were just

v

) gives the wrong final relation — you would miss the outer log and get

v = cx

instead of

\log v = cx

It's cx, not cy — check which variable the constant multiplies

The correct answer is

\log\dfrac{y}{x} = c x

(the

dx/x

side integrated to

\log x

). Decoys swap it to

cy

or flip the ratio to

\log\dfrac{x}{y}

. Track which side became

\log x

so you land on

cx

with the ratio

y/x

v = x + y gives dv/dx = 1 + dy/dx — keep the +1

Differentiating

v = x + y

yields

\dfrac{dv}{dx} = 1 + \dfrac{dy}{dx}

, so

\dfrac{dy}{dx} = \dfrac{dv}{dx} - 1

. Dropping the

1

(writing

\dfrac{dy}{dx} = \dfrac{dv}{dx}

) is the signature error and makes the equation fail to separate.

For v = ax + by, the coefficient rides through

For

\dfrac{dy}{dx} = (x + 9y)^2

, use

u = x + 9y

\dfrac{du}{dx} = 1 + 9\dfrac{dy}{dx}

. The integral picks up a

\dfrac13

factor:

\dfrac13\tan^{-1}(3u) = x + c

. Forgetting the

9

(or the resulting

\tfrac13

) loses the constant when you fit an initial condition.

Substitute v = x + y back at the end

Just like y = vx, the final line must be in

x, y

\tan^{-1}v = x + c

is only finished once you write

\tan^{-1}(x + y) = x + c

. Leaving

v

in the answer will not match any option.

Linear Differential Equations — the Integrating Factor

Learn this subtopic in the notes

Recognizing the Standard Linear Form

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

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

$IF$ the integrating factor e to the integral of P
$c$ the single arbitrary constant, fixed by an initial condition

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

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

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

Common traps

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.

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.

$e^{\log f(x)} = f(x)$ — 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.

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.

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.

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.

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.

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.

Growth, Decay, and Continuous Models

Learn this subtopic in the notes

The Modelling Step — Rate Proportional to Quantity

Rate proportional to quantity

\dfrac{dP}{dt} = kP \quad\Longrightarrow\quad \log P = kt + c

$P$ the changing quantity (mass, population, amount)
$k$ proportionality constant — positive for growth, negative for decay
$t$ time

The Exponential Solution P = P0 e^{kt} and Finding k

Exponential growth/decay solution

P = P_0\,e^{kt}, \qquad k = \dfrac{1}{t_1}\log\!\dfrac{P_1}{P_0}

$P_0$ value at $t=0$
$k$ rate constant, found from a second data point

Population and Bacteria — Doubling Time and Percentage Growth

Doubling growth

e^{kT} = 2 \quad\Longrightarrow\quad P = P_0\,2^{\,t/T}

$T$ doubling time
$t/T$ number of doubling periods elapsed

Radioactive Decay and Half-Life

Half-life rate constant

k = \dfrac{\log 2}{h}, \qquad m = m_0\left(\tfrac12\right)^{t/h}

$h$ half-life — time to lose half the mass
$m_0$ initial mass at $t=0$

Continuous Compounding of Money

Continuous compounding

A = P\,e^{rt}, \qquad e^{rT} = 2 \text{ if the principal doubles in } T

$P$ principal invested at $t=0$
$r$ annual rate as a decimal ( $x\% \to x/100$ )

Moisture Loss and General First-Order Rate Models

Fraction-lost time (pure decay)

\log\dfrac{P}{N} = -kt, \qquad k = \log 2 \text{ if half is lost in unit time}

$N$ initial content at $t=0$
$P/N$ fraction remaining

Special-Rate Models — Square-Root and Surface-Area Decay

Square-root and surface-area models

\dfrac{dx}{dt} = -k\sqrt{x} \Rightarrow 2\sqrt{x} = -kt + c; \qquad \dfrac{dV}{dt} = -kS \Rightarrow \dfrac{dr}{dt} = -k

$2\sqrt{x} = -kt+c$ the integrated square-root law — linear in $\sqrt{x}$
$dr/dt = -k$ surface-area evaporation ⇒ radius shrinks at a constant rate

Common traps

Decay carries a negative sign

'Rate of reduction / decay / loss' means

\dfrac{dP}{dt} = -kP

, not

+kP

. Dropping the minus makes the quantity grow — the answer then heads the wrong way. Keep

k>0

and put the sign in front explicitly.

'Proportional to' is not 'equal to'

'Rate proportional to P' introduces a constant

k

; it is

\dfrac{dP}{dt}=kP

, not

\dfrac{dP}{dt}=P

. You must determine

k

later from a data point — never assume

k=1

Cancel $P_0$ by dividing — don't solve for k first

For 'grows from A to B in time

T

, find value after another

T

', you never need

k

P_0

numerically.

e^{kT} = B/A

, and after

2T

the value is

A(B/A)^2

. Chasing

k = \tfrac1T\log(B/A)

and re-exponentiating wastes time and invites arithmetic slips.

The extra time is measured from the start

'In ANOTHER 40 years' when the first stage was already 40 years means the total elapsed time is

t=80

, so

P=P_0 e^{80k}

. Reading it as

t=40

again halves the exponent and drops a doubling.

'Doubles' means the ratio is 2, not '+2'

'The population doubles' sets

\dfrac{P}{P_0}=2

(so

e^{kT}=2

) — it does not mean

P = P_0 + 2

. Similarly 'triples' is a factor 3. Always translate the word into a multiplying ratio.

Turn a percentage into a factor before touching k

A '20% increase' is a factor of

\tfrac65

(i.e.

1.2

), NOT

0.20

. Write

e^{kT} = 1 + \tfrac{p}{100}

; using

0.20

instead of

1.20

corrupts

k

and every later value.

The initial decay rate is negative

Because mass decreases,

\dfrac{dm}{dt}=-km_0 = -\dfrac{m_0\log 2}{h}

carries a MINUS sign. The positive-looking distractor

\dfrac{m_0}{h}\log 2

is the classic trap — decay rates are negative.

Count half-lives only when time is a whole multiple

The shortcut

m = m_0(\tfrac12)^{t/h}

is easiest when

t/h

is an integer (3 half-lives, etc.). If

t

is not a clean multiple (e.g. 27 g → 8 g in 3 h, then 1 more hour), fall back to

m = m_0 e^{-kt}

with

e^{-k}

found from the given step.

Convert the % rate to a decimal

10% per year is

r = 0.10

, not

r = 10

. Using 10 blows the exponent up by a factor of 100. Write

r = x/100

every time.

Continuous compounding uses $e^{rt}$ , not $(1+r)^t$

The word 'continuously' means

A = Pe^{rt}

. Reaching for the annual-compound formula

A = P(1+r)^t

or simple interest is the intended distractor — it gives a different (wrong) number.

'99% lost' means the fraction LEFT is 0.01

Losing 99% leaves 1%: use

P/N = 0.01

, giving

\log 100

on top. Plugging in 0.99 (the fraction lost) instead of 0.01 (the fraction remaining) inverts the ratio and the answer.

The constant term needs factoring before you separate

\dfrac{dP}{dt}=0.5P-450

is NOT

\dfrac{dP}{P}=0.5\,dt

. Factor to

0.5(P-900)

first, so the variable that integrates cleanly is

P-900

, not

P

$\int \dfrac{dx}{\sqrt x} = 2\sqrt x$ , not $\log\sqrt x$

The square-root rate separates to

x^{-1/2}\,dx

, whose integral is

2\sqrt x

— a power-rule integral, NOT a logarithm. Reflexively writing

\log

(as for

dx/dt = -kx

) is the number-one error in these problems.

Surface-area evaporation makes the RADIUS linear

For a raindrop, 'proportional to surface area' plus

V=\tfrac43\pi r^3

forces

\dfrac{dr}{dt}=-k

, so

r = c - kt

is linear in

t

. Trying to make the volume or radius exponential misses the cancellation of

4\pi r^2

Newton's Law of Cooling

Learn this subtopic in the notes

The Cooling Model — Rate Proportional to Temperature Excess

Newton's law of cooling

\dfrac{d\theta}{dt} = -k(\theta - \theta_s), \qquad k > 0

$\theta$ temperature of the body at time t
$\theta_s$ surrounding (ambient) temperature — constant
$k$ positive cooling constant

Solving the Cooling Equation — Log Form and Exponential Form

Log form and its exponential solution

\log(\theta - \theta_s) = -kt + c \qquad\Longleftrightarrow\qquad \theta - \theta_s = (\theta_0 - \theta_s)\,e^{-kt}

$\theta_0$ initial temperature of the body (at t = 0)
$c$ constant of integration $= \log(\theta_0 - \theta_s)$

Two-Stage Cooling — Fix the Rate, Then Predict

Ratio of excesses over two intervals

\dfrac{\theta_2 - \theta_s}{\theta_1 - \theta_s} = e^{-k(t_2 - t_1)}

$\theta_1, \theta_2$ temperatures at times t₁, t₂
$\theta_s$ surrounding temperature — subtracted from both

The (Ratio)ⁿ Shortcut for Equal Time-Steps

Geometric decay of the excess over n equal steps

\theta_n - \theta_s = (\theta_0 - \theta_s)\,r^{\,n}, \qquad r = e^{-k\Delta t}

$r$ one-step ratio of excesses (constant for equal Δt)
$n$ number of equal time-steps = total time ÷ Δt

Common traps

Always work with the excess $\theta - \theta_s$ , not $\theta$

The quantity that obeys clean exponential decay is the temperature EXCESS

\theta - \theta_s

, not the raw temperature

\theta

. A body at

90^\circ

in a

20^\circ

room does not decay toward

0

— it decays toward

20

. Subtract the surrounding temperature before doing anything else.

The minus sign and $k>0$ together mean cooling

Write the model as

\dfrac{d\theta}{dt} = -k(\theta - \theta_s)

with

k>0

. The minus sign is what makes a hot body cool. Absorbing the sign into

k

(letting

k<0

) and then also writing a minus is a common double-negative slip.

Take the log of the EXCESS, not the temperature

The integral of

\dfrac{d\theta}{\theta - \theta_s}

\log(\theta - \theta_s)

, never

\log\theta

. Feeding the raw temperature into the log (writing

\log 370

instead of

\log 80

) is the most common wrong start on these questions.

Here $\log$ means natural log

Throughout this model

\log = \log_e

. The base cancels out anyway because you always take a ratio of two logs (or a ratio of excesses), so you never actually need its numerical value — but keep the notation consistent.

Subtract the surrounding temperature BEFORE forming the ratio

The ratio that stays constant is of the EXCESSES, not the raw temperatures. For

80\to50

in a

25^\circ

room the multiplier is

\dfrac{50-25}{80-25} = \dfrac{25}{55}

, not

\dfrac{50}{80}

. Using the bare temperatures is the number-one error and gives a wrong answer every time.

Match the exponent to the number of equal intervals

If the first interval is

30

min and the target time is

60

min, that is

60/30 = 2

steps, so the multiplier is SQUARED. For

20

minutes after a

5

-minute interval it is

20/5 = 4

steps — the ratio to the FOURTH power. Miscounting the number of steps changes the exponent.

Equal steps ⇒ geometric ratio of the EXCESSES

The excess is multiplied by the same ratio each equal step, so it decays geometrically — NOT linearly. Between

100\to60

the drop was

40^\circ

; the next equal step is not another

40^\circ

but a HALVING of the excess (

80\to40\to20\to10\to5

). Treating cooling as a constant per-step drop overshoots badly.

Count n as total time ÷ interval, then raise the ratio to that power

The exponent

n

is the number of equal intervals, not the number of minutes. For a

15

-min interval and a

60

-min target,

n = 60/15 = 4

, so use

r^4

. Plugging

n = 60

(the minutes) instead of

4

(the steps) is a fatal slip.

Differential Equations formulas

Order, Degree, Formation, and Verification

Differential Equation Terminology

Order = Order of the Highest Derivative Present

Degree = Power of the Highest Derivative After Clearing Radicals

When Degree Is Undefined (Derivative Inside a Transcendental)

Collapse Redundant Arbitrary Constants Before Counting Order

Formation: n Independent Constants ⇒ Order-n Differential Equation

Forming the Differential Equation of a Curve Family

Forming the Differential Equation of Circles and Parabolas

Verifying a Solution and Identifying Its Family

Common traps

Variable-Separable Differential Equations

The Separate-Then-Integrate Idea

Basic Separation and Integrating Both Sides

Applying an Initial Condition (Particular Solutions)

Separables in Disguise — Logs and Exponential Right Sides

Trigonometric-Product Separables

Rational Separables — arctan, arcsin, and Families of Circles

Direct Integration — dy/dx = f(x) and Slope-of-Curve Problems

Common traps

Homogeneous and Reducible Differential Equations

Recognizing a Homogeneous Differential Equation

The y = vx Substitution

Worked Homogeneous Equations and Initial Conditions

Homogeneous Curves Through a Point (Trig Ratio Slopes)

Log-Form Homogeneous Equations (v = y/x)

Reducible to Separable via v = x + y (or v = ax + by)

Common traps

Linear Differential Equations — the Integrating Factor

Recognizing the Standard Linear Form

The Integrating Factor and the Solution Formula

Simple Integrating Factors

Tricky Integrating Factors

Linear in x — Swap the Roles of x and y

Bernoulli Equations — Substitute to Linearize

Exact Equations by d(·)-Grouping

Direct Integration and Reduction of Order

Common traps

Growth, Decay, and Continuous Models

The Modelling Step — Rate Proportional to Quantity

The Exponential Solution P = P0 e^{kt} and Finding k

Population and Bacteria — Doubling Time and Percentage Growth

Radioactive Decay and Half-Life

Continuous Compounding of Money

Moisture Loss and General First-Order Rate Models

Special-Rate Models — Square-Root and Surface-Area Decay

Common traps

Newton's Law of Cooling

The Cooling Model — Rate Proportional to Temperature Excess

Solving the Cooling Equation — Log Form and Exponential Form

Two-Stage Cooling — Fix the Rate, Then Predict

The (Ratio)ⁿ Shortcut for Equal Time-Steps

Common traps

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