NDA Mathematics · Formula sheet

Differential Equations formulas

5 formulas and 13 common traps for NDA Mathematics Differential Equations, grouped by subtopic.

Order, Degree and Solutions

Order and degree of a differential equation

Order and degree

\text{order} = \text{order of the highest derivative present} \qquad \text{degree} = \text{power of the highest-order derivative, after making it polynomial in the derivatives}

Common traps

Clear fractional powers BEFORE reading the degree

The degree is NOT the fractional exponent you see. For

\big(2-(y')^2\big)^{0.6} = y''

, raise to the 5th power to get

\big(2-(y')^2\big)^3 = (y'')^5

: the degree is 5, not 0.6. Make it polynomial first.

Degree is undefined when a derivative sits inside a transcendental

For

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

the order is 2 but the degree does NOT exist — you can never make it polynomial in

\tfrac{dy}{dx}

. Writing "degree 1" because you see a first power is the trap; a derivative inside

\sin

\cos

\ln

e^{(\cdot)}

kills the degree.

Count INDEPENDENT constants

y=A[\sin(x+C)+\cos(x+C)]

looks like two constants, but it collapses to

B\sin(x+D)

— still two independent constants, so order 2 (giving

y''+y=0

). Combine first; constants that merge don't each count.

Forming an ODE from a Family of Curves

Learn this subtopic in the notes

Common traps

Differentiate as many times as there are constants

A one-constant family needs one differentiation (order 1); a two-constant family like

y^2=4a(x-b)

needs two (order 2, giving

yy''+(y')^2=0

). Differentiating too few times leaves a constant stranded in the answer.

The order of the resulting ODE equals the number of constants

Before you differentiate, the family

y=Ae^{2x}+Be^{-3x}

has 2 arbitrary constants, so the eliminated ODE is order 2 — guaranteed. Reading the order off the highest derivative you happen to reach mid-working (or stopping early) gives the wrong order; count the independent constants first and that IS the order.

A circle needs equal squared-term coefficients

After integrating,

\frac{a}{2}x^2 - \frac{b}{2}y^2

is a circle only if those coefficients match in magnitude (giving

a=-b

); otherwise it is an ellipse or hyperbola. Don't assume any separable solution is a circle.

Solving ODEs — Separable, Substitution, Integrating Factor

Learn this subtopic in the notes

Separation of variables

\int \frac{dy}{g(y)} = \int f(x)\,dx + C

Reducible to separable by substitution

Exact differentials to recognise

x\,dy + y\,dx = d(xy) \qquad \frac{x\,dy - y\,dx}{y^2} = d\!\left(\frac{x}{y}\right) \qquad x\,dx + y\,dy = \tfrac12\,d(x^2+y^2)

Linear equations and the integrating factor

Integrating factor

\mu = e^{\int P(x)\,dx},\qquad \frac{d}{dx}(\mu y) = \mu Q,\qquad y\,\mu = \int Q\,\mu\,dx + c,\qquad \frac{dx}{dy}+P(y)\,x=Q(y):\ \mu=e^{\int P(y)\,dy},\qquad \frac{dy}{dx}+Py=Qy^{n}:\ v=y^{1-n}

Initial-value problems and growth/decay

Growth/decay and order

\frac{dN}{dt} = kN \;\Rightarrow\; N = N_0 e^{kt} \qquad \text{number of arbitrary constants} = \text{order}

Common traps

Take logs / exponentials to unlock separation

Equations like

\ln(dy/dx)=ax+by

look non-separable until you exponentiate:

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

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

One arbitrary constant only — and add it at the integration step

After separating, integrating BOTH sides produces just ONE arbitrary constant for a first-order equation, not one per side.

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

gives

e^{y}=e^{x}+c

, never

e^{y}+c_1=e^{x}+c_2

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

Spot the glued combination first

If you cannot separate directly, look for

x+y

y-x

appearing as a unit — that is the signal to substitute

v

for it. Trying to force separation without the substitution leads nowhere.

Memorise the exact differentials with the right sign

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 the denominator — e.g. writing

\tfrac{y\,dx-x\,dy}{y^2}=d(x/y)

— gives

-d(x/y)

and the wrong answer.

Put the equation in STANDARD form before reading off P

The integrating factor is

e^{\int P\,dx}

only when the equation is written as

\dfrac{dy}{dx}+Py=Q

with coefficient

+1

\tfrac{dy}{dx}

. For

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

you must first divide by

x

to get

P=-\tfrac1x

(so

\mu=\tfrac1x

); using

P=-1

or forgetting the sign gives the wrong factor.

Bernoulli substitution is $v=y^{1-n}$ , not $y^{n-1}$

To linearise

\dfrac{dy}{dx}+Py=Qy^{n}

, substitute

v=y^{1-n}

(exponent

1-n

). For

n=2

that is

v=y^{-1}

, NOT

v=y^{1}

. Getting the exponent backwards (

y^{n-1}

) flips the sign of the resulting linear equation.

Apply the initial condition to the GENERAL solution

Solve fully (keeping the arbitrary constant) BEFORE substituting the initial value. Plugging the condition in too early — before integrating — loses the constant you are trying to determine.

Differential Equations formulas

Order, Degree and Solutions

Order and degree of a differential equation

Common traps

Forming an ODE from a Family of Curves

Common traps

Solving ODEs — Separable, Substitution, Integrating Factor

Reducible to separable by substitution

Linear equations and the integrating factor

Initial-value problems and growth/decay

Common traps

More NDA Mathematics formula sheets