NDA Maths · Differential Equations

Solving ODEs — Separable, Substitution, Integrating Factor

First-order ODEs are solved by a small toolkit: separate the variables, reduce a tangled one by substitution, or use an integrating factor for the linear case — then fit any initial condition.

All Differential Equations notes NDA Maths strategy

Why this matters

29 PYQs, the biggest subtopic and the home of most HARD questions. The whole skill is reading the equation's shape to pick the method: separable if the variables come apart, a substitution v = x ± y if they don't, an integrating factor if it is linear. Applications add growth/decay and particle-motion initial-value problems.

Test yourself — a quick 13-question Differential Equations recall quiz, instant score.

Concept 1 of 4: Separation of variables

The first thing to try: get all the y's (with dy) on one side and all the x's (with dx) on the other, then integrate both sides. Most first-order NDA equations either separate directly or do so after a small rewrite (taking logs, or recognising an exponential).

Definition

The separable method:

Write the equation as $g(y)\,dy = f(x)\,dx$ , then integrate both sides — don't forget the single arbitrary constant.
Exponentials separate: $\ln\big(\tfrac{dy}{dx}\big) = ax+by \Rightarrow \tfrac{dy}{dx}=e^{ax}e^{by} \Rightarrow e^{-by}\,dy=e^{ax}\,dx$ .
A constant derivative integrates trivially: $\cos\big(\tfrac{dy}{dx}\big)=p \Rightarrow \tfrac{dy}{dx}=\cos^{-1}p \Rightarrow y = x\cos^{-1}p + C$ .

Separation of variables

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

Worked example

Solve

\dfrac{dy}{dx} = (1+y^2)

— i.e.

dy = (1+y^2)\,dx

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Apr 2021 · Q99Moderate

Example 1 · Differential Equations · Solving and Verifying ODEs — Separable, IVP, and Applications

What is the solution of the following differential equation?

\ln\left(\frac{dy}{dx}\right)+y=x

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.

Drill 9 more on separation of variables

Concept 2 of 4: Reducible to separable by substitution

When x and y appear glued together as a combination like

x+y

y-x

, substitute a new variable for that combination. The substitution turns a tangled equation into a separable one in the new variable. Also learn to spot exact differentials like

d(xy)

and

d(x/y)

Definition

Two reduction tricks:

Substitute the combination: if the equation depends on $x+y$ (or $y-x$ ), set $v=x+y$ , so $\tfrac{dv}{dx}=1+\tfrac{dy}{dx}$ , and the equation becomes separable in $v$ .
Recognise exact differentials: $x\,dy+y\,dx=d(xy)$ ; $\dfrac{x\,dy-y\,dx}{y^2}=d\!\big(\tfrac{x}{y}\big)$ ; $x\,dx+y\,dy=\tfrac12 d(x^2+y^2)$ .

Antiderivatives the substitution step lands on (stated here so the last line is not a guess): $\int\sec v\,dv=\ln|\sec v+\tan v|+C$ , $\int\tan v\,dv=\ln|\sec v|+C=-\ln|\cos v|+C$ , $\int\cot v\,dv=\ln|\sin v|+C$ , $\int\csc v\,dv=\ln|\csc v-\cot v|+C$ . And the half-angle route for a cosine ratio: $\dfrac{1-\cos x}{1+\cos x}=\tan^2\tfrac x2=\sec^2\tfrac x2-1$ , so $\int\dfrac{1-\cos x}{1+\cos x}\,dx=2\tan\tfrac x2-x+C$ — the equation $\dfrac{dy}{dx}=\dfrac{1-\cos x}{1+\cos x}$ is solved in one line.

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)

Worked example

Solve

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

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

NDA · Sep 2018 · Q99Hard

Example 2 · Differential Equations · Solving and Verifying ODEs — Separable, IVP, and Applications

What is the solution of the differential equation

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

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.

Drill 8 more on reducible to separable by substitution

Concept 3 of 4: Linear equations and the integrating factor

A first-order LINEAR equation — y appears only to the first power, multiplied by functions of x — is solved by multiplying through by an integrating factor that turns the left side into the derivative of a product. Sometimes the equation is linear in x instead of y, so flip it to dx/dy first.

Definition

The integrating-factor method:

Standard form: $\dfrac{dy}{dx} + P(x)\,y = Q(x)$ .
Integrating factor $\mu = e^{\int P\,dx}$ ; then $\dfrac{d}{dx}(\mu y) = \mu Q$ , so $\mu y = \int \mu Q\,dx + c$ .
If the equation is linear in $x$ , use $\dfrac{dx}{dy} + P(y)\,x = Q(y)$ with $\mu = e^{\int P\,dy}$ .
Bernoulli $\dfrac{dy}{dx}+Py = Qy^{n}$ : substitute $v=y^{1-n}$ to make it linear.

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}

Worked example

Solve

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

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

NDA · Apr 2017 · Q88Moderate

Example 3 · Differential Equations · Solving and Verifying ODEs — Separable, IVP, and Applications

What is the general solution of the differential equation

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

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.

Drill 2 more on linear equations and the integrating factor

Concept 4 of 4: Initial-value problems and growth/decay

An initial value pins down the arbitrary constant: solve the ODE generally, then plug in the given point. The classic application is exponential growth and decay — anything whose rate of change is proportional to its current amount. Verifying a proposed solution just means substituting it back.

Definition

Applications and verification:

Growth/decay: $\dfrac{dy}{dt} = ky$ has solution $y = y_0 e^{kt}$ ( $k>0$ growth, $k<0$ decay — radioactivity, cooling).
IVP: find the general solution, then use the condition (e.g. $y(0)=y_0$ ) to fix the constant.
Verify a candidate by substituting it into the ODE; a factored equation like $(y')^2 - x\,y' = 0$ splits into $y'=0$ and $y'=x$ , giving two solution families.

Growth/decay and order

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

Worked example

A particle starts at the origin with

\dfrac{dx}{dt} = x + 1

. How long to travel 24 m?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

NDA · Apr 2021 · Q77Moderate

Example 4 · Differential Equations · Solving and Verifying ODEs — Separable, IVP, and Applications

A particle starts from origin with a velocity (in m/s) given by the equation

\frac{dx}{dt}=x+1

. The time (in second) taken by the particle to traverse a distance of 24 m is

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.

Drill 8 more on initial-value problems and growth/decay

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 (4)

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

Watch out for (7)

Take logs / exponentials to unlock separation
→ Separation of variables
One arbitrary constant only — and add it at the integration step
→ Separation of variables
Spot the glued combination first
→ Reducible to separable by substitution
Memorise the exact differentials with the right sign
→ Reducible to separable by substitution
Put the equation in STANDARD form before reading off P
→ Linear equations and the integrating factor
Bernoulli substitution is $v=y^{1-n}$ , not $y^{n-1}$
→ Linear equations and the integrating factor
Apply the initial condition to the GENERAL solution
→ Initial-value problems and growth/decay

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