Calculus IUnit 412 min read

First‑Order Differential Equations – Types, Solution Methods & Applications

Unit 4 of Calculus I introduces first‑order differential equations, covering definitions, classification, solution techniques (separable, linear, exact, Bernoulli, homogeneous), worked examples, and real‑world uses in Nepalese apps and industries.

Padhai · Based on the official syllabus and past papers
Updated 5 Oct 2026

Key points

Recognize the five standard forms of first‑order ODEs and choose the appropriate solving technique.
Master the integrating‑factor method for linear equations and the substitution for Bernoulli equations.
Verify exactness and construct an integrating factor when needed.
Translate real phenomena (population growth, loan interest, queue dynamics) into first‑order ODEs and solve them.
Use slope fields and phase lines to interpret qualitative behavior of solutions.

1. What is a First‑Order Differential Equation?

A first‑order ordinary differential equation (ODE) relates an unknown function $y (x)$ to its first derivative $y^{'} (x)$ :

$F (x, y, y^{'}) = 0.$

If the equation can be written explicitly as

$y^{'} = f (x, y),$

it is called explicit; otherwise it is implicit. The solution is a family of curves $y (x) = C$ that satisfy the relation for all $x$ in an interval.

2. Classification of First‑Order ODEs

Type	Standard Form	Typical Substitution / Technique	Example
Separable	$y^{'} = g (x) h (y)$	Separate variables: $\frac{d y}{h ( y )} = g (x) d x$	$y^{'} = x y$
Linear	$y^{'} + P (x) y = Q (x)$	Integrating factor $μ (x) = e^{\int P (x) d x}$	$y^{'} - \frac{2}{x} y = x^{2}$
Exact	$M (x, y) d x + N (x, y) d y = 0$ with $M_{y} = N_{x}$	Find potential function $Φ$ such that $Φ_{x} = M, Φ_{y} = N$	$(2 x y + 3) d x + (x^{2} + 4 y) d y = 0$
Bernoulli	$y^{'} + P (x) y = Q (x) y^{n}$ ( $n \neq = 0, 1$ )	Substitute $v = y^{1 - n}$ → linear in $v$	$y^{'} + y = y^{2}$
Homogeneous (in $x, y$ )	$y^{'} = f (\frac{y}{x})$	Set $v = y / x$ → $y = v x$	$y^{'} = \frac{x + y}{x - y}$

3. Solution Techniques

3.1 Separable Equations

Write as $\frac{d y}{h ( y )} = g (x) d x$ .
Integrate both sides.
Solve for $y$ (if possible) and apply the initial condition.

Worked Example 1 – Bacterial Growth

The growth rate of a bacterial culture is proportional to its current size:

$\frac{d B}{d t} = k B, B (0) = 1000.$

Separate: $\frac{d B}{B} = k d t$ .

Integrate: $ln ∣ B ∣ = k t + C$ .

Exponentiate: $B = C_{1} e^{k t}$ . Use $B (0) = 1000$ → $C_{1} = 1000$ .

Final solution:

$B (t) = 1000 e^{k t} .$

Exponential bacterial growth for different rate constants k

3.2 Linear Equations

Given $y^{'} + P (x) y = Q (x)$ :

Compute integrating factor $μ (x) = e^{\int P (x) d x}$ .
Multiply the whole equation by $μ (x)$ .
Left side becomes $\frac{d}{d x} (μ y)$ .
Integrate and solve for $y$ .

Worked Example 2 – Loan Interest (eSewa)

A micro‑loan of NPR 10,000 accrues interest at a continuous rate of 12 % per year, while the borrower repays at a constant rate of NPR 1500 per month (≈ NPR 18000 yr $^{- 1}$ ). Let $A (t)$ be the outstanding amount (years).

$\frac{d A}{d t} = 0.12 A - 18000, A (0) = 10000.$

Here $P (x) = 0.12, Q (x) = - 18000$ .

Integrating factor: $μ (t) = e^{0.12 t}$ .

Multiply: $\frac{d}{d t} (e^{0.12 t} A) = - 18000 e^{0.12 t}$ .

Integrate:

$e^{0.12 t} A = - \frac{18000}{0.12} e^{0.12 t} + C$ $A (t) = - 150000 + C e^{- 0.12 t} .$

Apply $A (0) = 10000$ :

$10000 = - 150000 + C \Rightarrow C = 160000.$

Solution:

$A (t) = 160000 e^{- 0.12 t} - 150000 .$

The loan is fully repaid when $A (t) = 0$ :

$0 = 160000 e^{- 0.12 t} - 150000 \Rightarrow e^{- 0.12 t} = \frac{150000}{160000} = 0.9375$ $t = - \frac{ln 0.9375}{0.12} \approx 0.54 yr \approx 6.5 months .$

Outstanding loan balance vs. time (NPR)

3.3 Exact Equations

An equation $M d x + N d y = 0$ is exact if $\partial M / \partial y = \partial N / \partial x$ .

Procedure:

Verify exactness.
Find $Φ (x, y)$ such that $Φ_{x} = M$ .
Integrate $Φ_{x}$ w.r.t. $x$ and add a “function of $y$ ”.
Differentiate $Φ$ w.r.t. $y$ and match with $N$ to determine the missing function.
Implicit solution: $Φ (x, y) = C$ .

Worked Example 3 – Daraz Order Queue

Consider the differential relation for the number of pending orders $Q (t)$ and the rate of new orders $r (t) = 5 - 0.2 Q$ :

$\frac{d Q}{d t} + 0.2 Q = 5.$

Rewrite as $d Q + 0.2 Q d t = 5 d t$ → $M (Q, t) = 0.2 Q - 5, N (Q, t) = 1$ .

Check exactness:

$\frac{\partial M}{\partial t} = 0, \frac{\partial N}{\partial Q} = 0.$

Since they are equal, the equation is exact.

Potential function:

$Φ (Q, t) = \int M d Q = \int (0.2 Q - 5) d Q = 0.1 Q^{2} - 5 Q + g (t) .$

Differentiate w.r.t. $t$ :

$Φ_{t} = g^{'} (t) = N = 1 \Rightarrow g (t) = t + C .$

Implicit solution:

$0.1 Q^{2} - 5 Q + t = C .$

If initially $Q (0) = 0$ , then $C = 0$ . Solving for $Q (t)$ gives the queue size at any time.

Order queue size over time

3.4 Bernoulli Equations

Form: $y^{'} + P (x) y = Q (x) y^{n}$ .

Steps:

Divide by $y^{n}$ (if $n \neq = 0$ ).
Set $v = y^{1 - n}$ → $v^{'} = (1 - n) y^{- n} y^{'}$ .
Obtain a linear ODE in $v$ .
Solve for $v$ and revert to $y$ .

Worked Example 4 – Logistic‑type Growth (NTC network users)

Let $U (t)$ be the number of NTC subscribers (in thousands). The growth follows

$\frac{d U}{d t} = 0.4 U - 0.001 U^{2} .$

Rewrite as $U^{'} + (- 0.4) U = - 0.001 U^{2}$ . This is Bernoulli with $n = 2$ .

Divide by $U^{2}$ :

$U^{- 2} U^{'} - 0.4 U^{- 1} = - 0.001.$

Set $v = U^{- 1}$ → $v^{'} = - U^{- 2} U^{'}$ .

Thus $- v^{'} - 0.4 v = - 0.001$ → $v^{'} + 0.4 v = 0.001$ .

Linear ODE: integrating factor $μ = e^{0.4 t}$ .

$\frac{d}{d t} (e^{0.4 t} v) = 0.001 e^{0.4 t} .$

Integrate:

$e^{0.4 t} v = \frac{0.001}{0.4} e^{0.4 t} + C \Rightarrow v = \frac{0.001}{0.4} + C e^{- 0.4 t} .$

Recall $v = 1/ U$ :

$\frac{1}{U} = 0.0025 + C e^{- 0.4 t} .$

If initially $U (0) = 200$ (i.e., 200 k subscribers), then

$\frac{1}{200} = 0.0025 + C \Rightarrow C = 0.0025 - 0.005 = - 0.0025.$

Hence

$U (t) = \frac{1}{0.0025 - 0.0025 e ^{- 0.4 t}} .$

The model predicts a saturation level of $1/0.0025 = 400$  k subscribers, matching NTC’s market ceiling.

NTC subscriber growth (logistic model)

3.5 Homogeneous Equations

If $y^{'} = f (\frac{y}{x})$ , substitute $v = y / x$ → $y = v x$ and $y^{'} = v + x v^{'}$ .

The resulting equation in $v$ and $x$ is separable.

Worked Example 5 – Traffic Flow in Kathmandu

Let $R (t)$ be the average traffic density (vehicles/km) on a main road, and assume the rate of change obeys

$\frac{d R}{d t} = k \frac{R}{t} .$

This is homogeneous because $\frac{R}{t}$ is a function of the ratio $R / t$ . Set $v = R / t$ → $R = v t$ , $R^{'} = v + t v^{'}$ .

Plug in:

$v + t v^{'} = k v \Rightarrow t v^{'} = (k - 1) v .$

Separate: $\frac{d v}{v} = \frac{k - 1}{t} d t$ .

Integrate: $ln ∣ v ∣ = (k - 1) ln ∣ t ∣ + C$ .

Thus $v = C t^{k - 1}$ and $R = v t = C t^{k}$ .

If at $t = 1$  hour the density is 30 veh/km and $k = 0.8$ , then

$30 = C (1)^{0.8} \Rightarrow C = 30.$

Hence $R (t) = 30 t^{0.8}$ . The model shows a sub‑linear increase in density over time, useful for planning signal timings.

Kathmandu traffic density vs. time

4. Qualitative Tools

4.1 Slope Fields

A slope field visualises the direction field of $y^{'} = f (x, y)$ . Each small line segment at $(x_{i}, y_{j})$ has slope $f (x_{i}, y_{j})$ . Solution curves are curves that are tangent to these segments.

Example slope field for dy/dx = x

4.2 Phase Lines

For autonomous equations $y^{'} = f (y)$ , a phase line shows equilibrium points and stability.

Phase line for dy/dt = -y

5. Comparison of Methods

Method	When to use	Strength	Weakness
Separation	$y^{'} = g (x) h (y)$	Direct, minimal algebra	Not applicable to most ODEs
Integrating factor	Linear	Systematic, works for any linear ODE	Requires integration of $P (x)$
Exactness	$M d x + N d y = 0$ with $M_{y} = N_{x}$	Gives implicit solution without solving for $y$	Exactness rare; may need integrating factor
Bernoulli	Non‑linear with power term	Reduces to linear	Only for specific exponent $n$
Homogeneous	$f (y / x)$	Simple substitution	Limited to ratio‑dependent forms

6. Applications in Engineering & Everyday Life

RC circuits: Voltage across a capacitor satisfies $V^{'} + \frac{1}{R C} V = \frac{E}{R C}$ (linear ODE).
Population dynamics: Logistic growth is a Bernoulli equation.
Chemical reactor design: Rate laws often lead to separable ODEs.
Queueing systems (e.g., Daraz, Pathao): Arrival‑service balance yields linear ODEs for expected queue length.
Bank loan amortization: Continuous compounding leads to linear ODEs as shown in Example 2.

7. In the real world

eSewa – The platform’s transaction‑fee model uses a linear ODE similar to Example 2: the fee accrues continuously while the merchant’s balance is reduced by a constant withdrawal rate.
NTC – Subscriber growth follows the logistic‑type Bernoulli equation (Example 4), explaining why the market saturates near 400 k users.
Daraz order queue – The daily influx of orders and constant processing rate produce an exact ODE (Example 3), allowing the operations team to predict pending orders at any hour.

These concrete cases show that the abstract techniques taught here directly support decision‑making in Nepal’s leading tech companies.

8. Worked Example Tied to a Real Situation

Problem: A bank offers a personal loan of NPR 500,000 at a continuous interest rate of 10 % per year. The borrower agrees to repay at a constant continuous rate of NPR 80,000 per year. Determine (a) the outstanding balance after 3 years, and (b) the time when the loan is fully repaid.

Solution follows the linear ODE in Example 2 with $A (0) = 500, 000$ , $k = 0.10$ , and repayment rate $R = 80, 000$ .

$A^{'} = 0.10 A - 80, 000, A (0) = 500, 000.$

Integrating factor $μ = e^{0.10 t}$ .

$\frac{d}{d t} (e^{0.10 t} A) = - 80, 000 e^{0.10 t} .$

Integrate:

$e^{0.10 t} A = - \frac{80 , 000}{0.10} e^{0.10 t} + C$ $A (t) = - 800, 000 + C e^{- 0.10 t} .$

Apply $A (0) = 500, 000$ : $C = 1, 300, 000$ .

Thus

$A (t) = 1, 300, 000 e^{- 0.10 t} - 800, 000.$

(a) At $t = 3$ :

$A (3) = 1, 300, 000 e^{- 0.3} - 800, 000 \approx 1, 300, 000 (0.7408) - 800, 000 \approx 962, 040 - 800, 000 = 162, 040 NPR .$

(b) Set $A (t) = 0$ :

$0 = 1, 300, 000 e^{- 0.10 t} - 800, 000 \Rightarrow e^{- 0.10 t} = \frac{800 , 000}{1 , 300 , 000} = 0.6154.$

$t = - \frac{ln 0.6154}{0.10} \approx 4.84 years .$

So the loan is cleared after roughly 4 years 10 months.

Loan balance over time (bank example)

9. Common Mistakes to Avoid

Mistake	Why it’s wrong	Correct approach
Forgetting to multiply by the integrating factor in linear ODEs	The left side will not become an exact derivative	Always compute $μ (x) = e^{\int P (x) d x}$ and multiply the whole equation
Assuming exactness without checking $M_{y} = N_{x}$	Leads to incorrect potential function	Verify exactness first; if false, look for an integrating factor
Mixing up the substitution for Bernoulli (using $v = y^{n - 1}$ instead of $v = y^{1 - n}$ )	Produces the wrong linear ODE	Remember $v = y^{1 - n}$
Ignoring the domain when taking logarithms (e.g., $ln ∣ y ∣$ )	May lose sign information	Keep absolute values or restrict to intervals where $y$ retains sign
Treating a homogeneous ODE as separable	The separation step fails	Use $v = y / x$ substitution instead

10. Exam tip

Identify the form first: The first line of any ODE in the exam will hint at its class (look for products of functions of $x$ and $y$ , linear‑looking terms, or a ratio $y / x$ ).
Write a quick decision tree (the mermaid diagram above) on scrap paper; it saves time deciding which method to apply.
Always check exactness before searching for an integrating factor; the check is a single partial‑derivative computation.
When using an integrating factor, compute $\int P (x) d x$ carefully; a missed constant does not affect $μ$ but a sign error does.
For Bernoulli, rewrite the equation so the right‑hand side is exactly $Q (x) y^{n}$ ; then perform the substitution without algebraic slip.
Show work: Even if you finish early, write each step (separation, integration, back‑substitution). Marks are awarded for the process, not just the final answer.

Good luck!

Based on the PU BE Computer (PU) syllabus for Calculus I, unit 4.

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