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Linear Differential Equations & Applications

Master Linear Differential Equations for JEE Main & Advanced. Covers Integrating Factors, Bernoulli, orthogonal trajectories, mixing, and verified practice.

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The Linear Differential Equation is the workhorse of applied mathematics. A first-order linear DE has a beautifully systematic solution method — the Integrating Factor (IF) — that converts any such equation into an exact derivative, making integration straightforward. Beyond the pure mathematics, linear DEs describe real-world phenomena: population growth and decay, Newton's law of cooling, mixing problems, electrical circuits. The Bernoulli Equation — a non-linear DE that reduces to linear via a clever substitution — rounds out this topic. For JEE Main and Advanced, linear DEs are among the highest-yield topics: expect 2–3 questions per exam, including IF computation, particular solutions with ICs, and Bernoulli reduction.

1. Linear Differential Equation — Standard Form

A first-order linear DE is of the form:

$\frac{d y}{d x} + P (x) y = Q (x)$

Integrating Factor (IF):

$IF = e^{\int P (x) d x}$

General Solution:

$y \cdot (IF) = \int Q (x) \cdot (IF) d x + C$

The key insight: multiplying the DE by IF converts the left side into $\frac{d}{d x} [y \cdot IF]$ exactly.

Linear in $x$ ( $d x / d y$ form)

If the equation is linear in x: $\frac{d x}{d y} + P (y) x = Q (y)$

IF = $e^{\int P (y) d y}$ , and $x \cdot IF = \int Q (y) \cdot IF d y + C$

2. Worked Examples — Building Up to JEE Level

Example 1 — Standard Form (JEE Main)

Solve: $\frac{d y}{d x} + \frac{y}{x} = x^{2}$

Here, $P = \frac{1}{x}, Q = x^{2}$ .

$IF = e^{\int \frac{d x}{x}} = e^{\ln x} = x$

$\frac{d}{d x} (x y) = x \cdot x^{2} = x^{3} ⟹ x y = \int x^{3} d x = \frac{x^{4}}{4} + C$

$y = \frac{x^{3}}{4} + \frac{C}{x}$

Example 2 — Gaussian Integrating Factor (JEE Main/Advanced)

Solve: $\frac{d y}{d x} + 2 x y = x$ , given $y (0) = 0$ .

Here, $P = 2 x, Q = x$ .

$IF = e^{\int 2 x d x} = e^{x^{2}}$

$e^{x^{2}} y = \int x e^{x^{2}} d x = \frac{e^{x^{2}}}{2} + C$

$y = \frac{1}{2} + C e^{- x^{2}}$

Apply IC $y (0) = 0$ : $0 = \frac{1}{2} + C ⟹ C = - \frac{1}{2}$

$y = \frac{1 - e^{- x^{2}}}{2}$

Example 3 — Trigonometric $Q (x)$ (JEE Main)

Solve: $(1 + x^{2}) \frac{d y}{d x} + 2 x y = \cos x$

Divide by $(1 + x^{2})$ : $\frac{d y}{d x} + (\frac{2 x}{1 + x^{2}}) y = \frac{\cos x}{1 + x^{2}}$

$IF = e^{\int \frac{2 x}{1 + x^{2}} d x} = e^{\ln (1 + x^{2})} = 1 + x^{2}$

$(1 + x^{2}) y = \int \cos x d x = \sin x + C$

$y = \frac{\sin x + C}{1 + x^{2}}$

Example 4 — Reverse Linear ( $d x / d y$ form) (JEE Main)

Solve: $(y + 1) \frac{d x}{d y} - x = e^{3 y} (y + 1)^{2}$

Divide by $(y + 1)$ : $\frac{d x}{d y} - \frac{x}{y + 1} = e^{3 y} (y + 1)$

Here, $P (y) = - \frac{1}{y + 1}$ .

$IF = e^{\int - \frac{d y}{y + 1}} = e^{- \ln (y + 1)} = \frac{1}{y + 1}$

$\frac{x}{y + 1} = \int e^{3 y} d y = \frac{e^{3 y}}{3} + C$

$x = (y + 1) (\frac{e^{3 y}}{3} + C)$

Example 5 — JEE Advanced: Non-trivial IF

Solve: $x \frac{d y}{d x} + (x - 1) y = x^{2} e^{x}$

Divide by $x$ : $\frac{d y}{d x} + (\frac{x - 1}{x}) y = x e^{x}$

Here, $P = \frac{x - 1}{x} = 1 - \frac{1}{x}$ .

$IF = e^{\int (1 - \frac{1}{x}) d x} = e^{x - \ln x} = \frac{e^{x}}{x}$

$(\frac{e^{x}}{x}) y = \int (x e^{x}) \cdot (\frac{e^{x}}{x}) d x = \int e^{2 x} d x = \frac{e^{2 x}}{2} + C$

$y = \frac{x e^{x}}{2} + C x e^{- x}$

3. Bernoulli's Differential Equation

The Bernoulli equation is:

$\frac{d y}{d x} + P (x) y = Q (x) y^{n} (n \neq 0, 1)$

Reduction: Divide by yⁿ, then substitute $v = y^{1 - n}$ :

$\frac{d v}{d x} + (1 - n) P (x) v = (1 - n) Q (x)$

This is now a linear DE in v.

Example (JEE Main)

Solve: $\frac{d y}{d x} + \frac{y}{x} = x^{2} y^{3}$

n = 3. Divide by y³: $y^{- 3} \frac{d y}{d x} + y^{- 2} / x = x^{2}$

Let v = y^(1−3) = y^(−2): $\frac{d v}{d x} = - 2 y^{- 3} \frac{d y}{d x}$

Equation becomes: $- \frac{1}{2} \frac{d v}{d x} + \frac{v}{x} = x^{2} \Rightarrow \frac{d v}{d x} - \frac{2 v}{x} = - 2 x^{2}$

Linear in v. IF = e^(∫−2/x dx) = x^(−2) = 1/x²

$\frac{v}{x^{2}} = \int (- 2 x^{2}) \cdot \frac{1}{x^{2}} d x = \int - 2 d x = - 2 x + C$

$v = - 2 x^{3} + C x^{2}$

Back-sub v = 1/y²: $\frac{1}{y^{2}} = C x^{2} - 2 x^{3}$

4. Applications of Differential Equations

A. Exponential Growth and Decay

$\frac{d N}{d t} = k N \Rightarrow N = N_{0} e^{k t}$

k > 0: growth (population, compound interest). k < 0 (write k = −λ): decay (radioactivity, drug concentration).

Worked: Population growth problem (JEE Main)

A population doubles in 5 years. After how many years does it triple?

At t=5: 2P₀ = P₀e^(5k) → k = (ln 2)/5

Triple: 3P₀ = P₀e^(kt) → t = (ln 3)/k = 5 ln 3 / ln 2 ≈ 7.93 years

B. Newton's Law of Cooling

$\frac{d T}{d t} = - k (T - T_{s}) \Rightarrow T - T_{s} = (T_{0} - T_{s}) e^{- k t}$

T = temperature of object, Tₛ = surrounding temperature, T₀ = initial temperature.

C. Mixing Problems (JEE Advanced)

A tank contains V litres of brine. Brine flows in at rate r_in (concentration c_in) and flows out at rate r_out. If S = amount of salt at time t:

$\frac{d S}{d t} = r_{i n} \cdot c_{i n} - r_{o u t} \cdot \frac{S}{V (t)}$

This is a linear DE in S. Solve via the IF method.

Worked Mixing Problem (JEE Advanced type)

A tank initially contains 100L of pure water. Brine with a salt concentration of 2 g/L enters at 3 L/min and drains at 3 L/min. Find $S (t)$ (salt at time $t$ ).

Since inflow equals outflow, volume stays constant at 100L.
Rate equation: $\frac{d S}{d t} = 3 (2) - 3 (\frac{S}{100}) = 6 - \frac{3 S}{100}$

Rearrange to standard linear form: $\frac{d S}{d t} + \frac{3 S}{100} = 6$

$IF = e^{\frac{3 t}{100}}$ .
Solution: $S e^{\frac{3 t}{100}} = \int 6 e^{\frac{3 t}{100}} d t = 200 e^{\frac{3 t}{100}} + C$

$S = 200 + C e^{- \frac{3 t}{100}}$ . With $S (0) = 0$ , we get $C = - 200$ .

$S (t) = 200 (1 - e^{- \frac{3 t}{100}}) grams$

D. Orthogonal Trajectories (JEE Advanced)

Two families of curves are orthogonal trajectories of each other if every curve in one family intersects every curve in the other family at right angles.

Method: Find the DE of family F; replace $\frac{d y}{d x}$ with $- \frac{d x}{d y}$ (slope of perpendicular); solve the new DE.

Example: Find orthogonal trajectories of $y = C e^{x}$ .

From $y = C e^{x} ⟹ y^{'} = C e^{x} = y$ .
For orthogonal trajectories: $\frac{d y}{d x} = - \frac{1}{y} ⟹ y d y = - d x ⟹ \frac{y^{2}}{2} = - x + K$

Orthogonal family: $x + \frac{y^{2}}{2} = K$ (a family of downward-opening parabolas).

Linear DE — Master Method Reference

Type	Standard Form	IF	General Solution
Linear in y	y' + P(x)y = Q(x)	e^(∫P dx)	y·IF = ∫Q·IF dx + C
Linear in x	x' + P(y)x = Q(y)	e^(∫P dy)	x·IF = ∫Q·IF dy + C
Bernoulli (n≠0,1)	y' + Py = Qyⁿ	Sub v = y^(1−n); IF for resulting linear DE	v·IF = ∫(1−n)Q·IF dx + C
Growth/Decay	dN/dt = kN	—	N = N₀e^(kt)
Newton's cooling	dT/dt = −k(T−Ts)	—	T−Ts = (T₀−Ts)e^(−kt)

JEE Power Tips — Linear DEs

Always write the DE in standard form dy/dx + Py = Q before finding IF. If you read P directly from an unstandard form (e.g., you forget to divide), the IF will be wrong and the entire solution fails. Make this the first step, always.
IF = e^(∫P dx) — no constant of integration inside the exponent. The constant of integration in the IF cancels out in the final solution, so we always omit it when computing the IF. Omit it deliberately, not by mistake.
For Bernoulli: the sign flips in the linear equation for v. If y' + Py = Qyⁿ, dividing by yⁿ and substituting v = y^(1−n) gives: dv/dx + (1−n)Pv = (1−n)Q. The factor (1−n) appears in front of both P and Q.
JEE ADV Orthogonal trajectories: Replace dy/dx with −dx/dy in the DE of the original family. This exploits the fact that perpendicular slopes multiply to −1. A common JEE Advanced question type.
JEE ADV Second-order linear DEs with constant coefficients — if y₁ and y₂ are two independent solutions, then y = C₁y₁ + C₂y₂ is the general solution (Superposition principle). E.g., y'' − 3y' + 2y = 0 has characteristic roots 1 and 2 → general solution y = C₁eˣ + C₂e²ˣ.
In mixing problems, the volume may change if inflow rate ≠ outflow rate. If inflow = a L/min and outflow = b L/min with initial volume V₀: V(t) = V₀ + (a−b)t. Include this in the DE before solving.

Practice Questions

Q1 (JEE Main): Solve $\frac{d y}{d x} + 2 y \tan x = \sin x$ , given $y (π / 3) = 0$ .

Explanation:

This is a linear differential equation of the form $\frac{d y}{d x} + P y = Q$ , where $P = 2 \tan x$ and $Q = \sin x$ .

Integrating Factor (IF):
$IF = e^{\int 2 \tan x d x} = e^{2 \ln | \sec x |} = \sec^{2} x$

Multiply the DE by the IF to write it as an exact derivative:
$\frac{d}{d x} (y \cdot \sec^{2} x) = \sin x \cdot \sec^{2} x$
$\frac{d}{d x} (y \cdot \sec^{2} x) = \frac{\sin x}{\cos^{2} x} = \tan x \cdot \sec x$

Integrate both sides:
$y \cdot \sec^{2} x = \int \tan x \cdot \sec x d x$
$y \cdot \sec^{2} x = \sec x + C$
$y = \cos x + C \cos^{2} x$

Apply the initial condition $y (π / 3) = 0$ :
$0 = \cos (π / 3) + C \cos^{2} (π / 3)$
$0 = \frac{1}{2} + C (\frac{1}{4}) ⟹ \frac{C}{4} = - \frac{1}{2} ⟹ C = - 2$

Final Solution: $y = \cos x - 2 \cos^{2} x$

Q2 (JEE Advanced): Solve the Bernoulli equation $\frac{d y}{d x} + \frac{y}{x} = y^{2} \ln x$ .

Explanation:

Divide the entire equation by $y^{2}$ ( $n = 2$ ):
$y^{- 2} \frac{d y}{d x} + \frac{y^{- 1}}{x} = \ln x$

Let $v = y^{- 1}$ . Differentiating gives $\frac{d v}{d x} = - y^{- 2} \frac{d y}{d x}$ .
Substitute into the equation:
$- \frac{d v}{d x} + \frac{v}{x} = \ln x ⟹ \frac{d v}{d x} - \frac{v}{x} = - \ln x$

This is now a linear DE in $v$ , with $P = - 1 / x$ and $Q = - \ln x$ .
$IF = e^{\int - 1 / x d x} = e^{- \ln x} = \frac{1}{x}$

Multiply by the IF and integrate:
$v \cdot (\frac{1}{x}) = \int (\frac{- \ln x}{x}) d x$
Let $u = \ln x, d u = \frac{1}{x} d x$ , so $\int - u d u = - \frac{u^{2}}{2} + C$ .
$\frac{v}{x} = - \frac{(\ln x)^{2}}{2} + C$
$v = C x - \frac{x (\ln x)^{2}}{2}$

Substitute back $v = y^{- 1}$ :
Final Solution: $\frac{1}{y} = C x - \frac{x (\ln x)^{2}}{2}$

Q3 (JEE Main): A radioactive substance decays at a rate proportional to the amount present. If 30% disintegrates in 15 years, what percentage remains after 60 years?

Explanation:

Let $N$ be the amount present. The DE is $\frac{d N}{d t} = - k N$ , which solves to:
$N (t) = N_{0} e^{- k t}$

Given that 30% disintegrates in 15 years, 70% remains:
$N (15) = 0.7 N_{0}$
$0.7 N_{0} = N_{0} e^{- 15 k} ⟹ e^{- 15 k} = 0.7$

We need to find the amount remaining at $t = 60$ years:
$N (60) = N_{0} e^{- 60 k} = N_{0} (e^{- 15 k})^{4}$
Substitute $e^{- 15 k} = 0.7$ :
$N (60) = N_{0} (0.7)^{4} = N_{0} (0.2401)$

Answer: Approximately 24.01% remains.

Q4 (JEE Main): If $y = y (x)$ is the solution of $\frac{d y}{d x} + 2 y = e^{x}$ with $y (0) = 0$ , find $y (\ln 2)$ .

Explanation:

This is a linear DE with $P = 2$ and $Q = e^{x}$ .
$IF = e^{\int 2 d x} = e^{2 x}$

The solution is given by:
$y \cdot e^{2 x} = \int e^{x} \cdot e^{2 x} d x$
$y \cdot e^{2 x} = \int e^{3 x} d x = \frac{e^{3 x}}{3} + C$
$y = \frac{e^{x}}{3} + C e^{- 2 x}$

Apply the initial condition $y (0) = 0$ :
$0 = \frac{1}{3} + C ⟹ C = - \frac{1}{3}$
$y = \frac{e^{x} - e^{- 2 x}}{3}$

Calculate $y (\ln 2)$ :
$y (\ln 2) = \frac{e^{\ln 2} - e^{- 2 \ln 2}}{3} = \frac{2 - 2^{- 2}}{3} = \frac{2 - 1 / 4}{3} = \frac{7 / 4}{3} = \frac{7}{12}$

Q5 (JEE Advanced): Find the orthogonal trajectories of the family of parabolas $y^{2} = 4 a x$ .

Explanation:

First, find the differential equation of the given family by eliminating the parameter $a$ .
Differentiate $y^{2} = 4 a x$ with respect to $x$ :
$2 y \frac{d y}{d x} = 4 a$
From the original equation, $4 a = \frac{y^{2}}{x}$ . Substitute this back:
$2 y \frac{d y}{d x} = \frac{y^{2}}{x} ⟹ \frac{d y}{d x} = \frac{y}{2 x}$

For orthogonal trajectories, replace $\frac{d y}{d x}$ with $- \frac{d x}{d y}$ (since $m_{1} \cdot m_{2} = - 1$ ):
$- \frac{d x}{d y} = \frac{y}{2 x} ⟹ \frac{d y}{d x} = - \frac{2 x}{y}$

Separate variables and integrate:
$y d y = - 2 x d x$
$\int y d y = \int - 2 x d x$
$\frac{y^{2}}{2} = - x^{2} + C_{1}$
$2 x^{2} + y^{2} = 2 C_{1}$

Final Solution: $2 x^{2} + y^{2} = C$ (which is a family of ellipses).

Q6 (JEE Advanced Integer type): Solve $\frac{d y}{d x} - \frac{y}{x} = \frac{1}{x^{2}}$ and find the value of $y (1)$ , given $y (2) = 0$ .

Explanation:

This is a linear DE where $P = - \frac{1}{x}$ and $Q = \frac{1}{x^{2}}$ .
$IF = e^{\int - 1 / x d x} = e^{- \ln x} = \frac{1}{x}$

The solution is given by:
$y \cdot (\frac{1}{x}) = \int (\frac{1}{x^{2}}) (\frac{1}{x}) d x$
$\frac{y}{x} = \int x^{- 3} d x$
$\frac{y}{x} = - \frac{x^{- 2}}{2} + C = - \frac{1}{2 x^{2}} + C$
$y = - \frac{1}{2 x} + C x$

Apply the initial condition $y (2) = 0$ :
$0 = - \frac{1}{4} + 2 C ⟹ 2 C = \frac{1}{4} ⟹ C = \frac{1}{8}$
$y = - \frac{1}{2 x} + \frac{x}{8}$

Find $y (1)$ :
$y (1) = - \frac{1}{2 (1)} + \frac{1}{8} = - \frac{4}{8} + \frac{1}{8}$
$= - \frac{3}{8}$

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Linear Differential Equations & Applications

1. Linear Differential Equation — Standard Form

Linear in $x$ ( $d x / d y$ form)

2. Worked Examples — Building Up to JEE Level

Example 1 — Standard Form (JEE Main)

Example 2 — Gaussian Integrating Factor (JEE Main/Advanced)

Example 3 — Trigonometric $Q (x)$ (JEE Main)

Example 4 — Reverse Linear ( $d x / d y$ form) (JEE Main)

Example 5 — JEE Advanced: Non-trivial IF

3. Bernoulli's Differential Equation

Example (JEE Main)

4. Applications of Differential Equations

A. Exponential Growth and Decay

Worked: Population growth problem (JEE Main)

B. Newton's Law of Cooling

C. Mixing Problems (JEE Advanced)

Worked Mixing Problem (JEE Advanced type)

D. Orthogonal Trajectories (JEE Advanced)

Linear DE — Master Method Reference

Practice Questions

Related Study Material

Practice Questions

Topic Information

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Target Subjects

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Subject Mastery

Linear Differential Equations & Applications

1. Linear Differential Equation — Standard Form

Linear in x (dx/dy form)

2. Worked Examples — Building Up to JEE Level

Example 1 — Standard Form (JEE Main)

Example 2 — Gaussian Integrating Factor (JEE Main/Advanced)

Example 3 — Trigonometric Q(x) (JEE Main)

Example 4 — Reverse Linear (dx/dy form) (JEE Main)

Example 5 — JEE Advanced: Non-trivial IF

3. Bernoulli's Differential Equation

Example (JEE Main)

4. Applications of Differential Equations

A. Exponential Growth and Decay

Worked: Population growth problem (JEE Main)

B. Newton's Law of Cooling

C. Mixing Problems (JEE Advanced)

Worked Mixing Problem (JEE Advanced type)

D. Orthogonal Trajectories (JEE Advanced)

Linear DE — Master Method Reference

Practice Questions

Related Study Material

Practice Questions

Topic Information

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Linear in $x$ ( $d x / d y$ form)

Example 3 — Trigonometric $Q (x)$ (JEE Main)

Example 4 — Reverse Linear ( $d x / d y$ form) (JEE Main)